What's the difference between &C::c and &(C::c)？

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up vote
8
down vote

favorite

Test code in below and I put the output info in comment.
I was using gcc 4.8.5 and Centos7.2.

#include <iostream>
#include <cstdio>

class C 

 public:
 void foo() 
 printf("%p, %pn", &C::c, &(C::c)); // output value is 0x4, 0x7ffc2e7f52e8
 std::cout << &C::c << std::endl; // output value is 1
 
 int a;
 int c;
;

int main(void)

 C co;

 printf("%pn", &C::c); // output value is 0x4
 std::cout << &C::c << std::endl; // output value is 1

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

 co.foo();

 return 0;

1. According to C++ operator Precedence，:: operator have higher precedence than & operator. I think &C::c is equal to &(C::c), but the output says no. Why are they different?


2. 
   &(C::c) causes compile error in main but not in foo function，why is that？


3. The value of &C::c is different in printf and std::cout, why is that？

c++

share|improve this question

edited 1 hour ago

melpomene

54.1k54084

asked 3 hours ago

Xianggang ZENG

483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You cannot printf a pointer to member value. %p is for regular pointers only.
  – n.m.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @n.m. Even worse, %p is for void * only.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
8
down vote

favorite

Test code in below and I put the output info in comment.
I was using gcc 4.8.5 and Centos7.2.

#include <iostream>
#include <cstdio>

class C 

 public:
 void foo() 
 printf("%p, %pn", &C::c, &(C::c)); // output value is 0x4, 0x7ffc2e7f52e8
 std::cout << &C::c << std::endl; // output value is 1
 
 int a;
 int c;
;

int main(void)

 C co;

 printf("%pn", &C::c); // output value is 0x4
 std::cout << &C::c << std::endl; // output value is 1

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

 co.foo();

 return 0;

1. According to C++ operator Precedence，:: operator have higher precedence than & operator. I think &C::c is equal to &(C::c), but the output says no. Why are they different?


2. 
   &(C::c) causes compile error in main but not in foo function，why is that？


3. The value of &C::c is different in printf and std::cout, why is that？

c++

share|improve this question

edited 1 hour ago

melpomene

54.1k54084

asked 3 hours ago

Xianggang ZENG

483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You cannot printf a pointer to member value. %p is for regular pointers only.
  – n.m.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @n.m. Even worse, %p is for void * only.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
8
down vote

favorite

up vote
8
down vote

favorite

Test code in below and I put the output info in comment.
I was using gcc 4.8.5 and Centos7.2.

#include <iostream>
#include <cstdio>

class C 

 public:
 void foo() 
 printf("%p, %pn", &C::c, &(C::c)); // output value is 0x4, 0x7ffc2e7f52e8
 std::cout << &C::c << std::endl; // output value is 1
 
 int a;
 int c;
;

int main(void)

 C co;

 printf("%pn", &C::c); // output value is 0x4
 std::cout << &C::c << std::endl; // output value is 1

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

 co.foo();

 return 0;

1. According to C++ operator Precedence，:: operator have higher precedence than & operator. I think &C::c is equal to &(C::c), but the output says no. Why are they different?


2. 
   &(C::c) causes compile error in main but not in foo function，why is that？


3. The value of &C::c is different in printf and std::cout, why is that？

c++

share|improve this question

edited 1 hour ago

melpomene

54.1k54084

asked 3 hours ago

Xianggang ZENG

483

Test code in below and I put the output info in comment.
I was using gcc 4.8.5 and Centos7.2.

#include <iostream>
#include <cstdio>

class C 

 public:
 void foo() 
 printf("%p, %pn", &C::c, &(C::c)); // output value is 0x4, 0x7ffc2e7f52e8
 std::cout << &C::c << std::endl; // output value is 1
 
 int a;
 int c;
;

int main(void)

 C co;

 printf("%pn", &C::c); // output value is 0x4
 std::cout << &C::c << std::endl; // output value is 1

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

 co.foo();

 return 0;

1. According to C++ operator Precedence，:: operator have higher precedence than & operator. I think &C::c is equal to &(C::c), but the output says no. Why are they different?


