rvalue qualified method and const expression
0
How can I make a const rvalue in a natural way?
Here is a simple example :
struct A {
void f() &&{} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
Why does the first static_assert() call 1 instead of 3?
c++ c++11 c++17 constexpr
edited Nov 27 '18 at 2:45
max66
36.8k74166
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
|
show 1 more comment
0
How can I make a const rvalue in a natural way?
Here is a simple example :
struct A {
void f() &&{} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
Why does the first static_assert() call 1 instead of 3?
c++ c++11 c++17 constexpr
edited Nov 27 '18 at 2:45
max66
36.8k74166
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
What's the question, exactly?
– T.C.
Nov 26 '18 at 22:24
Are you asking how to declare aconst rvalue?
– NathanOliver
Nov 26 '18 at 22:24
1
sorry but... you're checkingstatic_assert()s withconstexprreturningvoid?
– max66
Nov 26 '18 at 22:24
The question is, how can I make rvalue object works on a const expression
– Antoine Morrier
Nov 26 '18 at 22:24
@max66 it is a simple question, maybe the static_assert is not useful here
– Antoine Morrier
Nov 26 '18 at 22:25
|
show 1 more comment
0
0
0
How can I make a const rvalue in a natural way?
Here is a simple example :
struct A {
void f() &&{} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
Why does the first static_assert() call 1 instead of 3?
c++ c++11 c++17 constexpr
edited Nov 27 '18 at 2:45
max66
36.8k74166
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
How can I make a const rvalue in a natural way?
Here is a simple example :
struct A {
void f() &&{} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
Why does the first static_assert() call 1 instead of 3?
c++ c++11 c++17 constexpr
c++ c++11 c++17 constexpr
edited Nov 27 '18 at 2:45
max66
36.8k74166
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
edited Nov 27 '18 at 2:45
max66
36.8k74166
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
edited Nov 27 '18 at 2:45
max66
36.8k74166
edited Nov 27 '18 at 2:45
max66
36.8k74166
edited Nov 27 '18 at 2:45
max66
36.8k74166
36.8k74166
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
asked Nov 26 '18 at 22:22
Antoine MorrierAntoine Morrier
2,065719
2,065719
What's the question, exactly?
– T.C.
Nov 26 '18 at 22:24
Are you asking how to declare aconst rvalue?
– NathanOliver
Nov 26 '18 at 22:24
1
sorry but... you're checkingstatic_assert()s withconstexprreturningvoid?
– max66
Nov 26 '18 at 22:24
The question is, how can I make rvalue object works on a const expression
– Antoine Morrier
Nov 26 '18 at 22:24
@max66 it is a simple question, maybe the static_assert is not useful here
– Antoine Morrier
Nov 26 '18 at 22:25
|
show 1 more comment
What's the question, exactly?
– T.C.
Nov 26 '18 at 22:24
Are you asking how to declare aconst rvalue?
– NathanOliver
Nov 26 '18 at 22:24
1
sorry but... you're checkingstatic_assert()s withconstexprreturningvoid?
– max66
Nov 26 '18 at 22:24
The question is, how can I make rvalue object works on a const expression
– Antoine Morrier
Nov 26 '18 at 22:24
@max66 it is a simple question, maybe the static_assert is not useful here
– Antoine Morrier
Nov 26 '18 at 22:25
What's the question, exactly?
– T.C.
Nov 26 '18 at 22:24
What's the question, exactly?
– T.C.
Nov 26 '18 at 22:24
Are you asking how to declare a
const rvalue?– NathanOliver
Nov 26 '18 at 22:24
Are you asking how to declare a
const rvalue?– NathanOliver
Nov 26 '18 at 22:24
1
1
sorry but... you're checking
static_assert()s with constexpr returning void ?– max66
Nov 26 '18 at 22:24
sorry but... you're checking
static_assert()s with constexpr returning void ?– max66
Nov 26 '18 at 22:24
The question is, how can I make rvalue object works on a const expression
– Antoine Morrier
Nov 26 '18 at 22:24
The question is, how can I make rvalue object works on a const expression
– Antoine Morrier
Nov 26 '18 at 22:24
@max66 it is a simple question, maybe the static_assert is not useful here
– Antoine Morrier
Nov 26 '18 at 22:25
@max66 it is a simple question, maybe the static_assert is not useful here
– Antoine Morrier
Nov 26 '18 at 22:25
|
show 1 more comment
3 Answers
3
active
oldest
votes
4
The issue here is A{} doesn't give you a const A, it just gives you an A so 2 gets called since it is callable on a non const rvalue.
