Why isn't abs constexpr?
In <cinttypes>
, since C++11, there are the following two overloads:
std::intmax_t abs( std::intmax_t n );
std::intmax_t imaxabs( std::intmax_t n );
Why aren't those two functions constexpr
?
c++ c++11 constexpr
|
show 8 more comments
In <cinttypes>
, since C++11, there are the following two overloads:
std::intmax_t abs( std::intmax_t n );
std::intmax_t imaxabs( std::intmax_t n );
Why aren't those two functions constexpr
?
c++ c++11 constexpr
5
@BasileStarynkevitch: Can you explain one of these compatibility reasons?
– Lightness Races in Orbit
Dec 30 '14 at 16:14
5
@BasileStarynkevitch I don't buy that. Given a compatible compiler, a constexpr piece of code will work the same as it used, i.e. when passed in non-constexpr values it will return a non-constexpr. If it preferable to compatible to not make these functions constexpr (or alteast with constexpr overloads) then it would be preferable to not include move instructors for vector. Being that this uses namespaces (so is not C) and is a header provided only in C++11, I would like to see constexpr overlaods.
– IdeaHat
Dec 30 '14 at 16:16
7
abs
is a C99 function. And C99 has no idea ofconstexpr
– Basile Starynkevitch
Dec 30 '14 at 16:32
17
There are a lot of functions that could be constexpr and are not (yet). More functions will slowly become constexpr with each revision of the standard. Feel free to participate in that process (instructions on isocpp.org).
– Marc Glisse
Dec 30 '14 at 16:40
3
abs
is notconstexpr
because the standard decided so. I think this is the wrong place to ask why. What I can say is that if you really need aconstexpr abs
, write your own one. You can't go wrong.
– edmz
Dec 30 '14 at 17:02
|
show 8 more comments
In <cinttypes>
, since C++11, there are the following two overloads:
std::intmax_t abs( std::intmax_t n );
std::intmax_t imaxabs( std::intmax_t n );
Why aren't those two functions constexpr
?
c++ c++11 constexpr
In <cinttypes>
, since C++11, there are the following two overloads:
std::intmax_t abs( std::intmax_t n );
std::intmax_t imaxabs( std::intmax_t n );
Why aren't those two functions constexpr
?
c++ c++11 constexpr
c++ c++11 constexpr
asked Dec 30 '14 at 16:05
ShoeShoe
59.2k26123215
59.2k26123215
5
@BasileStarynkevitch: Can you explain one of these compatibility reasons?
– Lightness Races in Orbit
Dec 30 '14 at 16:14
5
@BasileStarynkevitch I don't buy that. Given a compatible compiler, a constexpr piece of code will work the same as it used, i.e. when passed in non-constexpr values it will return a non-constexpr. If it preferable to compatible to not make these functions constexpr (or alteast with constexpr overloads) then it would be preferable to not include move instructors for vector. Being that this uses namespaces (so is not C) and is a header provided only in C++11, I would like to see constexpr overlaods.
– IdeaHat
Dec 30 '14 at 16:16
7
abs
is a C99 function. And C99 has no idea ofconstexpr
– Basile Starynkevitch
Dec 30 '14 at 16:32
17
There are a lot of functions that could be constexpr and are not (yet). More functions will slowly become constexpr with each revision of the standard. Feel free to participate in that process (instructions on isocpp.org).
– Marc Glisse
Dec 30 '14 at 16:40
3
abs
is notconstexpr
because the standard decided so. I think this is the wrong place to ask why. What I can say is that if you really need aconstexpr abs
, write your own one. You can't go wrong.
– edmz
Dec 30 '14 at 17:02
|
show 8 more comments
5
@BasileStarynkevitch: Can you explain one of these compatibility reasons?
– Lightness Races in Orbit
Dec 30 '14 at 16:14
5
@BasileStarynkevitch I don't buy that. Given a compatible compiler, a constexpr piece of code will work the same as it used, i.e. when passed in non-constexpr values it will return a non-constexpr. If it preferable to compatible to not make these functions constexpr (or alteast with constexpr overloads) then it would be preferable to not include move instructors for vector. Being that this uses namespaces (so is not C) and is a header provided only in C++11, I would like to see constexpr overlaods.
– IdeaHat
Dec 30 '14 at 16:16
7
abs
is a C99 function. And C99 has no idea ofconstexpr
– Basile Starynkevitch
Dec 30 '14 at 16:32
17
There are a lot of functions that could be constexpr and are not (yet). More functions will slowly become constexpr with each revision of the standard. Feel free to participate in that process (instructions on isocpp.org).