2. 
   &(C::c) causes compile error in main but not in foo function，why is that？


3. The value of &C::c is different in printf and std::cout, why is that？

c++

c++

share|improve this question

edited 1 hour ago

melpomene

54.1k54084

asked 3 hours ago

Xianggang ZENG

483

share|improve this question

edited 1 hour ago

melpomene

54.1k54084

asked 3 hours ago

Xianggang ZENG

483

share|improve this question

share|improve this question

edited 1 hour ago

melpomene

54.1k54084

edited 1 hour ago

melpomene

54.1k54084

edited 1 hour ago

melpomene

54.1k54084

54.1k54084

asked 3 hours ago

Xianggang ZENG

483

asked 3 hours ago

Xianggang ZENG

483

asked 3 hours ago

Xianggang ZENG

483

483

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You cannot printf a pointer to member value. %p is for regular pointers only.
  – n.m.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @n.m. Even worse, %p is for void * only.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You cannot printf a pointer to member value. %p is for regular pointers only.
  – n.m.
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @n.m. Even worse, %p is for void * only.
  – melpomene
  2 hours ago
  

  
  
  
  

You cannot printf a pointer to member value. %p is for regular pointers only.
– n.m.
2 hours ago

You cannot printf a pointer to member value. %p is for regular pointers only.
– n.m.
2 hours ago

1

1

@n.m. Even worse, %p is for void * only.
– melpomene
2 hours ago

@n.m. Even worse, %p is for void * only.
– melpomene
2 hours ago

add a comment | 

4 Answers
4

active

oldest

votes

up vote
5
down vote



accepted

C++ distinguishes two forms of operands to the & operator, lvalues in general and (qualified) identifiers specifically. In &C::c the operand of & is a qualified identifier (i.e. just a name) whereas in &(C::c) the operand is a general expression (because ( cannot be part of a name).

The qualified identifier form has a special case: If it refers to a non-static member of a class (like your C::c), & returns a special value known as a "pointer to member of C". See here for more information about member pointers.

In &(C::c) there is no special case. C::c is resolved normally and fails because there is no object to get a c member of. At least that's what happens in main; in methods of C (like your foo) there is an implicit this object, so C::c actually means this->c there.

As for why the output is different for printf vs. cout: When you try to print a member pointer with <<, it is implicitly converted to a bool, yielding false if it's a null pointer and true otherwise. false is printed as 0; true is printed as 1. Your member pointer is not null, so you get 1. This is different from normal pointers, which are implicitly converted to void * and printed as addresses, but member pointers cannot be converted to void * so the only applicable overload of operator<< is the one for bool. See https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt#Notes.

Note that technically your printf calls have undefined behavior. %p takes a void * and you're passing it pointers of different types. In normal function calls the automatic conversion from T * to void * would kick in, but printf is a variable-arguments function that provides no type context to its argument list, so you need a manual conversion:

printf("%pn", static_cast<void *>(&(C::c)));

---

The relevant part of the standard is [expr.unary.op], saying:

The result of the unary & operator is a pointer to its operand.
The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C​::​m.
Otherwise, if the type of the expression is T, the result has type “pointer to T” [...]

share|improve this answer

edited 1 hour ago

answered 2 hours ago

melpomene

54.1k54084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M Not if you do it in foo. See the rest of my answer.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
  – Xianggang ZENG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @XianggangZENG Thanks, edited.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
  – melpomene
  1 hour ago
  

  
  
  
  

 | 
show 2 more comments

up vote
1
down vote

While the expression &C::c result in a pointer to member c the expression &(C::c) yield the address of the member variable c. The difference you're seeing in the output is that std::cout involve a bool implicit conversion that tells you whether the pointer is null or not.

Since &C::c is actually not null is implicitly converted to bool with value true or 1

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jans

5,01111926

add a comment | 

up vote
0
down vote

First, you can not access the int c by using &(C::c) outside a class. &(C::c) means the memory address of "c of instance C", somewhat &(this->c) here. However your c is not a static member of class C and there is no C instance. You can not access int x = C::c outside, either.

So you see an error with:

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

If you have a static int c, then C::c outside the class is OK, because no instance is needed here.

