Constexpr operator new

Is it possible to overload the operator new to be constexpr function? Something like:

constexpr void * operator new( std::size_t count );

The reason why would be to execute constexpr function within the overloaded operator body where count argument value would be an input data... As the operator is invoked by:

SomeClass * foo = new SomeClass();

The size of the data type is know at compile time, isn’t it? (count== sizeof(SomeClass)) So the count can be considered as compile time constant?

constexpr void * operator new( std::size_t count )

{

  if constexpr ( count >= 10 ) { /* do some compile-time business */ }

}

Many thanks in advance to anyone willing to help!

edited Jan 26 at 15:17

marc_s

577k12911151260

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

4

constexpr cannot have side effects, thus this would be contradictory

– OznOg
Nov 21 '18 at 19:19

4

There is no such thing as compile-time dynamic memory allocation. Allocating memory for your program at compile time is by definition just specifying a quantity of static memory necessary at runtime. If you can calculate at compile time you need N bytes at runtime for your task, then just have a properly aligned static buffer of size N.

– François Andrieux
Nov 21 '18 at 19:31

5

It doesn't matter anyway. [expr.const] prohibits new-expressions in constant expressions across the board.

– T.C.
Nov 21 '18 at 20:04

5

You want to look at the C++2a papers trying to make everything constexpr.

– Marc Glisse
Nov 21 '18 at 20:10

1

@FrançoisAndrieux Read this: open-std.org/JTC1/SC22/WG21/docs/papers/2018/p0784r1.html

– Oliv
Nov 21 '18 at 21:27

|
show 6 more comments

Is it possible to overload the operator new to be constexpr function? Something like:

constexpr void * operator new( std::size_t count );

The reason why would be to execute constexpr function within the overloaded operator body where count argument value would be an input data... As the operator is invoked by:

SomeClass * foo = new SomeClass();

The size of the data type is know at compile time, isn’t it? (count== sizeof(SomeClass)) So the count can be considered as compile time constant?

constexpr void * operator new( std::size_t count )

{

  if constexpr ( count >= 10 ) { /* do some compile-time business */ }

}

Many thanks in advance to anyone willing to help!

edited Jan 26 at 15:17

marc_s

577k12911151260

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

4

constexpr cannot have side effects, thus this would be contradictory

– OznOg
Nov 21 '18 at 19:19

4

There is no such thing as compile-time dynamic memory allocation. Allocating memory for your program at compile time is by definition just specifying a quantity of static memory necessary at runtime. If you can calculate at compile time you need N bytes at runtime for your task, then just have a properly aligned static buffer of size N.

– François Andrieux
Nov 21 '18 at 19:31

5

It doesn't matter anyway. [expr.const] prohibits new-expressions in constant expressions across the board.

– T.C.
Nov 21 '18 at 20:04

5

You want to look at the C++2a papers trying to make everything constexpr.

– Marc Glisse
Nov 21 '18 at 20:10

1

@FrançoisAndrieux Read this: open-std.org/JTC1/SC22/WG21/docs/papers/2018/p0784r1.html

– Oliv
Nov 21 '18 at 21:27

|
show 6 more comments

Is it possible to overload the operator new to be constexpr function? Something like:

constexpr void * operator new( std::size_t count );

The reason why would be to execute constexpr function within the overloaded operator body where count argument value would be an input data... As the operator is invoked by:

SomeClass * foo = new SomeClass();

The size of the data type is know at compile time, isn’t it? (count== sizeof(SomeClass)) So the count can be considered as compile time constant?

constexpr void * operator new( std::size_t count )

{

  if constexpr ( count >= 10 ) { /* do some compile-time business */ }

}

Many thanks in advance to anyone willing to help!

edited Jan 26 at 15:17

marc_s

577k12911151260

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

Is it possible to overload the operator new to be constexpr function? Something like:

constexpr void * operator new( std::size_t count );

The reason why would be to execute constexpr function within the overloaded operator body where count argument value would be an input data... As the operator is invoked by:

SomeClass * foo = new SomeClass();

The size of the data type is know at compile time, isn’t it? (count== sizeof(SomeClass)) So the count can be considered as compile time constant?

constexpr void * operator new( std::size_t count )

{

  if constexpr ( count >= 10 ) { /* do some compile-time business */ }

}

Many thanks in advance to anyone willing to help!

c++ c++17 constexpr if-constexpr

edited Jan 26 at 15:17

marc_s

577k12911151260

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

edited Jan 26 at 15:17

marc_s

577k12911151260

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

edited Jan 26 at 15:17

marc_s

577k12911151260

edited Jan 26 at 15:17

marc_s

577k12911151260

edited Jan 26 at 15:17

marc_s

577k12911151260

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

asked Nov 21 '18 at 19:18

Martin Kopecký

1866

4

constexpr cannot have side effects, thus this would be contradictory

– OznOg
Nov 21 '18 at 19:19

4

There is no such thing as compile-time dynamic memory allocation. Allocating memory for your program at compile time is by definition just specifying a quantity of static memory necessary at runtime. If you can calculate at compile time you need N bytes at runtime for your task, then just have a properly aligned static buffer of size N.