– Marc Glisse
Dec 30 '14 at 16:40
3
abs
is notconstexpr
because the standard decided so. I think this is the wrong place to ask why. What I can say is that if you really need aconstexpr abs
, write your own one. You can't go wrong.
– edmz
Dec 30 '14 at 17:02
5
5
@BasileStarynkevitch: Can you explain one of these compatibility reasons?
– Lightness Races in Orbit
Dec 30 '14 at 16:14
@BasileStarynkevitch: Can you explain one of these compatibility reasons?
– Lightness Races in Orbit
Dec 30 '14 at 16:14
5
5
@BasileStarynkevitch I don't buy that. Given a compatible compiler, a constexpr piece of code will work the same as it used, i.e. when passed in non-constexpr values it will return a non-constexpr. If it preferable to compatible to not make these functions constexpr (or alteast with constexpr overloads) then it would be preferable to not include move instructors for vector. Being that this uses namespaces (so is not C) and is a header provided only in C++11, I would like to see constexpr overlaods.
– IdeaHat
Dec 30 '14 at 16:16
@BasileStarynkevitch I don't buy that. Given a compatible compiler, a constexpr piece of code will work the same as it used, i.e. when passed in non-constexpr values it will return a non-constexpr. If it preferable to compatible to not make these functions constexpr (or alteast with constexpr overloads) then it would be preferable to not include move instructors for vector. Being that this uses namespaces (so is not C) and is a header provided only in C++11, I would like to see constexpr overlaods.
– IdeaHat
Dec 30 '14 at 16:16
7
7
abs
is a C99 function. And C99 has no idea of constexpr
– Basile Starynkevitch
Dec 30 '14 at 16:32
abs
is a C99 function. And C99 has no idea of constexpr
– Basile Starynkevitch
Dec 30 '14 at 16:32
17
17
There are a lot of functions that could be constexpr and are not (yet). More functions will slowly become constexpr with each revision of the standard. Feel free to participate in that process (instructions on isocpp.org).
– Marc Glisse
Dec 30 '14 at 16:40
There are a lot of functions that could be constexpr and are not (yet). More functions will slowly become constexpr with each revision of the standard. Feel free to participate in that process (instructions on isocpp.org).
– Marc Glisse
Dec 30 '14 at 16:40
3
3
abs
is not constexpr
because the standard decided so. I think this is the wrong place to ask why. What I can say is that if you really need a constexpr abs
, write your own one. You can't go wrong.– edmz
Dec 30 '14 at 17:02
abs
is not constexpr
because the standard decided so. I think this is the wrong place to ask why. What I can say is that if you really need a constexpr abs
, write your own one. You can't go wrong.– edmz
Dec 30 '14 at 17:02
|
show 8 more comments
3 Answers
3
active
oldest
votes
I can't give a good reason for why abs
couldn't be constexpr
and apparently neither can gcc
. When I use gcc 4.9.2
with this program:
#include <cstdlib>
#include <cinttypes>
#include <cassert>
constexpr intmax_t abs3 = std::abs(3);
constexpr intmax_t absneg3 = std::abs(-3);
int main()
{
assert(abs3 == absneg3);
}
it compiles and runs to completion with no warnings or errors. You can try it here. However, clang++
(version 3.5.0) throws a compile-time error:
abs.cpp:6:20: error: constexpr variable 'abs3' must be initialized by a constant expression.
I think that clang++
actually gets it right here, because in section 27.9.2 [c.files] of the 2011 standard, it says:
The contents of header are the same as the Standard C Library header , with the following changes:
— the header includes the header instead of , and
— if and only if the type intmax_t designates an extended integer type (3.9.1), the following function
signatures are added:
intmax_t abs(intmax_t);
imaxdiv_t div(intmax_t, intmax_t);
which shall have the same semantics as the function signatures intmax_t imaxabs(intmax_t) and
imaxdiv_t imaxdiv(intmax_t, intmax_t), respectively.
In the current working draft of the C++ standard, as in the published 2014 version, it says in section 17.6.5.6 [constexpr.functions]:
This standard explicitly requires that certain standard library functions are constexpr (7.1.5). An implementation shall not declare any standard library function signature as constexpr except for those where it
is explicitly required.
So the result, for now, is that these functions are still not constexpr
according to the standard (which you knew) but they could be, as demonstrated by the gcc
compiler.
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in definingconstexpr
functions in<cmath>
.
– leek
Nov 22 '18 at 23:04
add a comment |
The simple explanation is probably that nobody went through every function listed as part of the standard library, and updated the entry to make it constexpr
.