And let's run

#include <iostream>
#include <cstdio>

class C

 public:
 void foo() 
 printf("%p, %p, this=%pn", &C::c, &(C::c), this);
 
 int a;
 int c;
;

int main(void)

 C co;
 co.foo();
 return 0;

The output is:

0x4, 0x7ffee78e47f4, this=0x7ffee78e47f0
// 0x4 + this == 0x7ffee78e47f4
// see reference

And for std::out: the << &C::c is implicit-casted to bool, so true is 1 you saw. You can treat &C::c as an offset of c in C, it is unusable without a C instance.

That's all.

Some reference: C++: Pointer to class data member "::*"

Full description: https://en.cppreference.com/w/cpp/language/pointer

share|improve this answer

edited 2 hours ago

answered 2 hours ago

shawn

2,420616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In C::c, the C can only be a class or namespace name (never an instance)
  – M.M
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
-2
down vote

Q1: There is special meaning to the syntax of & followed by an unparenthesized qualified-id. It means to form a pointer-to-member. Furthermore, there is no other way to form a pointer-to-member.

Note: The section C++17 [expr.prim.paren]/1 says that redundant parentheses can be inserted around any expression "except otherwise indicated", however this situation seems to be intended as "otherwise indicated".

---

Q3: In both cases where you write printf("%pn", &C::c);, &C::c is a pointer-to-member. The %p format specifier is only for void * so this causes undefined behaviour, and the program output is meaningless.

The code cout << &C::c; outputs a pointer-to-member via operator<<(bool val), since there is implicit conversion from pointer-to-member to bool (with result true in all cases), see [conv.bool]/1.

For further discussion of how to print a pointer-to-member, see this answer.

---

Q2: The code &(C::c) does not form a pointer-to-member as explained above.

Now, the code C::c is in the grammatical category id-expression. (Which is qualified-id and unqualified-id). An id-expression has some restrictions on its use, [expr.prim.id]/1:

An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member’s class or a class derived from that class, or


• to form a pointer to member (7.6.2.1), or


• if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

When we are inside the C::foo function, the first of those bullet points applies. The code is the same as &c but with unnecessary qualification. This has type int *. You could output this with std::cout << &(C::c); which would show a memory address, the address of this->c.

When we are in the main function , none of the three bullet points apply and therefore the &(C::c) is ill-formed.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

M.M

102k10106224

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
  – melpomene
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

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4 Answers
4

active

oldest

votes

4 Answers
4

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

C++ distinguishes two forms of operands to the & operator, lvalues in general and (qualified) identifiers specifically. In &C::c the operand of & is a qualified identifier (i.e. just a name) whereas in &(C::c) the operand is a general expression (because ( cannot be part of a name).

The qualified identifier form has a special case: If it refers to a non-static member of a class (like your C::c), & returns a special value known as a "pointer to member of C". See here for more information about member pointers.

In &(C::c) there is no special case. C::c is resolved normally and fails because there is no object to get a c member of. At least that's what happens in main; in methods of C (like your foo) there is an implicit this object, so C::c actually means this->c there.

As for why the output is different for printf vs. cout: When you try to print a member pointer with <<, it is implicitly converted to a bool, yielding false if it's a null pointer and true otherwise. false is printed as 0; true is printed as 1. Your member pointer is not null, so you get 1. This is different from normal pointers, which are implicitly converted to void * and printed as addresses, but member pointers cannot be converted to void * so the only applicable overload of operator<< is the one for bool. See https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt#Notes.

Note that technically your printf calls have undefined behavior. %p takes a void * and you're passing it pointers of different types. In normal function calls the automatic conversion from T * to void * would kick in, but printf is a variable-arguments function that provides no type context to its argument list, so you need a manual conversion:

printf("%pn", static_cast<void *>(&(C::c)));

---

The relevant part of the standard is [expr.unary.op], saying:

The result of the unary & operator is a pointer to its operand.
The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C​::​m.
Otherwise, if the type of the expression is T, the result has type “pointer to T” [...]

share|improve this answer

edited 1 hour ago

answered 2 hours ago

melpomene

54.1k54084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M Not if you do it in foo. See the rest of my answer.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
  – Xianggang ZENG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @XianggangZENG Thanks, edited.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
  – melpomene
  1 hour ago
  

  
  
  
  

 | 
show 2 more comments

up vote
5
down vote



accepted

C++ distinguishes two forms of operands to the & operator, lvalues in general and (qualified) identifiers specifically. In &C::c the operand of & is a qualified identifier (i.e. just a name) whereas in &(C::c) the operand is a general expression (because ( cannot be part of a name).