If you want 4 to be called you need to make a const A and you can do that using an alias declaration. If you have using A_const = const A; then A_const{} gives you a const A and A_const{}.f() will call 4 instead of 2.
Essentially what it does is static_assert(const A{}.f());, but since syntactically you can't write it that way we need to using declaration to give us a single word type that is a const A.
Additionally you could rewrite
static_assert(A{}.f());
as
static_assert(std::add_const_t<A>{}.f());
and also get a const A rvalue.
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
@AntoineMorrier No, The C++ syntax does not allow you to writestatic_assert(const A{}.f());
– NathanOliver
Nov 26 '18 at 22:32
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
1
@AntoineMorrier Just remember you can't move something that isconst/constexprso havingconst &&is basically useless.
– NathanOliver
Nov 26 '18 at 22:47
|
show 4 more comments
2
The question is, how can I make rvalue object works on a const expression
Another way can be through a constexpr function returning an A const
#include <iostream>
struct A
{
std::size_t f() & { return 1u; }
std::size_t f() && { return 2u; }
constexpr std::size_t f() const & {return 3u; }
constexpr std::size_t f() const && {return 4u; }
};
constexpr A const foo ()
{ return {}; }
int main()
{
static_assert( foo().f() == 4u, "!" );
}
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
add a comment |
2
Maybe this does what you want?
struct A {
constexpr bool f() &&{return true;} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
@AntoineMorrier - well... (1) isconstexpr. I suppose doesn't works in C++11 (where aconstexprmethod is automatically alsoconst) but should works in C++14 and C++17.
– max66
Nov 26 '18 at 22:50
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
4
The issue here is A{} doesn't give you a const A, it just gives you an A so 2 gets called since it is callable on a non const rvalue.
If you want 4 to be called you need to make a const A and you can do that using an alias declaration. If you have using A_const = const A; then A_const{} gives you a const A and A_const{}.f() will call 4 instead of 2.
Essentially what it does is static_assert(const A{}.f());, but since syntactically you can't write it that way we need to using declaration to give us a single word type that is a const A.
Additionally you could rewrite
static_assert(A{}.f());
as
static_assert(std::add_const_t<A>{}.f());
and also get a const A rvalue.
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
@AntoineMorrier No, The C++ syntax does not allow you to writestatic_assert(const A{}.f());
– NathanOliver
Nov 26 '18 at 22:32
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
1
@AntoineMorrier Just remember you can't move something that isconst/constexprso havingconst &&is basically useless.
– NathanOliver
Nov 26 '18 at 22:47
|
show 4 more comments
4
The issue here is A{} doesn't give you a const A, it just gives you an A so 2 gets called since it is callable on a non const rvalue.
If you want 4 to be called you need to make a const A and you can do that using an alias declaration. If you have using A_const = const A; then A_const{} gives you a const A and A_const{}.f() will call 4 instead of 2.
Essentially what it does is static_assert(const A{}.f());, but since syntactically you can't write it that way we need to using declaration to give us a single word type that is a const A.
Additionally you could rewrite
static_assert(A{}.f());
as
static_assert(std::add_const_t<A>{}.f());
and also get a const A rvalue.
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
@AntoineMorrier No, The C++ syntax does not allow you to writestatic_assert(const A{}.f());
– NathanOliver
Nov 26 '18 at 22:32
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
1
@AntoineMorrier Just remember you can't move something that isconst/constexprso havingconst &&is basically useless.
