Can the return type of the function be obtained from within the function?
Clash Royale CLAN TAG#URR8PPP
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Can the return type of a function be obtained in a simple way within the function?
For example, given:
template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type
typename std::remove_reference<decltype(*p)>::type f; // <-- here
...
In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?
c++ c++11 trailing-return-type
asked Sep 5 at 17:33
BeeOnRope
24k873162
add a comment |Â
up vote
38
down vote
favorite
3
Can the return type of a function be obtained in a simple way within the function?
For example, given:
template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type
typename std::remove_reference<decltype(*p)>::type f; // <-- here
...
In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?
c++ c++11 trailing-return-type
asked Sep 5 at 17:33
BeeOnRope
24k873162
1
In C++14, you can remove the trailing return type instead, and usereturn f;to deduce the return type.
– jingyu9575
Sep 6 at 4:42
add a comment |Â
up vote
38
down vote
favorite
3
up vote
38
down vote
favorite
3
3
Can the return type of a function be obtained in a simple way within the function?
For example, given:
template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type
typename std::remove_reference<decltype(*p)>::type f; // <-- here
...
In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?
c++ c++11 trailing-return-type
asked Sep 5 at 17:33
BeeOnRope
24k873162
Can the return type of a function be obtained in a simple way within the function?
For example, given:
template <typename P>
static inline auto foo(P p) -> typename std::remove_reference<decltype(*p)>::type
typename std::remove_reference<decltype(*p)>::type f; // <-- here
...
In C++11 can I refer to the big nasty return type of foo, within foo itself, without repeating it, at the line marked // <-- here?
c++ c++11 trailing-return-type
c++ c++11 trailing-return-type
asked Sep 5 at 17:33
BeeOnRope
24k873162
asked Sep 5 at 17:33
BeeOnRope
24k873162
edited Sep 5 at 18:24
asked Sep 5 at 17:33
BeeOnRope
24k873162
asked Sep 5 at 17:33
BeeOnRope
24k873162
asked Sep 5 at 17:33
BeeOnRope
24k873162
24k873162
1
In C++14, you can remove the trailing return type instead, and usereturn f;to deduce the return type.
– jingyu9575
Sep 6 at 4:42
add a comment |Â
1
In C++14, you can remove the trailing return type instead, and usereturn f;to deduce the return type.
– jingyu9575
Sep 6 at 4:42
1
1
In C++14, you can remove the trailing return type instead, and use
return f; to deduce the return type.– jingyu9575
Sep 6 at 4:42
In C++14, you can remove the trailing return type instead, and use
return f; to deduce the return type.– jingyu9575
Sep 6 at 4:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
50
down vote
accepted
Call the function with a decltype.
decltype(foo(p)) f;
answered Sep 5 at 17:35
Rakete1111
32.7k975110
6
life made easy bydecltype;)
– JeJo
Sep 5 at 18:18
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
50
down vote
accepted
Call the function with a decltype.
decltype(foo(p)) f;
answered Sep 5 at 17:35
Rakete1111
32.7k975110
6
life made easy bydecltype;)
– JeJo
Sep 5 at 18:18
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
add a comment |Â
up vote
50
down vote
accepted
Call the function with a decltype.
decltype(foo(p)) f;
answered Sep 5 at 17:35
Rakete1111
32.7k975110
6
life made easy bydecltype;)
– JeJo
Sep 5 at 18:18
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
add a comment |Â
up vote
50
down vote
accepted
up vote
50
down vote
accepted
Call the function with a decltype.
decltype(foo(p)) f;
answered Sep 5 at 17:35
Rakete1111
32.7k975110
Call the function with a decltype.
decltype(foo(p)) f;
answered Sep 5 at 17:35
Rakete1111
32.7k975110
answered Sep 5 at 17:35
Rakete1111
32.7k975110
answered Sep 5 at 17:35
Rakete1111
32.7k975110
answered Sep 5 at 17:35
Rakete1111
32.7k975110
32.7k975110
6
life made easy bydecltype;)
– JeJo
Sep 5 at 18:18
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
add a comment |Â
6
life made easy bydecltype;)
– JeJo
Sep 5 at 18:18
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
6
6
life made easy by
decltype ;)– JeJo
Sep 5 at 18:18
life made easy by
decltype ;)– JeJo
Sep 5 at 18:18
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
Works well in this case where there are few parameters...
– Max Langhof
Sep 6 at 8:01
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
@MaxLanghof Yes. The more general solution would be to define an alias and use that. :)
– Rakete1111
Sep 6 at 8:02
add a comment |Â
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1
In C++14, you can remove the trailing return type instead, and use
return f;to deduce the return type.– jingyu9575
Sep 6 at 4:42