The qualified identifier form has a special case: If it refers to a non-static member of a class (like your C::c), & returns a special value known as a "pointer to member of C". See here for more information about member pointers.

In &(C::c) there is no special case. C::c is resolved normally and fails because there is no object to get a c member of. At least that's what happens in main; in methods of C (like your foo) there is an implicit this object, so C::c actually means this->c there.

As for why the output is different for printf vs. cout: When you try to print a member pointer with <<, it is implicitly converted to a bool, yielding false if it's a null pointer and true otherwise. false is printed as 0; true is printed as 1. Your member pointer is not null, so you get 1. This is different from normal pointers, which are implicitly converted to void * and printed as addresses, but member pointers cannot be converted to void * so the only applicable overload of operator<< is the one for bool. See https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt#Notes.

Note that technically your printf calls have undefined behavior. %p takes a void * and you're passing it pointers of different types. In normal function calls the automatic conversion from T * to void * would kick in, but printf is a variable-arguments function that provides no type context to its argument list, so you need a manual conversion:

printf("%pn", static_cast<void *>(&(C::c)));

---

The relevant part of the standard is [expr.unary.op], saying:

The result of the unary & operator is a pointer to its operand.
The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C​::​m.
Otherwise, if the type of the expression is T, the result has type “pointer to T” [...]

share|improve this answer

edited 1 hour ago

answered 2 hours ago

melpomene

54.1k54084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M Not if you do it in foo. See the rest of my answer.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
  – Xianggang ZENG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @XianggangZENG Thanks, edited.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
  – melpomene
  1 hour ago
  

  
  
  
  

 | 
show 2 more comments

up vote
5
down vote



accepted

up vote
5
down vote



accepted

C++ distinguishes two forms of operands to the & operator, lvalues in general and (qualified) identifiers specifically. In &C::c the operand of & is a qualified identifier (i.e. just a name) whereas in &(C::c) the operand is a general expression (because ( cannot be part of a name).

The qualified identifier form has a special case: If it refers to a non-static member of a class (like your C::c), & returns a special value known as a "pointer to member of C". See here for more information about member pointers.

In &(C::c) there is no special case. C::c is resolved normally and fails because there is no object to get a c member of. At least that's what happens in main; in methods of C (like your foo) there is an implicit this object, so C::c actually means this->c there.

As for why the output is different for printf vs. cout: When you try to print a member pointer with <<, it is implicitly converted to a bool, yielding false if it's a null pointer and true otherwise. false is printed as 0; true is printed as 1. Your member pointer is not null, so you get 1. This is different from normal pointers, which are implicitly converted to void * and printed as addresses, but member pointers cannot be converted to void * so the only applicable overload of operator<< is the one for bool. See https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt#Notes.

Note that technically your printf calls have undefined behavior. %p takes a void * and you're passing it pointers of different types. In normal function calls the automatic conversion from T * to void * would kick in, but printf is a variable-arguments function that provides no type context to its argument list, so you need a manual conversion:

printf("%pn", static_cast<void *>(&(C::c)));

---

The relevant part of the standard is [expr.unary.op], saying:

The result of the unary & operator is a pointer to its operand.
The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C​::​m.
Otherwise, if the type of the expression is T, the result has type “pointer to T” [...]

share|improve this answer

edited 1 hour ago

answered 2 hours ago

melpomene

54.1k54084

C++ distinguishes two forms of operands to the & operator, lvalues in general and (qualified) identifiers specifically. In &C::c the operand of & is a qualified identifier (i.e. just a name) whereas in &(C::c) the operand is a general expression (because ( cannot be part of a name).

The qualified identifier form has a special case: If it refers to a non-static member of a class (like your C::c), & returns a special value known as a "pointer to member of C". See here for more information about member pointers.

In &(C::c) there is no special case. C::c is resolved normally and fails because there is no object to get a c member of. At least that's what happens in main; in methods of C (like your foo) there is an implicit this object, so C::c actually means this->c there.