– NathanOliver
Nov 26 '18 at 22:47
|
show 4 more comments
4
4
4
The issue here is A{} doesn't give you a const A, it just gives you an A so 2 gets called since it is callable on a non const rvalue.
If you want 4 to be called you need to make a const A and you can do that using an alias declaration. If you have using A_const = const A; then A_const{} gives you a const A and A_const{}.f() will call 4 instead of 2.
Essentially what it does is static_assert(const A{}.f());, but since syntactically you can't write it that way we need to using declaration to give us a single word type that is a const A.
Additionally you could rewrite
static_assert(A{}.f());
as
static_assert(std::add_const_t<A>{}.f());
and also get a const A rvalue.
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
The issue here is A{} doesn't give you a const A, it just gives you an A so 2 gets called since it is callable on a non const rvalue.
If you want 4 to be called you need to make a const A and you can do that using an alias declaration. If you have using A_const = const A; then A_const{} gives you a const A and A_const{}.f() will call 4 instead of 2.
Essentially what it does is static_assert(const A{}.f());, but since syntactically you can't write it that way we need to using declaration to give us a single word type that is a const A.
Additionally you could rewrite
static_assert(A{}.f());
as
static_assert(std::add_const_t<A>{}.f());
and also get a const A rvalue.
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
edited Nov 26 '18 at 22:37
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
answered Nov 26 '18 at 22:30
NathanOliverNathanOliver
92.6k16129195
92.6k16129195
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
@AntoineMorrier No, The C++ syntax does not allow you to writestatic_assert(const A{}.f());
– NathanOliver
Nov 26 '18 at 22:32
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
1
@AntoineMorrier Just remember you can't move something that isconst/constexprso havingconst &&is basically useless.
– NathanOliver
Nov 26 '18 at 22:47
|
show 4 more comments
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
@AntoineMorrier No, The C++ syntax does not allow you to writestatic_assert(const A{}.f());
– NathanOliver
Nov 26 '18 at 22:32
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
1
@AntoineMorrier Just remember you can't move something that isconst/constexprso havingconst &&is basically useless.
– NathanOliver
Nov 26 '18 at 22:47
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
So there is no way to write it in a natural way?
– Antoine Morrier
Nov 26 '18 at 22:31
@AntoineMorrier No, The C++ syntax does not allow you to write
static_assert(const A{}.f());– NathanOliver
Nov 26 '18 at 22:32
@AntoineMorrier No, The C++ syntax does not allow you to write
static_assert(const A{}.f());– NathanOliver
Nov 26 '18 at 22:32
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
So note: This is not a language limitation, it's a parsing limitation. You can solve it using an alias.
– N00byEdge
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
I think I should try to not use ref qualified methods, making an alias is not a good solution for me :). Thanks ;)
– Antoine Morrier
Nov 26 '18 at 22:33
1
1
@AntoineMorrier Just remember you can't move something that is
const/constexpr so having const && is basically useless.– NathanOliver
Nov 26 '18 at 22:47
@AntoineMorrier Just remember you can't move something that is
const/constexpr so having const && is basically useless.– NathanOliver
Nov 26 '18 at 22:47
|
show 4 more comments
2
The question is, how can I make rvalue object works on a const expression
Another way can be through a constexpr function returning an A const
#include <iostream>
struct A
{
std::size_t f() & { return 1u; }
std::size_t f() && { return 2u; }
constexpr std::size_t f() const & {return 3u; }
constexpr std::size_t f() const && {return 4u; }
};
constexpr A const foo ()
{ return {}; }
int main()
{
static_assert( foo().f() == 4u, "!" );
}
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
add a comment |
2
The question is, how can I make rvalue object works on a const expression
Another way can be through a constexpr function returning an A const
#include <iostream>
struct A
{
std::size_t f() & { return 1u; }
std::size_t f() && { return 2u; }
constexpr std::size_t f() const & {return 3u; }
constexpr std::size_t f() const && {return 4u; }
};
constexpr A const foo ()
{ return {}; }
int main()
{
static_assert( foo().f() == 4u, "!" );
}
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
add a comment |
2
2
2
The question is, how can I make rvalue object works on a const expression
Another way can be through a constexpr function returning an A const
#include <iostream>
struct A
{
std::size_t f() & { return 1u; }
std::size_t f() && { return 2u; }
constexpr std::size_t f() const & {return 3u; }
constexpr std::size_t f() const && {return 4u; }
};
constexpr A const foo ()
{ return {}; }
int main()
{
static_assert( foo().f() == 4u, "!" );
}
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
The question is, how can I make rvalue object works on a const expression
Another way can be through a constexpr function returning an A const
#include <iostream>
struct A
{
std::size_t f() & { return 1u; }
std::size_t f() && { return 2u; }
constexpr std::size_t f() const & {return 3u; }
constexpr std::size_t f() const && {return 4u; }
};
constexpr A const foo ()
{ return {}; }
int main()
{
static_assert( foo().f() == 4u, "!" );
}
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
answered Nov 26 '18 at 22:35
max66max66
36.8k74166
36.8k74166
add a comment |
add a comment |
2
Maybe this does what you want?
