Is â€œsizeof new int;â€ UB?

up vote
7
down vote

favorite

code:

#include<iostream>

using namespace std;

int main() 

 size_t i = sizeof new int;

 cout<<i;

In GCC compiler, working fine without any warning or error and printed output 8.

But, in clang compiler, i got following warning:

warning: expression with side effects has no effect in an unevaluated context [-Wunevaluated-expression]
 size_t i = sizeof new int;

Which one is true?

Is sizeof new int; UB?

asked 33 mins ago

M.S Chaudhari

2,65211030

1

I see nothing wrong. This warning is more of a reminder, everything is well defined here.
â€“Â YSC
32 mins ago

This question is related to the C counterpart Why does sizeof(x++) not increment x?, but it is a bit clearer in C++.
â€“Â Bartek Banachewicz
18 mins ago

@BartekBanachewicz - a closer equivalent in C would be sizeof malloc(some_value) which would not call malloc() and, if performed after #include <stdlib.h>, would give a result equal to sizeof(void *).
â€“Â Peter
11 mins ago

add a commentÂ |Â

up vote
7
down vote

favorite

code:

#include<iostream>

using namespace std;

int main() 

 size_t i = sizeof new int;

 cout<<i;

In GCC compiler, working fine without any warning or error and printed output 8.

But, in clang compiler, i got following warning:

warning: expression with side effects has no effect in an unevaluated context [-Wunevaluated-expression]
 size_t i = sizeof new int;

Which one is true?

Is sizeof new int; UB?

asked 33 mins ago

M.S Chaudhari

2,65211030

1

I see nothing wrong. This warning is more of a reminder, everything is well defined here.
â€“Â YSC
32 mins ago

This question is related to the C counterpart Why does sizeof(x++) not increment x?, but it is a bit clearer in C++.
â€“Â Bartek Banachewicz
18 mins ago

@BartekBanachewicz - a closer equivalent in C would be sizeof malloc(some_value) which would not call malloc() and, if performed after #include <stdlib.h>, would give a result equal to sizeof(void *).
â€“Â Peter
11 mins ago

add a commentÂ |Â

up vote
7
down vote

favorite

code:

#include<iostream>

using namespace std;

int main() 

 size_t i = sizeof new int;

 cout<<i;

In GCC compiler, working fine without any warning or error and printed output 8.

But, in clang compiler, i got following warning:

warning: expression with side effects has no effect in an unevaluated context [-Wunevaluated-expression]
 size_t i = sizeof new int;

Which one is true?

Is sizeof new int; UB?

asked 33 mins ago

M.S Chaudhari

2,65211030

code:

#include<iostream>

using namespace std;

int main() 

 size_t i = sizeof new int;

 cout<<i;

In GCC compiler, working fine without any warning or error and printed output 8.

But, in clang compiler, i got following warning:

warning: expression with side effects has no effect in an unevaluated context [-Wunevaluated-expression]
 size_t i = sizeof new int;

Which one is true?

Is sizeof new int; UB?

c++ c++11 g++ sizeof clang++

asked 33 mins ago

M.S Chaudhari

2,65211030

asked 33 mins ago

M.S Chaudhari

2,65211030

asked 33 mins ago

M.S Chaudhari

2,65211030

asked 33 mins ago

M.S Chaudhari

2,65211030

asked 33 mins ago

M.S Chaudhari

2,65211030

1

I see nothing wrong. This warning is more of a reminder, everything is well defined here.
â€“Â YSC
32 mins ago

This question is related to the C counterpart Why does sizeof(x++) not increment x?, but it is a bit clearer in C++.
â€“Â Bartek Banachewicz
18 mins ago

@BartekBanachewicz - a closer equivalent in C would be sizeof malloc(some_value) which would not call malloc() and, if performed after #include <stdlib.h>, would give a result equal to sizeof(void *).
â€“Â Peter
11 mins ago

add a commentÂ |Â

1

I see nothing wrong. This warning is more of a reminder, everything is well defined here.
â€“Â YSC
32 mins ago

This question is related to the C counterpart Why does sizeof(x++) not increment x?, but it is a bit clearer in C++.
â€“Â Bartek Banachewicz
18 mins ago

@BartekBanachewicz - a closer equivalent in C would be sizeof malloc(some_value) which would not call malloc() and, if performed after #include <stdlib.h>, would give a result equal to sizeof(void *).
â€“Â Peter
11 mins ago

I see nothing wrong. This warning is more of a reminder, everything is well defined here.
â€“Â YSC
32 mins ago

This question is related to the C counterpart Why does sizeof(x++) not increment x?, but it is a bit clearer in C++.
â€“Â Bartek Banachewicz
18 mins ago

@BartekBanachewicz - a closer equivalent in C would be sizeof malloc(some_value) which would not call malloc() and, if performed after #include <stdlib.h>, would give a result equal to sizeof(void *).
â€“Â Peter
11 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
15
down vote

accepted

The warning doesn't state that it's UB; it merely says that the context of use, namely sizeof, won't trigger the side effects (which in case of new is allocating memory).

[expr.sizeof]
The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand. The operand is either an expression, which is an unevaluated operand ([expr.prop]), or a parenthesized type-id.

The standard also helpfully explains what that means:

[expr.context] (...) An unevaluated operand is not evaluated.