As for why the output is different for printf vs. cout: When you try to print a member pointer with <<, it is implicitly converted to a bool, yielding false if it's a null pointer and true otherwise. false is printed as 0; true is printed as 1. Your member pointer is not null, so you get 1. This is different from normal pointers, which are implicitly converted to void * and printed as addresses, but member pointers cannot be converted to void * so the only applicable overload of operator<< is the one for bool. See https://en.cppreference.com/w/cpp/io/basic_ostream/operator_ltlt#Notes.

Note that technically your printf calls have undefined behavior. %p takes a void * and you're passing it pointers of different types. In normal function calls the automatic conversion from T * to void * would kick in, but printf is a variable-arguments function that provides no type context to its argument list, so you need a manual conversion:

printf("%pn", static_cast<void *>(&(C::c)));

---

The relevant part of the standard is [expr.unary.op], saying:

The result of the unary & operator is a pointer to its operand.
The operand shall be an lvalue or a qualified-id. If the operand is a qualified-id naming a non-static or variant member m of some class C with type T, the result has type “pointer to member of class C of type T” and is a prvalue designating C​::​m.
Otherwise, if the type of the expression is T, the result has type “pointer to T” [...]

share|improve this answer

edited 1 hour ago

answered 2 hours ago

melpomene

54.1k54084

share|improve this answer

share|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 2 hours ago

melpomene

54.1k54084

answered 2 hours ago

melpomene

54.1k54084

answered 2 hours ago

melpomene

54.1k54084

54.1k54084

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M Not if you do it in foo. See the rest of my answer.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
  – Xianggang ZENG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @XianggangZENG Thanks, edited.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
  – melpomene
  1 hour ago
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M Not if you do it in foo. See the rest of my answer.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  ”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
  – Xianggang ZENG
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @XianggangZENG Thanks, edited.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
  – melpomene
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
  – melpomene
  1 hour ago
  

  
  
  
  

@M.M Not if you do it in foo. See the rest of my answer.
– melpomene
1 hour ago

@M.M Not if you do it in foo. See the rest of my answer.
– melpomene
1 hour ago

”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
– Xianggang ZENG
1 hour ago

”try to print a pointer with <<，it's implicitly converted to a bool"。 I try int b = 1; std::cout << &b << std::endl;， it output an address not just 1.
– Xianggang ZENG
1 hour ago

@XianggangZENG Thanks, edited.
– melpomene
1 hour ago

@XianggangZENG Thanks, edited.
– melpomene
1 hour ago

@M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
– melpomene
1 hour ago

@M.M The one in main is commented out ... or at least it was before you edited OP's code. Now it no longer matches the question. Sigh.
– melpomene
1 hour ago

@M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
– melpomene
1 hour ago

@M.M That's the second question. The first question is why the output is different for &C::c vs &(C::c). (The third question is why the output is different for printf vs cout.)
– melpomene
1 hour ago

 | 
show 2 more comments

up vote
1
down vote

While the expression &C::c result in a pointer to member c the expression &(C::c) yield the address of the member variable c. The difference you're seeing in the output is that std::cout involve a bool implicit conversion that tells you whether the pointer is null or not.

Since &C::c is actually not null is implicitly converted to bool with value true or 1

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jans

5,01111926

add a comment | 

up vote
1
down vote

While the expression &C::c result in a pointer to member c the expression &(C::c) yield the address of the member variable c. The difference you're seeing in the output is that std::cout involve a bool implicit conversion that tells you whether the pointer is null or not.

Since &C::c is actually not null is implicitly converted to bool with value true or 1

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jans

5,01111926

add a comment | 

up vote
1
down vote

up vote
1
down vote

While the expression &C::c result in a pointer to member c the expression &(C::c) yield the address of the member variable c. The difference you're seeing in the output is that std::cout involve a bool implicit conversion that tells you whether the pointer is null or not.

Since &C::c is actually not null is implicitly converted to bool with value true or 1

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jans

5,01111926

While the expression &C::c result in a pointer to member c the expression &(C::c) yield the address of the member variable c. The difference you're seeing in the output is that std::cout involve a bool implicit conversion that tells you whether the pointer is null or not.