Updating a standard is a process involving human effort to propose changes and get those proposals reviewed and accepted. And that means, when a new feature is added - and constexpr
is a new feature - not everything else is automagically updated to use that feature. This particular case may well be updated in a future evolution of the standard - if some people take on the - often thankless - task of proposing the set of changes, some other people take on the task of reviewing it, and convincing those with voting power on the committee to accept it.
add a comment |
It has been proposed in P0533:
A function in
<cmath>
shall be declared
constexpr
if and only if:
- When taken to act on the set of rational numbers,
the function is closed (excluding division by zero);
- The function does not modify any of its arguments
which have external visibility;
- The function is not strongly dependent on the
rounding mode.
By means of a brief illustration,
abs
satisfies all three criteria; however, functions such asexp
,sqrt
,cos
,sin
fall foul of the first criterion and so are excluded asconstexpr
candidates. Finally, as discussed above,nearbyint
fails the final criterion.
add a comment |
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3 Answers
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3 Answers
3
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oldest
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oldest
votes
I can't give a good reason for why abs
couldn't be constexpr
and apparently neither can gcc
. When I use gcc 4.9.2
with this program:
#include <cstdlib>
#include <cinttypes>
#include <cassert>
constexpr intmax_t abs3 = std::abs(3);
constexpr intmax_t absneg3 = std::abs(-3);
int main()
{
assert(abs3 == absneg3);
}
it compiles and runs to completion with no warnings or errors. You can try it here. However, clang++
(version 3.5.0) throws a compile-time error:
abs.cpp:6:20: error: constexpr variable 'abs3' must be initialized by a constant expression.
I think that clang++
actually gets it right here, because in section 27.9.2 [c.files] of the 2011 standard, it says:
The contents of header are the same as the Standard C Library header , with the following changes:
— the header includes the header instead of , and
— if and only if the type intmax_t designates an extended integer type (3.9.1), the following function
signatures are added:
intmax_t abs(intmax_t);
imaxdiv_t div(intmax_t, intmax_t);
which shall have the same semantics as the function signatures intmax_t imaxabs(intmax_t) and
imaxdiv_t imaxdiv(intmax_t, intmax_t), respectively.
In the current working draft of the C++ standard, as in the published 2014 version, it says in section 17.6.5.6 [constexpr.functions]:
This standard explicitly requires that certain standard library functions are constexpr (7.1.5). An implementation shall not declare any standard library function signature as constexpr except for those where it
is explicitly required.
So the result, for now, is that these functions are still not constexpr
according to the standard (which you knew) but they could be, as demonstrated by the gcc
compiler.
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in definingconstexpr
functions in<cmath>
.
– leek
Nov 22 '18 at 23:04
add a comment |
I can't give a good reason for why abs
couldn't be constexpr
and apparently neither can gcc
. When I use gcc 4.9.2
with this program:
#include <cstdlib>
#include <cinttypes>
#include <cassert>
constexpr intmax_t abs3 = std::abs(3);
constexpr intmax_t absneg3 = std::abs(-3);
int main()
{
assert(abs3 == absneg3);
}
it compiles and runs to completion with no warnings or errors. You can try it here. However, clang++
(version 3.5.0) throws a compile-time error:
abs.cpp:6:20: error: constexpr variable 'abs3' must be initialized by a constant expression.
I think that clang++
actually gets it right here, because in section 27.9.2 [c.files] of the 2011 standard, it says:
The contents of header are the same as the Standard C Library header , with the following changes:
— the header includes the header instead of , and
— if and only if the type intmax_t designates an extended integer type (3.9.1), the following function
signatures are added:
intmax_t abs(intmax_t);
imaxdiv_t div(intmax_t, intmax_t);
which shall have the same semantics as the function signatures intmax_t imaxabs(intmax_t) and
imaxdiv_t imaxdiv(intmax_t, intmax_t), respectively.
In the current working draft of the C++ standard, as in the published 2014 version, it says in section 17.6.5.6 [constexpr.functions]:
This standard explicitly requires that certain standard library functions are constexpr (7.1.5). An implementation shall not declare any standard library function signature as constexpr except for those where it
is explicitly required.
So the result, for now, is that these functions are still not constexpr
according to the standard (which you knew) but they could be, as demonstrated by the gcc
compiler.
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in definingconstexpr
functions in<cmath>
.
– leek
Nov 22 '18 at 23:04
add a comment |
I can't give a good reason for why abs
couldn't be constexpr
and apparently neither can gcc
. When I use gcc 4.9.2
with this program:
#include <cstdlib>
#include <cinttypes>
#include <cassert>
constexpr intmax_t abs3 = std::abs(3);
constexpr intmax_t absneg3 = std::abs(-3);
int main()
{
assert(abs3 == absneg3);
}
it compiles and runs to completion with no warnings or errors. You can try it here. However, clang++
(version 3.5.0) throws a compile-time error:
abs.cpp:6:20: error: constexpr variable 'abs3' must be initialized by a constant expression.