struct A {
constexpr bool f() &&{return true;} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
@AntoineMorrier - well... (1) isconstexpr. I suppose doesn't works in C++11 (where aconstexprmethod is automatically alsoconst) but should works in C++14 and C++17.
– max66
Nov 26 '18 at 22:50
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
add a comment |
2
Maybe this does what you want?
struct A {
constexpr bool f() &&{return true;} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
@AntoineMorrier - well... (1) isconstexpr. I suppose doesn't works in C++11 (where aconstexprmethod is automatically alsoconst) but should works in C++14 and C++17.
– max66
Nov 26 '18 at 22:50
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
add a comment |
2
2
2
Maybe this does what you want?
struct A {
constexpr bool f() &&{return true;} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
Maybe this does what you want?
struct A {
constexpr bool f() &&{return true;} // 1
constexpr bool f() const &{return true;} // 2
constexpr bool f() const &&{return true;} // 3 really usefull?
};
int main() {
static_assert(A{}.f()); // does not compile because call 1 instead of 3
constexpr A a;
static_assert(a.f());
}
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
answered Nov 26 '18 at 22:38
RichardRichard
8,84721423
8,84721423
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
@AntoineMorrier - well... (1) isconstexpr. I suppose doesn't works in C++11 (where aconstexprmethod is automatically alsoconst) but should works in C++14 and C++17.
– max66
Nov 26 '18 at 22:50
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
add a comment |
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
@AntoineMorrier - well... (1) isconstexpr. I suppose doesn't works in C++11 (where aconstexprmethod is automatically alsoconst) but should works in C++14 and C++17.
– max66
Nov 26 '18 at 22:50
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
No, because in your case, A{}.f still call 1, that is not constexpr. However, it could be good for my use, because making constexpr f() && is not that absurd since I am dealing with template. So thanks
– Antoine Morrier
Nov 26 '18 at 22:41
@AntoineMorrier - well... (1) is
constexpr. I suppose doesn't works in C++11 (where a constexpr method is automatically also const) but should works in C++14 and C++17.– max66
Nov 26 '18 at 22:50
@AntoineMorrier - well... (1) is
constexpr. I suppose doesn't works in C++11 (where a constexpr method is automatically also const) but should works in C++14 and C++17.– max66
Nov 26 '18 at 22:50
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
Thanks for the information :). It works perfectly well. I remember have seen an article where it was explained that constexpr does not mean const method in C++14 yes. Thanks for the remaining :)
– Antoine Morrier
Nov 26 '18 at 22:59
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What's the question, exactly?
– T.C.
Nov 26 '18 at 22:24
Are you asking how to declare a
const rvalue?– NathanOliver
Nov 26 '18 at 22:24
1
sorry but... you're checking
static_assert()s withconstexprreturningvoid?– max66
Nov 26 '18 at 22:24
The question is, how can I make rvalue object works on a const expression
– Antoine Morrier
Nov 26 '18 at 22:24
@max66 it is a simple question, maybe the static_assert is not useful here
– Antoine Morrier
Nov 26 '18 at 22:25