Mixing
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C++11 constexpr function pass parameter
Consider the following code:
static constexpr int make_const(const int i){
return i;
}
void t1(const int i)
{
constexpr int ii = make_const(i); // error occurs here (i is not a constant expression)
std::cout<<ii;
}
int main()
{
t1(12);
}
Why I have an error on make_const call?
UPDATE
But this one works:
constexpr int t1(const int i)
{
return make_const(i);
}
However, this not:
template<int i>
constexpr bool do_something(){
return i;
}
constexpr int t1(const int i)
{
return do_something<make_const(i)>(); // error occurs here (i is not a constant expression)
}
c++ c++11 constexpr
asked May 9 '14 at 16:09
tower120tower120
1,84712251
|
show 5 more comments
Consider the following code:
static constexpr int make_const(const int i){
return i;
}
void t1(const int i)
{
constexpr int ii = make_const(i); // error occurs here (i is not a constant expression)
std::cout<<ii;
}
int main()
{
t1(12);
}
Why I have an error on make_const call?
UPDATE
But this one works:
constexpr int t1(const int i)
{
return make_const(i);
}
However, this not:
template<int i>
constexpr bool do_something(){
return i;
}
constexpr int t1(const int i)
{
return do_something<make_const(i)>(); // error occurs here (i is not a constant expression)
}
c++ c++11 constexpr
asked May 9 '14 at 16:09
tower120tower120
1,84712251
1
Well because in the general case,iisn'tconstexprinvoid t1(const int).
– stefan
May 9 '14 at 16:15
1
how can I make it constexpr, then?
– tower120
May 9 '14 at 16:21
making it a template argument is your only option
– stefan
May 9 '14 at 16:22
Your update does compile (though it does something completely different), what do you mean it doesn't work? (theconstis useless)
– Marc Glisse
May 9 '14 at 16:34
3
There is no direct way to do what you want to do. This feature/limitation is probably the most frequently asked question about constexpr.
– Marc Glisse
May 9 '14 at 21:06
|
show 5 more comments
Consider the following code:
static constexpr int make_const(const int i){
return i;
}
void t1(const int i)
{
constexpr int ii = make_const(i); // error occurs here (i is not a constant expression)
std::cout<<ii;
}
int main()
{
t1(12);
}
Why I have an error on make_const call?
UPDATE
But this one works:
constexpr int t1(const int i)
{
return make_const(i);
}
However, this not:
template<int i>
constexpr bool do_something(){
return i;
}
constexpr int t1(const int i)
{
return do_something<make_const(i)>(); // error occurs here (i is not a constant expression)
}
c++ c++11 constexpr
asked May 9 '14 at 16:09
tower120tower120
1,84712251
Consider the following code:
static constexpr int make_const(const int i){
return i;
}
void t1(const int i)
{
constexpr int ii = make_const(i); // error occurs here (i is not a constant expression)
std::cout<<ii;
}
int main()
{
t1(12);
}
Why I have an error on make_const call?
UPDATE
But this one works:
constexpr int t1(const int i)
{
return make_const(i);
}
However, this not:
template<int i>
constexpr bool do_something(){
return i;
}
constexpr int t1(const int i)
{
return do_something<make_const(i)>(); // error occurs here (i is not a constant expression)
}
c++ c++11 constexpr
c++ c++11 constexpr
asked May 9 '14 at 16:09
tower120tower120
1,84712251
asked May 9 '14 at 16:09
tower120tower120
1,84712251
edited May 9 '14 at 17:09
tower120
asked May 9 '14 at 16:09
tower120tower120
1,84712251
asked May 9 '14 at 16:09
tower120tower120
1,84712251
asked May 9 '14 at 16:09
tower120tower120
1,84712251
1,84712251
1
Well because in the general case,iisn'tconstexprinvoid t1(const int).
– stefan
May 9 '14 at 16:15
1
how can I make it constexpr, then?
– tower120
May 9 '14 at 16:21
making it a template argument is your only option
– stefan
May 9 '14 at 16:22
Your update does compile (though it does something completely different), what do you mean it doesn't work? (theconstis useless)
– Marc Glisse
May 9 '14 at 16:34
3
There is no direct way to do what you want to do. This feature/limitation is probably the most frequently asked question about constexpr.
– Marc Glisse
May 9 '14 at 21:06
|
show 5 more comments
1
Well because in the general case,iisn'tconstexprinvoid t1(const int).
– stefan
May 9 '14 at 16:15
1
how can I make it constexpr, then?
– tower120
May 9 '14 at 16:21
making it a template argument is your only option
– stefan
May 9 '14 at 16:22
Your update does compile (though it does something completely different), what do you mean it doesn't work? (theconstis useless)
– Marc Glisse
May 9 '14 at 16:34
3
There is no direct way to do what you want to do. This feature/limitation is probably the most frequently asked question about constexpr.