– François Andrieux
Nov 21 '18 at 19:31

5

It doesn't matter anyway. [expr.const] prohibits new-expressions in constant expressions across the board.

– T.C.
Nov 21 '18 at 20:04

5

You want to look at the C++2a papers trying to make everything constexpr.

– Marc Glisse
Nov 21 '18 at 20:10

1

@FrançoisAndrieux Read this: open-std.org/JTC1/SC22/WG21/docs/papers/2018/p0784r1.html

– Oliv
Nov 21 '18 at 21:27

|
show 6 more comments

4

constexpr cannot have side effects, thus this would be contradictory

– OznOg
Nov 21 '18 at 19:19

4

There is no such thing as compile-time dynamic memory allocation. Allocating memory for your program at compile time is by definition just specifying a quantity of static memory necessary at runtime. If you can calculate at compile time you need N bytes at runtime for your task, then just have a properly aligned static buffer of size N.

– François Andrieux
Nov 21 '18 at 19:31

5

It doesn't matter anyway. [expr.const] prohibits new-expressions in constant expressions across the board.

– T.C.
Nov 21 '18 at 20:04

5

You want to look at the C++2a papers trying to make everything constexpr.

– Marc Glisse
Nov 21 '18 at 20:10

1

@FrançoisAndrieux Read this: open-std.org/JTC1/SC22/WG21/docs/papers/2018/p0784r1.html

– Oliv
Nov 21 '18 at 21:27

constexpr cannot have side effects, thus this would be contradictory

– OznOg
Nov 21 '18 at 19:19

There is no such thing as compile-time dynamic memory allocation. Allocating memory for your program at compile time is by definition just specifying a quantity of static memory necessary at runtime. If you can calculate at compile time you need N bytes at runtime for your task, then just have a properly aligned static buffer of size N.

– François Andrieux
Nov 21 '18 at 19:31

It doesn't matter anyway. [expr.const] prohibits new-expressions in constant expressions across the board.

– T.C.
Nov 21 '18 at 20:04

You want to look at the C++2a papers trying to make everything constexpr.

– Marc Glisse
Nov 21 '18 at 20:10

@FrançoisAndrieux Read this: open-std.org/JTC1/SC22/WG21/docs/papers/2018/p0784r1.html

– Oliv
Nov 21 '18 at 21:27

|
show 6 more comments

1 Answer
1

active

oldest

votes

You can't overload operator new to be constexpr, the main problem is attributed to the C++ standard directive §9.1.5/1 The constexpr specifier [dcl.constexpr] (Emphasis Mine):

The constexpr specifier shall be applied only to the definition of a
variable or variable template or the declaration of a function or
function template. A function or static data member declared with the
constexpr specifier is implicitly an inline function or variable
(9.1.6). If any declaration of a function or function template has a
constexpr specifier, then all its declarations shall contain the
constexpr specifier.

That is, in order to overload operator new all its previous declarations must also be constexpr, which they aren't and thus overloading it as constexpr you get a compile time error.

edited Nov 21 '18 at 20:06

answered Nov 21 '18 at 19:58

101010

31.6k863124

I'm not 100% sure that this is correct. Doesn't this talk about the case when there are several declarations of the same function (with the same signature)? I don't really see the rationale behind, if one overload was constexpr, then all of them must be constexpr as well.

– geza
Nov 21 '18 at 20:19

Ok, Many thanks you all Guys for your replies.

– Martin Kopecký
Nov 21 '18 at 20:19

add a comment |

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1 Answer
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You can't overload operator new to be constexpr, the main problem is attributed to the C++ standard directive §9.1.5/1 The constexpr specifier [dcl.constexpr] (Emphasis Mine):

The constexpr specifier shall be applied only to the definition of a
variable or variable template or the declaration of a function or
function template. A function or static data member declared with the
constexpr specifier is implicitly an inline function or variable
(9.1.6). If any declaration of a function or function template has a
constexpr specifier, then all its declarations shall contain the
constexpr specifier.

