constexpr begin of a std::array

Clash Royale CLAN TAG#URR8PPP

up vote
6
down vote

favorite

1

I am having trouble understanding why both gcc-8.2.0 and clang-7.0.0 reject the following code (live code here):

#include <array>

int main() 

 constexpr std::array<int,3> v1,2,3;
 constexpr auto b = v.begin(); // error: not a constexpr 
 return 0;

with the error

error: '(std::array<int, 3>::const_pointer)(& v.std::array<int,3>::_M_elems)' 
is not a constant expression (constexpr auto b = v.begin();)

According to en.cppreference.com, the begin() member function is declared constexpr. Is this a compiler bug?

c++ c++17 constexpr

share|improve this question

edited 1 hour ago

Barry

172k18292537

asked 1 hour ago

linuxfever

1,7381923

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use .cbegin .
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @АлексейНеудачин makes no difference.
  – user2079303
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @АлексейНеудачин begin() and cbegin() do the same thing on const objects.
  – Barry
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so try auto b=... and then if it fails const . also you change auto to std::....::citerator
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

1

I am having trouble understanding why both gcc-8.2.0 and clang-7.0.0 reject the following code (live code here):

#include <array>

int main() 

 constexpr std::array<int,3> v1,2,3;
 constexpr auto b = v.begin(); // error: not a constexpr 
 return 0;

with the error

error: '(std::array<int, 3>::const_pointer)(& v.std::array<int,3>::_M_elems)' 
is not a constant expression (constexpr auto b = v.begin();)

According to en.cppreference.com, the begin() member function is declared constexpr. Is this a compiler bug?

c++ c++17 constexpr

share|improve this question

edited 1 hour ago

Barry

172k18292537

asked 1 hour ago

linuxfever

1,7381923

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use .cbegin .
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @АлексейНеудачин makes no difference.
  – user2079303
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @АлексейНеудачин begin() and cbegin() do the same thing on const objects.
  – Barry
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so try auto b=... and then if it fails const . also you change auto to std::....::citerator
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

1

up vote
6
down vote

favorite

1

1

I am having trouble understanding why both gcc-8.2.0 and clang-7.0.0 reject the following code (live code here):

#include <array>

int main() 

 constexpr std::array<int,3> v1,2,3;
 constexpr auto b = v.begin(); // error: not a constexpr 
 return 0;

with the error

error: '(std::array<int, 3>::const_pointer)(& v.std::array<int,3>::_M_elems)' 
is not a constant expression (constexpr auto b = v.begin();)

According to en.cppreference.com, the begin() member function is declared constexpr. Is this a compiler bug?

c++ c++17 constexpr

share|improve this question

edited 1 hour ago

Barry

172k18292537

asked 1 hour ago

linuxfever

1,7381923

I am having trouble understanding why both gcc-8.2.0 and clang-7.0.0 reject the following code (live code here):

#include <array>

int main() 

 constexpr std::array<int,3> v1,2,3;
 constexpr auto b = v.begin(); // error: not a constexpr 
 return 0;

with the error

error: '(std::array<int, 3>::const_pointer)(& v.std::array<int,3>::_M_elems)' 
is not a constant expression (constexpr auto b = v.begin();)

According to en.cppreference.com, the begin() member function is declared constexpr. Is this a compiler bug?

c++ c++17 constexpr

c++ c++17 constexpr

share|improve this question

edited 1 hour ago

Barry

172k18292537

asked 1 hour ago

linuxfever

1,7381923

share|improve this question

edited 1 hour ago

Barry

172k18292537

asked 1 hour ago

linuxfever

1,7381923

share|improve this question

share|improve this question

edited 1 hour ago

Barry

172k18292537

edited 1 hour ago

Barry

172k18292537

edited 1 hour ago

Barry

172k18292537

172k18292537

asked 1 hour ago

linuxfever

1,7381923

asked 1 hour ago

linuxfever

1,7381923

asked 1 hour ago

linuxfever

1,7381923

1,7381923

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use .cbegin .
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @АлексейНеудачин makes no difference.
  – user2079303
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @АлексейНеудачин begin() and cbegin() do the same thing on const objects.
  – Barry
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so try auto b=... and then if it fails const . also you change auto to std::....::citerator
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  use .cbegin .
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  @АлексейНеудачин makes no difference.
  – user2079303
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @АлексейНеудачин begin() and cbegin() do the same thing on const objects.
  – Barry
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  so try auto b=... and then if it fails const . also you change auto to std::....::citerator
  – ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
  1 hour ago
  
  

  
  