I think that clang++
actually gets it right here, because in section 27.9.2 [c.files] of the 2011 standard, it says:
The contents of header are the same as the Standard C Library header , with the following changes:
— the header includes the header instead of , and
— if and only if the type intmax_t designates an extended integer type (3.9.1), the following function
signatures are added:
intmax_t abs(intmax_t);
imaxdiv_t div(intmax_t, intmax_t);
which shall have the same semantics as the function signatures intmax_t imaxabs(intmax_t) and
imaxdiv_t imaxdiv(intmax_t, intmax_t), respectively.
In the current working draft of the C++ standard, as in the published 2014 version, it says in section 17.6.5.6 [constexpr.functions]:
This standard explicitly requires that certain standard library functions are constexpr (7.1.5). An implementation shall not declare any standard library function signature as constexpr except for those where it
is explicitly required.
So the result, for now, is that these functions are still not constexpr
according to the standard (which you knew) but they could be, as demonstrated by the gcc
compiler.
I can't give a good reason for why abs
couldn't be constexpr
and apparently neither can gcc
. When I use gcc 4.9.2
with this program:
#include <cstdlib>
#include <cinttypes>
#include <cassert>
constexpr intmax_t abs3 = std::abs(3);
constexpr intmax_t absneg3 = std::abs(-3);
int main()
{
assert(abs3 == absneg3);
}
it compiles and runs to completion with no warnings or errors. You can try it here. However, clang++
(version 3.5.0) throws a compile-time error:
abs.cpp:6:20: error: constexpr variable 'abs3' must be initialized by a constant expression.
I think that clang++
actually gets it right here, because in section 27.9.2 [c.files] of the 2011 standard, it says:
The contents of header are the same as the Standard C Library header , with the following changes:
— the header includes the header instead of , and
— if and only if the type intmax_t designates an extended integer type (3.9.1), the following function
signatures are added:
intmax_t abs(intmax_t);
imaxdiv_t div(intmax_t, intmax_t);
which shall have the same semantics as the function signatures intmax_t imaxabs(intmax_t) and
imaxdiv_t imaxdiv(intmax_t, intmax_t), respectively.
In the current working draft of the C++ standard, as in the published 2014 version, it says in section 17.6.5.6 [constexpr.functions]:
This standard explicitly requires that certain standard library functions are constexpr (7.1.5). An implementation shall not declare any standard library function signature as constexpr except for those where it
is explicitly required.
So the result, for now, is that these functions are still not constexpr
according to the standard (which you knew) but they could be, as demonstrated by the gcc
compiler.
answered Apr 5 '15 at 12:47
EdwardEdward
4,97111946
4,97111946
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in definingconstexpr
functions in<cmath>
.
– leek
Nov 22 '18 at 23:04
add a comment |
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in definingconstexpr
functions in<cmath>
.
– leek
Nov 22 '18 at 23:04
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
For more details re gcc and match functions treated as constexpr. gcc allowed this since they thought this was going to be allowed but it was not.
– Shafik Yaghmour
Nov 13 '15 at 20:29
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in defining
constexpr
functions in <cmath>
.– leek
Nov 22 '18 at 23:04
In Proposal P0533, it is stated that GCC has taken the lead as the "State of the Art" in defining
constexpr
functions in <cmath>
.– leek
Nov 22 '18 at 23:04
add a comment |
The simple explanation is probably that nobody went through every function listed as part of the standard library, and updated the entry to make it constexpr
.
Updating a standard is a process involving human effort to propose changes and get those proposals reviewed and accepted. And that means, when a new feature is added - and constexpr
is a new feature - not everything else is automagically updated to use that feature. This particular case may well be updated in a future evolution of the standard - if some people take on the - often thankless - task of proposing the set of changes, some other people take on the task of reviewing it, and convincing those with voting power on the committee to accept it.
add a comment |
The simple explanation is probably that nobody went through every function listed as part of the standard library, and updated the entry to make it constexpr
.
Updating a standard is a process involving human effort to propose changes and get those proposals reviewed and accepted. And that means, when a new feature is added - and constexpr
is a new feature - not everything else is automagically updated to use that feature. This particular case may well be updated in a future evolution of the standard - if some people take on the - often thankless - task of proposing the set of changes, some other people take on the task of reviewing it, and convincing those with voting power on the committee to accept it.
add a comment |
The simple explanation is probably that nobody went through every function listed as part of the standard library, and updated the entry to make it constexpr
.