Since &C::c is actually not null is implicitly converted to bool with value true or 1

share|improve this answer

edited 2 hours ago

answered 2 hours ago

Jans

5,01111926

share|improve this answer

share|improve this answer

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

answered 2 hours ago

Jans

5,01111926

answered 2 hours ago

Jans

5,01111926

answered 2 hours ago

Jans

5,01111926

5,01111926

add a comment | 

add a comment | 

up vote
0
down vote

First, you can not access the int c by using &(C::c) outside a class. &(C::c) means the memory address of "c of instance C", somewhat &(this->c) here. However your c is not a static member of class C and there is no C instance. You can not access int x = C::c outside, either.

So you see an error with:

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

If you have a static int c, then C::c outside the class is OK, because no instance is needed here.

And let's run

#include <iostream>
#include <cstdio>

class C

 public:
 void foo() 
 printf("%p, %p, this=%pn", &C::c, &(C::c), this);
 
 int a;
 int c;
;

int main(void)

 C co;
 co.foo();
 return 0;

The output is:

0x4, 0x7ffee78e47f4, this=0x7ffee78e47f0
// 0x4 + this == 0x7ffee78e47f4
// see reference

And for std::out: the << &C::c is implicit-casted to bool, so true is 1 you saw. You can treat &C::c as an offset of c in C, it is unusable without a C instance.

That's all.

Some reference: C++: Pointer to class data member "::*"

Full description: https://en.cppreference.com/w/cpp/language/pointer

share|improve this answer

edited 2 hours ago

answered 2 hours ago

shawn

2,420616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In C::c, the C can only be a class or namespace name (never an instance)
  – M.M
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

First, you can not access the int c by using &(C::c) outside a class. &(C::c) means the memory address of "c of instance C", somewhat &(this->c) here. However your c is not a static member of class C and there is no C instance. You can not access int x = C::c outside, either.

So you see an error with:

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

If you have a static int c, then C::c outside the class is OK, because no instance is needed here.

And let's run

#include <iostream>
#include <cstdio>

class C

 public:
 void foo() 
 printf("%p, %p, this=%pn", &C::c, &(C::c), this);
 
 int a;
 int c;
;

int main(void)

 C co;
 co.foo();
 return 0;

The output is:

0x4, 0x7ffee78e47f4, this=0x7ffee78e47f0
// 0x4 + this == 0x7ffee78e47f4
// see reference

And for std::out: the << &C::c is implicit-casted to bool, so true is 1 you saw. You can treat &C::c as an offset of c in C, it is unusable without a C instance.

That's all.

Some reference: C++: Pointer to class data member "::*"

Full description: https://en.cppreference.com/w/cpp/language/pointer

share|improve this answer

edited 2 hours ago

answered 2 hours ago

shawn

2,420616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In C::c, the C can only be a class or namespace name (never an instance)
  – M.M
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

First, you can not access the int c by using &(C::c) outside a class. &(C::c) means the memory address of "c of instance C", somewhat &(this->c) here. However your c is not a static member of class C and there is no C instance. You can not access int x = C::c outside, either.

So you see an error with:

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

If you have a static int c, then C::c outside the class is OK, because no instance is needed here.

And let's run

#include <iostream>
#include <cstdio>

class C

 public:
 void foo() 
 printf("%p, %p, this=%pn", &C::c, &(C::c), this);
 
 int a;
 int c;
;

int main(void)

 C co;
 co.foo();
 return 0;

The output is:

0x4, 0x7ffee78e47f4, this=0x7ffee78e47f0
// 0x4 + this == 0x7ffee78e47f4
// see reference

And for std::out: the << &C::c is implicit-casted to bool, so true is 1 you saw. You can treat &C::c as an offset of c in C, it is unusable without a C instance.

That's all.

Some reference: C++: Pointer to class data member "::*"

Full description: https://en.cppreference.com/w/cpp/language/pointer

share|improve this answer

edited 2 hours ago

answered 2 hours ago

shawn

2,420616

First, you can not access the int c by using &(C::c) outside a class. &(C::c) means the memory address of "c of instance C", somewhat &(this->c) here. However your c is not a static member of class C and there is no C instance. You can not access int x = C::c outside, either.