That is, in order to overload operator new all its previous declarations must also be constexpr, which they aren't and thus overloading it as constexpr you get a compile time error.

edited Nov 21 '18 at 20:06

answered Nov 21 '18 at 19:58

101010

31.6k863124

I'm not 100% sure that this is correct. Doesn't this talk about the case when there are several declarations of the same function (with the same signature)? I don't really see the rationale behind, if one overload was constexpr, then all of them must be constexpr as well.

– geza
Nov 21 '18 at 20:19

Ok, Many thanks you all Guys for your replies.

– Martin Kopecký
Nov 21 '18 at 20:19

add a comment |

You can't overload operator new to be constexpr, the main problem is attributed to the C++ standard directive §9.1.5/1 The constexpr specifier [dcl.constexpr] (Emphasis Mine):

The constexpr specifier shall be applied only to the definition of a
variable or variable template or the declaration of a function or
function template. A function or static data member declared with the
constexpr specifier is implicitly an inline function or variable
(9.1.6). If any declaration of a function or function template has a
constexpr specifier, then all its declarations shall contain the
constexpr specifier.

That is, in order to overload operator new all its previous declarations must also be constexpr, which they aren't and thus overloading it as constexpr you get a compile time error.

edited Nov 21 '18 at 20:06

answered Nov 21 '18 at 19:58

101010

31.6k863124

I'm not 100% sure that this is correct. Doesn't this talk about the case when there are several declarations of the same function (with the same signature)? I don't really see the rationale behind, if one overload was constexpr, then all of them must be constexpr as well.

– geza
Nov 21 '18 at 20:19

Ok, Many thanks you all Guys for your replies.

– Martin Kopecký
Nov 21 '18 at 20:19

add a comment |

You can't overload operator new to be constexpr, the main problem is attributed to the C++ standard directive §9.1.5/1 The constexpr specifier [dcl.constexpr] (Emphasis Mine):

The constexpr specifier shall be applied only to the definition of a
variable or variable template or the declaration of a function or
function template. A function or static data member declared with the
constexpr specifier is implicitly an inline function or variable
(9.1.6). If any declaration of a function or function template has a
constexpr specifier, then all its declarations shall contain the
constexpr specifier.

That is, in order to overload operator new all its previous declarations must also be constexpr, which they aren't and thus overloading it as constexpr you get a compile time error.

edited Nov 21 '18 at 20:06

answered Nov 21 '18 at 19:58

101010

31.6k863124

You can't overload operator new to be constexpr, the main problem is attributed to the C++ standard directive §9.1.5/1 The constexpr specifier [dcl.constexpr] (Emphasis Mine):

The constexpr specifier shall be applied only to the definition of a
variable or variable template or the declaration of a function or
function template. A function or static data member declared with the
constexpr specifier is implicitly an inline function or variable
(9.1.6). If any declaration of a function or function template has a
constexpr specifier, then all its declarations shall contain the
constexpr specifier.

That is, in order to overload operator new all its previous declarations must also be constexpr, which they aren't and thus overloading it as constexpr you get a compile time error.

edited Nov 21 '18 at 20:06

answered Nov 21 '18 at 19:58

101010

31.6k863124

edited Nov 21 '18 at 20:06

answered Nov 21 '18 at 19:58

101010

31.6k863124

answered Nov 21 '18 at 19:58

101010

31.6k863124

answered Nov 21 '18 at 19:58

101010

31.6k863124

I'm not 100% sure that this is correct. Doesn't this talk about the case when there are several declarations of the same function (with the same signature)? I don't really see the rationale behind, if one overload was constexpr, then all of them must be constexpr as well.

– geza
Nov 21 '18 at 20:19

Ok, Many thanks you all Guys for your replies.

– Martin Kopecký
Nov 21 '18 at 20:19

add a comment |

I'm not 100% sure that this is correct. Doesn't this talk about the case when there are several declarations of the same function (with the same signature)? I don't really see the rationale behind, if one overload was constexpr, then all of them must be constexpr as well.

– geza
Nov 21 '18 at 20:19

Ok, Many thanks you all Guys for your replies.

– Martin Kopecký
Nov 21 '18 at 20:19

I'm not 100% sure that this is correct. Doesn't this talk about the case when there are several declarations of the same function (with the same signature)? I don't really see the rationale behind, if one overload was constexpr, then all of them must be constexpr as well.

– geza
Nov 21 '18 at 20:19

Ok, Many thanks you all Guys for your replies.

– Martin Kopecký
Nov 21 '18 at 20:19

add a comment |

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