Updating a standard is a process involving human effort to propose changes and get those proposals reviewed and accepted. And that means, when a new feature is added - and constexpr
is a new feature - not everything else is automagically updated to use that feature. This particular case may well be updated in a future evolution of the standard - if some people take on the - often thankless - task of proposing the set of changes, some other people take on the task of reviewing it, and convincing those with voting power on the committee to accept it.
The simple explanation is probably that nobody went through every function listed as part of the standard library, and updated the entry to make it constexpr
.
Updating a standard is a process involving human effort to propose changes and get those proposals reviewed and accepted. And that means, when a new feature is added - and constexpr
is a new feature - not everything else is automagically updated to use that feature. This particular case may well be updated in a future evolution of the standard - if some people take on the - often thankless - task of proposing the set of changes, some other people take on the task of reviewing it, and convincing those with voting power on the committee to accept it.
answered Apr 5 '15 at 13:23
PeterPeter
27.5k32155
27.5k32155
add a comment |
add a comment |
It has been proposed in P0533:
A function in
<cmath>
shall be declared
constexpr
if and only if:
- When taken to act on the set of rational numbers,
the function is closed (excluding division by zero);
- The function does not modify any of its arguments
which have external visibility;
- The function is not strongly dependent on the
rounding mode.
By means of a brief illustration,
abs
satisfies all three criteria; however, functions such asexp
,sqrt
,cos
,sin
fall foul of the first criterion and so are excluded asconstexpr
candidates. Finally, as discussed above,nearbyint
fails the final criterion.
add a comment |
It has been proposed in P0533:
A function in
<cmath>
shall be declared
constexpr
if and only if:
- When taken to act on the set of rational numbers,
the function is closed (excluding division by zero);
- The function does not modify any of its arguments
which have external visibility;
- The function is not strongly dependent on the
rounding mode.
By means of a brief illustration,
abs
satisfies all three criteria; however, functions such asexp
,sqrt
,cos
,sin
fall foul of the first criterion and so are excluded asconstexpr
candidates. Finally, as discussed above,nearbyint
fails the final criterion.
add a comment |
It has been proposed in P0533:
A function in
<cmath>
shall be declared
constexpr
if and only if:
- When taken to act on the set of rational numbers,
the function is closed (excluding division by zero);
- The function does not modify any of its arguments
which have external visibility;
- The function is not strongly dependent on the
rounding mode.
By means of a brief illustration,
abs
satisfies all three criteria; however, functions such asexp
,sqrt
,cos
,sin
fall foul of the first criterion and so are excluded asconstexpr
candidates. Finally, as discussed above,nearbyint
fails the final criterion.
It has been proposed in P0533:
A function in
<cmath>
shall be declared
constexpr
if and only if:
- When taken to act on the set of rational numbers,
the function is closed (excluding division by zero);
- The function does not modify any of its arguments
which have external visibility;
- The function is not strongly dependent on the
rounding mode.
By means of a brief illustration,
abs
satisfies all three criteria; however, functions such asexp
,sqrt
,cos
,sin
fall foul of the first criterion and so are excluded asconstexpr
candidates. Finally, as discussed above,nearbyint
fails the final criterion.
answered Nov 22 '18 at 23:02
leekleek
499410
499410
add a comment |
add a comment |
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5
@BasileStarynkevitch: Can you explain one of these compatibility reasons?
– Lightness Races in Orbit
Dec 30 '14 at 16:14
5
@BasileStarynkevitch I don't buy that. Given a compatible compiler, a constexpr piece of code will work the same as it used, i.e. when passed in non-constexpr values it will return a non-constexpr. If it preferable to compatible to not make these functions constexpr (or alteast with constexpr overloads) then it would be preferable to not include move instructors for vector. Being that this uses namespaces (so is not C) and is a header provided only in C++11, I would like to see constexpr overlaods.
– IdeaHat
Dec 30 '14 at 16:16
7
abs
is a C99 function. And C99 has no idea ofconstexpr
– Basile Starynkevitch
Dec 30 '14 at 16:32
17
There are a lot of functions that could be constexpr and are not (yet). More functions will slowly become constexpr with each revision of the standard. Feel free to participate in that process (instructions on isocpp.org).
– Marc Glisse
Dec 30 '14 at 16:40
3
abs
is notconstexpr
because the standard decided so. I think this is the wrong place to ask why. What I can say is that if you really need aconstexpr abs
, write your own one. You can't go wrong.– edmz
Dec 30 '14 at 17:02