So you see an error with:

// printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c'

If you have a static int c, then C::c outside the class is OK, because no instance is needed here.

And let's run

#include <iostream>
#include <cstdio>

class C

 public:
 void foo() 
 printf("%p, %p, this=%pn", &C::c, &(C::c), this);
 
 int a;
 int c;
;

int main(void)

 C co;
 co.foo();
 return 0;

The output is:

0x4, 0x7ffee78e47f4, this=0x7ffee78e47f0
// 0x4 + this == 0x7ffee78e47f4
// see reference

And for std::out: the << &C::c is implicit-casted to bool, so true is 1 you saw. You can treat &C::c as an offset of c in C, it is unusable without a C instance.

That's all.

Some reference: C++: Pointer to class data member "::*"

Full description: https://en.cppreference.com/w/cpp/language/pointer

share|improve this answer

edited 2 hours ago

answered 2 hours ago

shawn

2,420616

share|improve this answer

share|improve this answer

edited 2 hours ago

edited 2 hours ago

edited 2 hours ago

answered 2 hours ago

shawn

2,420616

answered 2 hours ago

shawn

2,420616

answered 2 hours ago

shawn

2,420616

2,420616

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In C::c, the C can only be a class or namespace name (never an instance)
  – M.M
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  In C::c, the C can only be a class or namespace name (never an instance)
  – M.M
  1 hour ago
  

  
  
  
  

In C::c, the C can only be a class or namespace name (never an instance)
– M.M
1 hour ago

In C::c, the C can only be a class or namespace name (never an instance)
– M.M
1 hour ago

add a comment | 

up vote
-2
down vote

Q1: There is special meaning to the syntax of & followed by an unparenthesized qualified-id. It means to form a pointer-to-member. Furthermore, there is no other way to form a pointer-to-member.

Note: The section C++17 [expr.prim.paren]/1 says that redundant parentheses can be inserted around any expression "except otherwise indicated", however this situation seems to be intended as "otherwise indicated".

---

Q3: In both cases where you write printf("%pn", &C::c);, &C::c is a pointer-to-member. The %p format specifier is only for void * so this causes undefined behaviour, and the program output is meaningless.

The code cout << &C::c; outputs a pointer-to-member via operator<<(bool val), since there is implicit conversion from pointer-to-member to bool (with result true in all cases), see [conv.bool]/1.

For further discussion of how to print a pointer-to-member, see this answer.

---

Q2: The code &(C::c) does not form a pointer-to-member as explained above.

Now, the code C::c is in the grammatical category id-expression. (Which is qualified-id and unqualified-id). An id-expression has some restrictions on its use, [expr.prim.id]/1:

An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member’s class or a class derived from that class, or


• to form a pointer to member (7.6.2.1), or


• if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

When we are inside the C::foo function, the first of those bullet points applies. The code is the same as &c but with unnecessary qualification. This has type int *. You could output this with std::cout << &(C::c); which would show a memory address, the address of this->c.

When we are in the main function , none of the three bullet points apply and therefore the &(C::c) is ill-formed.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

M.M

102k10106224

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
  – melpomene
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
-2
down vote

Q1: There is special meaning to the syntax of & followed by an unparenthesized qualified-id. It means to form a pointer-to-member. Furthermore, there is no other way to form a pointer-to-member.

Note: The section C++17 [expr.prim.paren]/1 says that redundant parentheses can be inserted around any expression "except otherwise indicated", however this situation seems to be intended as "otherwise indicated".

---

Q3: In both cases where you write printf("%pn", &C::c);, &C::c is a pointer-to-member. The %p format specifier is only for void * so this causes undefined behaviour, and the program output is meaningless.

The code cout << &C::c; outputs a pointer-to-member via operator<<(bool val), since there is implicit conversion from pointer-to-member to bool (with result true in all cases), see [conv.bool]/1.

For further discussion of how to print a pointer-to-member, see this answer.

---

Q2: The code &(C::c) does not form a pointer-to-member as explained above.

Now, the code C::c is in the grammatical category id-expression. (Which is qualified-id and unqualified-id). An id-expression has some restrictions on its use, [expr.prim.id]/1:

An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member’s class or a class derived from that class, or


• to form a pointer to member (7.6.2.1), or


• if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

When we are inside the C::foo function, the first of those bullet points applies. The code is the same as &c but with unnecessary qualification. This has type int *. You could output this with std::cout << &(C::c); which would show a memory address, the address of this->c.