It's a fine, although a weird way to write sizeof(int*).

answered 31 mins ago

Bartek Banachewicz

28.8k460103

add a commentÂ |Â

up vote
4
down vote

new operator returns pointer to the allocated memory. So size_t i = sizeof new int; will not cause undefined behaviour.

Warning is legit and only warns about the effect of side-effect on the operand and that's because operands of sizeof is not evaluated.

For example:

int i = 1;
std::cout << i << 'n'; // Prints 1
size_t size = sizeof(i++); // i++ will not be evaluated
std::cout << i << 'n'; // Prints 1

edited 25 mins ago

answered 31 mins ago

haccks

84.2k20123215

Curiously, it's easy to make i++ be evaluated: cpp.sh/9774o
â€“Â Dan M.
9 mins ago

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
15
down vote

accepted

The warning doesn't state that it's UB; it merely says that the context of use, namely sizeof, won't trigger the side effects (which in case of new is allocating memory).

[expr.sizeof]
The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand. The operand is either an expression, which is an unevaluated operand ([expr.prop]), or a parenthesized type-id.

The standard also helpfully explains what that means:

[expr.context] (...) An unevaluated operand is not evaluated.

It's a fine, although a weird way to write sizeof(int*).

answered 31 mins ago

Bartek Banachewicz

28.8k460103

add a commentÂ |Â

up vote
15
down vote

accepted

The warning doesn't state that it's UB; it merely says that the context of use, namely sizeof, won't trigger the side effects (which in case of new is allocating memory).

[expr.sizeof]
The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand. The operand is either an expression, which is an unevaluated operand ([expr.prop]), or a parenthesized type-id.

The standard also helpfully explains what that means:

[expr.context] (...) An unevaluated operand is not evaluated.

It's a fine, although a weird way to write sizeof(int*).

answered 31 mins ago

Bartek Banachewicz

28.8k460103

add a commentÂ |Â

up vote
15
down vote

accepted

The warning doesn't state that it's UB; it merely says that the context of use, namely sizeof, won't trigger the side effects (which in case of new is allocating memory).

[expr.sizeof]
The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand. The operand is either an expression, which is an unevaluated operand ([expr.prop]), or a parenthesized type-id.

The standard also helpfully explains what that means:

[expr.context] (...) An unevaluated operand is not evaluated.

It's a fine, although a weird way to write sizeof(int*).

answered 31 mins ago

Bartek Banachewicz

28.8k460103

The warning doesn't state that it's UB; it merely says that the context of use, namely sizeof, won't trigger the side effects (which in case of new is allocating memory).

[expr.sizeof]
The sizeof operator yields the number of bytes occupied by a non-potentially-overlapping object of the type of its operand. The operand is either an expression, which is an unevaluated operand ([expr.prop]), or a parenthesized type-id.

The standard also helpfully explains what that means:

[expr.context] (...) An unevaluated operand is not evaluated.

It's a fine, although a weird way to write sizeof(int*).

answered 31 mins ago

Bartek Banachewicz

28.8k460103

answered 31 mins ago

Bartek Banachewicz

28.8k460103

answered 31 mins ago

Bartek Banachewicz

28.8k460103

answered 31 mins ago

Bartek Banachewicz

28.8k460103

add a commentÂ |Â

up vote
4
down vote

For example:

int i = 1;
std::cout << i << 'n'; // Prints 1
size_t size = sizeof(i++); // i++ will not be evaluated
std::cout << i << 'n'; // Prints 1

edited 25 mins ago

answered 31 mins ago

haccks

84.2k20123215

Curiously, it's easy to make i++ be evaluated: cpp.sh/9774o
â€“Â Dan M.
9 mins ago

add a commentÂ |Â

up vote
4
down vote

For example:

int i = 1;
std::cout << i << 'n'; // Prints 1
size_t size = sizeof(i++); // i++ will not be evaluated
std::cout << i << 'n'; // Prints 1

edited 25 mins ago

answered 31 mins ago

haccks

84.2k20123215

Curiously, it's easy to make i++ be evaluated: cpp.sh/9774o
â€“Â Dan M.
9 mins ago

add a commentÂ |Â

up vote
4
down vote

For example:

int i = 1;
std::cout << i << 'n'; // Prints 1
size_t size = sizeof(i++); // i++ will not be evaluated
std::cout << i << 'n'; // Prints 1

edited 25 mins ago

answered 31 mins ago

haccks

84.2k20123215

For example:

int i = 1;
std::cout << i << 'n'; // Prints 1
size_t size = sizeof(i++); // i++ will not be evaluated
std::cout << i << 'n'; // Prints 1

edited 25 mins ago

answered 31 mins ago

haccks

84.2k20123215

edited 25 mins ago

answered 31 mins ago

haccks

84.2k20123215

answered 31 mins ago

haccks

84.2k20123215

answered 31 mins ago

haccks

84.2k20123215

Curiously, it's easy to make i++ be evaluated: cpp.sh/9774o
â€“Â Dan M.
9 mins ago

add a commentÂ |Â

Curiously, it's easy to make i++ be evaluated: cpp.sh/9774o
â€“Â Dan M.
9 mins ago

Curiously, it's easy to make i++ be evaluated: cpp.sh/9774o
â€“Â Dan M.
9 mins ago

add a commentÂ |Â

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