  
  

use .cbegin .
– ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
1 hour ago

use .cbegin .
– ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
1 hour ago

2

2

@АлексейНеудачин makes no difference.
– user2079303
1 hour ago

@АлексейНеудачин makes no difference.
– user2079303
1 hour ago

@АлексейНеудачин begin() and cbegin() do the same thing on const objects.
– Barry
1 hour ago

@АлексейНеудачин begin() and cbegin() do the same thing on const objects.
– Barry
1 hour ago

so try auto b=... and then if it fails const . also you change auto to std::....::citerator
– ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
1 hour ago

so try auto b=... and then if it fails const . also you change auto to std::....::citerator
– ÐÐ»ÐµÐºÑÐµÐ¹ Неудачин
1 hour ago

add a comment | 

2 Answers
2

active

oldest

votes

up vote
19
down vote



accepted

So let's sidestep std::array to make this a bit easier:

template <typename T, size_t N>
struct array 
 T elems[N];

 constexpr T const* begin() const return elems; 
;

void foo() 
 constexpr array<int,3> v1, 2, 3;
 constexpr auto b = v.begin(); // error


constexpr array<int, 3> global_v1, 2, 3;
constexpr auto global_b = global_v.begin(); // ok

Why is b an error but global_b is okay? Likewise, why would b become okay if we declared v to be static constexpr? The problem is fundamentally about pointers. In order to have a constant expression that's a pointer, it has to point to one, known, constant thing, always. That doesn't really work for local variables without static storage duration, since they have fundamentally mutable address. But for function-local statics or globals, they do have one constant address, so you can take a constant pointer to them.

---

In standardese, from [expr.const]/6:

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

• if the value is an object of class type, [...]


• if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object ([expr.add]), the address of a function, or a null pointer value, and


• [...]

b is none of those things in the second bullet, so this fails. But global_b satisfies the bolded condition - as would b if v were declared static.

share|improve this answer

answered 1 hour ago

Barry

172k18292537

add a comment | 

up vote
-3
down vote

Because standard iterators cannot be used in constexpr expressions, there is a proposal for it though which I think is available in c++17 check with c++17 compatible compiler

share|improve this answer

answered 1 hour ago

Ogunleye Ayowale Pius

114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
  – Scheff
  49 mins ago
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
19
down vote



accepted

So let's sidestep std::array to make this a bit easier:

template <typename T, size_t N>
struct array 
 T elems[N];

 constexpr T const* begin() const return elems; 
;

void foo() 
 constexpr array<int,3> v1, 2, 3;
 constexpr auto b = v.begin(); // error


constexpr array<int, 3> global_v1, 2, 3;
constexpr auto global_b = global_v.begin(); // ok

Why is b an error but global_b is okay? Likewise, why would b become okay if we declared v to be static constexpr? The problem is fundamentally about pointers. In order to have a constant expression that's a pointer, it has to point to one, known, constant thing, always. That doesn't really work for local variables without static storage duration, since they have fundamentally mutable address. But for function-local statics or globals, they do have one constant address, so you can take a constant pointer to them.

---

In standardese, from [expr.const]/6:

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

• if the value is an object of class type, [...]


• if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object ([expr.add]), the address of a function, or a null pointer value, and


• [...]

b is none of those things in the second bullet, so this fails. But global_b satisfies the bolded condition - as would b if v were declared static.

share|improve this answer

answered 1 hour ago

Barry

172k18292537

add a comment | 

up vote
19
down vote



accepted

So let's sidestep std::array to make this a bit easier:

template <typename T, size_t N>
struct array 
 T elems[N];

 constexpr T const* begin() const return elems; 
;

void foo() 
 constexpr array<int,3> v1, 2, 3;
 constexpr auto b = v.begin(); // error


constexpr array<int, 3> global_v1, 2, 3;
constexpr auto global_b = global_v.begin(); // ok

Why is b an error but global_b is okay? Likewise, why would b become okay if we declared v to be static constexpr? The problem is fundamentally about pointers. In order to have a constant expression that's a pointer, it has to point to one, known, constant thing, always. That doesn't really work for local variables without static storage duration, since they have fundamentally mutable address. But for function-local statics or globals, they do have one constant address, so you can take a constant pointer to them.

---

In standardese, from [expr.const]/6:

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

• if the value is an object of class type, [...]