– Marc Glisse
May 9 '14 at 21:06
1
1
Well because in the general case,
i isn't constexpr in void t1(const int).– stefan
May 9 '14 at 16:15
Well because in the general case,
i isn't constexpr in void t1(const int).– stefan
May 9 '14 at 16:15
1
1
how can I make it constexpr, then?
– tower120
May 9 '14 at 16:21
how can I make it constexpr, then?
– tower120
May 9 '14 at 16:21
making it a template argument is your only option
– stefan
May 9 '14 at 16:22
making it a template argument is your only option
– stefan
May 9 '14 at 16:22
Your update does compile (though it does something completely different), what do you mean it doesn't work? (the
const is useless)– Marc Glisse
May 9 '14 at 16:34
Your update does compile (though it does something completely different), what do you mean it doesn't work? (the
const is useless)– Marc Glisse
May 9 '14 at 16:34
3
3
There is no direct way to do what you want to do. This feature/limitation is probably the most frequently asked question about constexpr.
– Marc Glisse
May 9 '14 at 21:06
There is no direct way to do what you want to do. This feature/limitation is probably the most frequently asked question about constexpr.
– Marc Glisse
May 9 '14 at 21:06
|
show 5 more comments
3 Answers
3
active
oldest
votes
A constexpr function and a constexpr variable are related, but different things.
A constexpr variable is a variable whose value is guaranteed to be available at compile time.
A constexpr function is a function that, if evaluated with constexpr arguments, and behaves "properly" during its execution, will be evaluated at compile time.
If you pass a non-constexpr int to a constexpr function, it will not magically make it evaluated at compile time. It will, however, be allowed to pass the constexprness of its input parameters through itself (normal functions cannot do this).
constexpr on functions is a mixture of documentation and restriction on how they are written and instructions to the compiler.
The reason behind this is to allow the same function to be evaluated both at compile time, and at run time. If passed runtime arguments, it is a runtime function. If passed constexpr arguments, it may be evaluated at compile time (and will be if used in certain contexts).
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
add a comment |
One important difference between const and constexpr is that a constexpr can be evaluated at compile time.
By writing constexpr int ii = make_const(i); you are telling the compiler that the expression is to be evaluted at compile time. Since i is evaluted at run-time, the compiler is unable to do this and gives you an error.
answered May 9 '14 at 16:23
TASTAS
1,8961015
add a comment |
Because t1() is not a constexpr function, the parameter i is a runtime variable... which you can't pass to a constexpr function. Constexpr expects the parameter to be known at compile time.
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
2
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
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
A constexpr function and a constexpr variable are related, but different things.
A constexpr variable is a variable whose value is guaranteed to be available at compile time.
A constexpr function is a function that, if evaluated with constexpr arguments, and behaves "properly" during its execution, will be evaluated at compile time.
If you pass a non-constexpr int to a constexpr function, it will not magically make it evaluated at compile time. It will, however, be allowed to pass the constexprness of its input parameters through itself (normal functions cannot do this).
constexpr on functions is a mixture of documentation and restriction on how they are written and instructions to the compiler.
The reason behind this is to allow the same function to be evaluated both at compile time, and at run time. If passed runtime arguments, it is a runtime function. If passed constexpr arguments, it may be evaluated at compile time (and will be if used in certain contexts).
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
add a comment |
A constexpr function and a constexpr variable are related, but different things.
A constexpr variable is a variable whose value is guaranteed to be available at compile time.
A constexpr function is a function that, if evaluated with constexpr arguments, and behaves "properly" during its execution, will be evaluated at compile time.
If you pass a non-constexpr int to a constexpr function, it will not magically make it evaluated at compile time. It will, however, be allowed to pass the constexprness of its input parameters through itself (normal functions cannot do this).
constexpr on functions is a mixture of documentation and restriction on how they are written and instructions to the compiler.
The reason behind this is to allow the same function to be evaluated both at compile time, and at run time. If passed runtime arguments, it is a runtime function. If passed constexpr arguments, it may be evaluated at compile time (and will be if used in certain contexts).
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
add a comment |
A constexpr function and a constexpr variable are related, but different things.
A constexpr variable is a variable whose value is guaranteed to be available at compile time.
A constexpr function is a function that, if evaluated with constexpr arguments, and behaves "properly" during its execution, will be evaluated at compile time.
If you pass a non-constexpr int to a constexpr function, it will not magically make it evaluated at compile time. It will, however, be allowed to pass the constexprness of its input parameters through itself (normal functions cannot do this).
constexpr on functions is a mixture of documentation and restriction on how they are written and instructions to the compiler.