When we are in the main function , none of the three bullet points apply and therefore the &(C::c) is ill-formed.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

M.M

102k10106224

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
  – melpomene
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

up vote
-2
down vote

up vote
-2
down vote

Q1: There is special meaning to the syntax of & followed by an unparenthesized qualified-id. It means to form a pointer-to-member. Furthermore, there is no other way to form a pointer-to-member.

Note: The section C++17 [expr.prim.paren]/1 says that redundant parentheses can be inserted around any expression "except otherwise indicated", however this situation seems to be intended as "otherwise indicated".

---

Q3: In both cases where you write printf("%pn", &C::c);, &C::c is a pointer-to-member. The %p format specifier is only for void * so this causes undefined behaviour, and the program output is meaningless.

The code cout << &C::c; outputs a pointer-to-member via operator<<(bool val), since there is implicit conversion from pointer-to-member to bool (with result true in all cases), see [conv.bool]/1.

For further discussion of how to print a pointer-to-member, see this answer.

---

Q2: The code &(C::c) does not form a pointer-to-member as explained above.

Now, the code C::c is in the grammatical category id-expression. (Which is qualified-id and unqualified-id). An id-expression has some restrictions on its use, [expr.prim.id]/1:

An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member’s class or a class derived from that class, or


• to form a pointer to member (7.6.2.1), or


• if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

When we are inside the C::foo function, the first of those bullet points applies. The code is the same as &c but with unnecessary qualification. This has type int *. You could output this with std::cout << &(C::c); which would show a memory address, the address of this->c.

When we are in the main function , none of the three bullet points apply and therefore the &(C::c) is ill-formed.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

M.M

102k10106224

Q1: There is special meaning to the syntax of & followed by an unparenthesized qualified-id. It means to form a pointer-to-member. Furthermore, there is no other way to form a pointer-to-member.

Note: The section C++17 [expr.prim.paren]/1 says that redundant parentheses can be inserted around any expression "except otherwise indicated", however this situation seems to be intended as "otherwise indicated".

---

Q3: In both cases where you write printf("%pn", &C::c);, &C::c is a pointer-to-member. The %p format specifier is only for void * so this causes undefined behaviour, and the program output is meaningless.

The code cout << &C::c; outputs a pointer-to-member via operator<<(bool val), since there is implicit conversion from pointer-to-member to bool (with result true in all cases), see [conv.bool]/1.

For further discussion of how to print a pointer-to-member, see this answer.

---

Q2: The code &(C::c) does not form a pointer-to-member as explained above.

Now, the code C::c is in the grammatical category id-expression. (Which is qualified-id and unqualified-id). An id-expression has some restrictions on its use, [expr.prim.id]/1:

An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

• as part of a class member access in which the object expression refers to the member’s class or a class derived from that class, or


• to form a pointer to member (7.6.2.1), or


• if that id-expression denotes a non-static data member and it appears in an unevaluated operand.

When we are inside the C::foo function, the first of those bullet points applies. The code is the same as &c but with unnecessary qualification. This has type int *. You could output this with std::cout << &(C::c); which would show a memory address, the address of this->c.

When we are in the main function , none of the three bullet points apply and therefore the &(C::c) is ill-formed.

share|improve this answer

edited 1 hour ago

answered 2 hours ago

M.M

102k10106224

share|improve this answer

share|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 2 hours ago

M.M

102k10106224

answered 2 hours ago

M.M

102k10106224

answered 2 hours ago

M.M

102k10106224

102k10106224

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
  – melpomene
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
  – melpomene
  2 hours ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
  – melpomene
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
  – melpomene
  2 hours ago
  

  
  
  
  

The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
– melpomene
2 hours ago

The compiler doesn't say "no", it says invalid use of non-static data member 'C::c'.
– melpomene
2 hours ago

Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
– melpomene
2 hours ago

Please remove the // from the line // printf("%pn", &(C::c)); // compile error, invalid use of non-static data member 'C::c' and try again.
– melpomene
2 hours ago

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