• if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object ([expr.add]), the address of a function, or a null pointer value, and


• [...]

b is none of those things in the second bullet, so this fails. But global_b satisfies the bolded condition - as would b if v were declared static.

share|improve this answer

answered 1 hour ago

Barry

172k18292537

add a comment | 

up vote
19
down vote



accepted

up vote
19
down vote



accepted

So let's sidestep std::array to make this a bit easier:

template <typename T, size_t N>
struct array 
 T elems[N];

 constexpr T const* begin() const return elems; 
;

void foo() 
 constexpr array<int,3> v1, 2, 3;
 constexpr auto b = v.begin(); // error


constexpr array<int, 3> global_v1, 2, 3;
constexpr auto global_b = global_v.begin(); // ok

Why is b an error but global_b is okay? Likewise, why would b become okay if we declared v to be static constexpr? The problem is fundamentally about pointers. In order to have a constant expression that's a pointer, it has to point to one, known, constant thing, always. That doesn't really work for local variables without static storage duration, since they have fundamentally mutable address. But for function-local statics or globals, they do have one constant address, so you can take a constant pointer to them.

---

In standardese, from [expr.const]/6:

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

• if the value is an object of class type, [...]


• if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object ([expr.add]), the address of a function, or a null pointer value, and


• [...]

b is none of those things in the second bullet, so this fails. But global_b satisfies the bolded condition - as would b if v were declared static.

share|improve this answer

answered 1 hour ago

Barry

172k18292537

So let's sidestep std::array to make this a bit easier:

template <typename T, size_t N>
struct array 
 T elems[N];

 constexpr T const* begin() const return elems; 
;

void foo() 
 constexpr array<int,3> v1, 2, 3;
 constexpr auto b = v.begin(); // error


constexpr array<int, 3> global_v1, 2, 3;
constexpr auto global_b = global_v.begin(); // ok

Why is b an error but global_b is okay? Likewise, why would b become okay if we declared v to be static constexpr? The problem is fundamentally about pointers. In order to have a constant expression that's a pointer, it has to point to one, known, constant thing, always. That doesn't really work for local variables without static storage duration, since they have fundamentally mutable address. But for function-local statics or globals, they do have one constant address, so you can take a constant pointer to them.

---

In standardese, from [expr.const]/6:

A constant expression is either a glvalue core constant expression that refers to an entity that is a permitted result of a constant expression (as defined below), or a prvalue core constant expression whose value satisfies the following constraints:

• if the value is an object of class type, [...]


• if the value is of pointer type, it contains the address of an object with static storage duration, the address past the end of such an object ([expr.add]), the address of a function, or a null pointer value, and


• [...]

b is none of those things in the second bullet, so this fails. But global_b satisfies the bolded condition - as would b if v were declared static.

share|improve this answer

answered 1 hour ago

Barry

172k18292537

share|improve this answer

share|improve this answer

answered 1 hour ago

Barry

172k18292537

answered 1 hour ago

Barry

172k18292537

answered 1 hour ago

Barry

172k18292537

172k18292537

add a comment | 

add a comment | 

up vote
-3
down vote

Because standard iterators cannot be used in constexpr expressions, there is a proposal for it though which I think is available in c++17 check with c++17 compatible compiler

share|improve this answer

answered 1 hour ago

Ogunleye Ayowale Pius

114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
  – Scheff
  49 mins ago
  
  

  
  
  
  

add a comment | 

up vote
-3
down vote

Because standard iterators cannot be used in constexpr expressions, there is a proposal for it though which I think is available in c++17 check with c++17 compatible compiler

share|improve this answer

answered 1 hour ago

Ogunleye Ayowale Pius

114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
  – Scheff
  49 mins ago
  
  

  
  
  
  

add a comment | 

up vote
-3
down vote

up vote
-3
down vote

Because standard iterators cannot be used in constexpr expressions, there is a proposal for it though which I think is available in c++17 check with c++17 compatible compiler

share|improve this answer

answered 1 hour ago

Ogunleye Ayowale Pius

114

Because standard iterators cannot be used in constexpr expressions, there is a proposal for it though which I think is available in c++17 check with c++17 compatible compiler

share|improve this answer

answered 1 hour ago

Ogunleye Ayowale Pius

114

share|improve this answer

share|improve this answer

answered 1 hour ago

Ogunleye Ayowale Pius

114

answered 1 hour ago

Ogunleye Ayowale Pius

114

answered 1 hour ago

Ogunleye Ayowale Pius

114

114

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
  – Scheff
  49 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
  – Scheff
  49 mins ago
  
  

  
  
  
  

1

1

check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
– Scheff
49 mins ago

check with c++17 compatible compiler This is what OP did. You can see it if you follow OP's here link. (She/he has explicitly set -std=c++17 in the options of godbolt.org.)
– Scheff
49 mins ago

add a comment | 

 

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