The reason behind this is to allow the same function to be evaluated both at compile time, and at run time. If passed runtime arguments, it is a runtime function. If passed constexpr arguments, it may be evaluated at compile time (and will be if used in certain contexts).
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
A constexpr function and a constexpr variable are related, but different things.
A constexpr variable is a variable whose value is guaranteed to be available at compile time.
A constexpr function is a function that, if evaluated with constexpr arguments, and behaves "properly" during its execution, will be evaluated at compile time.
If you pass a non-constexpr int to a constexpr function, it will not magically make it evaluated at compile time. It will, however, be allowed to pass the constexprness of its input parameters through itself (normal functions cannot do this).
constexpr on functions is a mixture of documentation and restriction on how they are written and instructions to the compiler.
The reason behind this is to allow the same function to be evaluated both at compile time, and at run time. If passed runtime arguments, it is a runtime function. If passed constexpr arguments, it may be evaluated at compile time (and will be if used in certain contexts).
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
answered May 9 '14 at 17:52
Yakk - Adam NevraumontYakk - Adam Nevraumont
188k21199384
188k21199384
add a comment |
add a comment |
One important difference between const and constexpr is that a constexpr can be evaluated at compile time.
By writing constexpr int ii = make_const(i); you are telling the compiler that the expression is to be evaluted at compile time. Since i is evaluted at run-time, the compiler is unable to do this and gives you an error.
answered May 9 '14 at 16:23
TASTAS
1,8961015
add a comment |
One important difference between const and constexpr is that a constexpr can be evaluated at compile time.
By writing constexpr int ii = make_const(i); you are telling the compiler that the expression is to be evaluted at compile time. Since i is evaluted at run-time, the compiler is unable to do this and gives you an error.
answered May 9 '14 at 16:23
TASTAS
1,8961015
add a comment |
One important difference between const and constexpr is that a constexpr can be evaluated at compile time.
By writing constexpr int ii = make_const(i); you are telling the compiler that the expression is to be evaluted at compile time. Since i is evaluted at run-time, the compiler is unable to do this and gives you an error.
answered May 9 '14 at 16:23
TASTAS
1,8961015
One important difference between const and constexpr is that a constexpr can be evaluated at compile time.
By writing constexpr int ii = make_const(i); you are telling the compiler that the expression is to be evaluted at compile time. Since i is evaluted at run-time, the compiler is unable to do this and gives you an error.
answered May 9 '14 at 16:23
TASTAS
1,8961015
answered May 9 '14 at 16:23
TASTAS
1,8961015
answered May 9 '14 at 16:23
TASTAS
1,8961015
answered May 9 '14 at 16:23
TASTAS
1,8961015
1,8961015
add a comment |
add a comment |
Because t1() is not a constexpr function, the parameter i is a runtime variable... which you can't pass to a constexpr function. Constexpr expects the parameter to be known at compile time.
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
2
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
add a comment |
Because t1() is not a constexpr function, the parameter i is a runtime variable... which you can't pass to a constexpr function. Constexpr expects the parameter to be known at compile time.
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
2
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
add a comment |
Because t1() is not a constexpr function, the parameter i is a runtime variable... which you can't pass to a constexpr function. Constexpr expects the parameter to be known at compile time.
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
Because t1() is not a constexpr function, the parameter i is a runtime variable... which you can't pass to a constexpr function. Constexpr expects the parameter to be known at compile time.
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
answered May 9 '14 at 16:16
paxos1977paxos1977
62.1k2278116
62.1k2278116
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
2
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
add a comment |
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
2
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Making t1 constexpr wouldn't change anything.
– Marc Glisse
May 9 '14 at 16:19
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
Yes, I test your suggestion. Same old story.
– tower120
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@MarcGlisse t1() is not correctly formed to be a constexpr function, I wasn't suggesting making it so would correct the problem.
– paxos1977
May 9 '14 at 16:20
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
@praxos1977 - see updated question. Your suggestion is wrong.
– tower120
May 9 '14 at 16:23
2
2
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
@tower120, I wasn't making a suggestion. I was pointing out that you can't pass a runtime variable to a constexpr function. TAS obviously worded his answer better than I did.
– paxos1977
May 9 '14 at 16:24
add a comment |
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1
Well because in the general case,
iisn'tconstexprinvoid t1(const int).– stefan
May 9 '14 at 16:15
1
how can I make it constexpr, then?
– tower120
May 9 '14 at 16:21
making it a template argument is your only option
– stefan
May 9 '14 at 16:22
Your update does compile (though it does something completely different), what do you mean it doesn't work? (the
constis useless)– Marc Glisse
May 9 '14 at 16:34
3
There is no direct way to do what you want to do. This feature/limitation is probably the most frequently asked question about constexpr.
– Marc Glisse
May 9 '14 at 21:06