Simplest way to determine return type of function

Given a very simple, but lengthy function, such as:

int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???

What is the most simple and concise way to determine the function's return type (ReturnTypeOfFoo, in this example: int) at compile time without repeating the function's parameter types (by name only, since it is known that the function does not have any additional overloads)?

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

asked Dec 7 '18 at 16:29

Cybran

1,000717

Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.

– Jay
Dec 7 '18 at 16:33

1

Maybe decltype(foo)::result_type?

– Thomas Lang
Dec 7 '18 at 16:33

maybe en.cppreference.com/w/cpp/types/result_of?

– bracco23
Dec 7 '18 at 16:35

1

@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?

– user463035818
Dec 7 '18 at 16:39

@user463035818 You may be right here, I was referring to the result_type member of a std::function type.

– Thomas Lang
Dec 7 '18 at 16:40

|
show 3 more comments

Given a very simple, but lengthy function, such as:

int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

asked Dec 7 '18 at 16:29

Cybran

1,000717

Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.

– Jay
Dec 7 '18 at 16:33

1

Maybe decltype(foo)::result_type?

– Thomas Lang
Dec 7 '18 at 16:33

maybe en.cppreference.com/w/cpp/types/result_of?

– bracco23
Dec 7 '18 at 16:35

1

@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?

– user463035818
Dec 7 '18 at 16:39

@user463035818 You may be right here, I was referring to the result_type member of a std::function type.

– Thomas Lang
Dec 7 '18 at 16:40

|
show 3 more comments

Given a very simple, but lengthy function, such as:

int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

asked Dec 7 '18 at 16:29

Cybran

1,000717

Given a very simple, but lengthy function, such as:

int foo(int a, int b, int c, int d) {

    return 1;

}



// using ReturnTypeOfFoo = ???

c++ function c++17 return-type compile-time

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

asked Dec 7 '18 at 16:29

Cybran

1,000717

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

asked Dec 7 '18 at 16:29

Cybran

1,000717

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

edited Dec 7 '18 at 17:44

Konrad Rudolph

395k1017821025

asked Dec 7 '18 at 16:29

Cybran

1,000717

asked Dec 7 '18 at 16:29

Cybran

1,000717

asked Dec 7 '18 at 16:29

Cybran

1,000717

Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.

– Jay
Dec 7 '18 at 16:33

1

Maybe decltype(foo)::result_type?

– Thomas Lang
Dec 7 '18 at 16:33

maybe en.cppreference.com/w/cpp/types/result_of?

– bracco23
Dec 7 '18 at 16:35

1

@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?

– user463035818
Dec 7 '18 at 16:39

@user463035818 You may be right here, I was referring to the result_type member of a std::function type.

– Thomas Lang
Dec 7 '18 at 16:40

|
show 3 more comments

Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.

– Jay
Dec 7 '18 at 16:33

1

Maybe decltype(foo)::result_type?

– Thomas Lang
Dec 7 '18 at 16:33

maybe en.cppreference.com/w/cpp/types/result_of?

– bracco23
Dec 7 '18 at 16:35

1

@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?

– user463035818
Dec 7 '18 at 16:39

@user463035818 You may be right here, I was referring to the result_type member of a std::function type.

– Thomas Lang
Dec 7 '18 at 16:40

Depending on the compiler or the support libraries available to you... docs.microsoft.com/en-us/cpp/cpp/attributes?view=vs-2017, Fairly sure that Concepts also has some APIs which allow you to deduce this information at run time or compile time.

– Jay
Dec 7 '18 at 16:33

Maybe decltype(foo)::result_type?

– Thomas Lang
Dec 7 '18 at 16:33

maybe en.cppreference.com/w/cpp/types/result_of?

– bracco23
Dec 7 '18 at 16:35

@ThomasLang where does this come from? decltype(foo) is a function that has no result_type member, or do I miss something?

– user463035818
Dec 7 '18 at 16:39

@user463035818 You may be right here, I was referring to the result_type member of a std::function type.

– Thomas Lang
Dec 7 '18 at 16:40

|
show 3 more comments

3 Answers
3

active

oldest

votes

You can leverage std::function here which will give you a typedef for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:

using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;

We can make this a little more generic like

template<typename Callable>

using return_type_of_t = 

    typename decltype(std::function{std::declval<Callable>()})::result_type;

which then lets you use it like

int foo(int a, int b, int c, int d) {

    return 1;

}



auto bar = (){ return 1; };



struct baz_ 

{ 

    double operator()(){ return 0; } 

} baz;



using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;

using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;

using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.

– Cybran
Dec 7 '18 at 19:08

add a comment |

Most simple and concise is probably:

template <typename R, typename... Args>

R return_type_of(R(*)(Args...));



using ReturnTypeOfFoo = decltype(return_type_of(foo));

Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.

But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.

answered Dec 7 '18 at 16:34

Barry

179k18310569

add a comment |

I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:

template <typename R, typename ... Args>

R getRetType (R(*)(Args...));

and using decltype()?

using ReturnTypeOfFoo = decltype( getRetType(&foo) );

Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

answered Dec 7 '18 at 16:34

max66

35.2k73763

add a comment |

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;

We can make this a little more generic like

template<typename Callable>

using return_type_of_t = 

    typename decltype(std::function{std::declval<Callable>()})::result_type;

which then lets you use it like

int foo(int a, int b, int c, int d) {

    return 1;

}



auto bar = (){ return 1; };



struct baz_ 

{ 

    double operator()(){ return 0; } 

} baz;



using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;

using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;

using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.

– Cybran
Dec 7 '18 at 19:08

add a comment |

using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;

We can make this a little more generic like

template<typename Callable>

using return_type_of_t = 

    typename decltype(std::function{std::declval<Callable>()})::result_type;

which then lets you use it like

int foo(int a, int b, int c, int d) {

    return 1;

}



auto bar = (){ return 1; };



struct baz_ 

{ 

    double operator()(){ return 0; } 

} baz;



using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;

using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;

using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.

– Cybran
Dec 7 '18 at 19:08

add a comment |

using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;

We can make this a little more generic like

template<typename Callable>

using return_type_of_t = 

    typename decltype(std::function{std::declval<Callable>()})::result_type;

which then lets you use it like

int foo(int a, int b, int c, int d) {

    return 1;

}



auto bar = (){ return 1; };



struct baz_ 

{ 

    double operator()(){ return 0; } 

} baz;



using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;

using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;

using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;

We can make this a little more generic like

template<typename Callable>

using return_type_of_t = 

    typename decltype(std::function{std::declval<Callable>()})::result_type;

which then lets you use it like

int foo(int a, int b, int c, int d) {

    return 1;

}



auto bar = (){ return 1; };



struct baz_ 

{ 

    double operator()(){ return 0; } 

} baz;



using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;

using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;

using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

edited Dec 8 '18 at 8:50

coldspeed

124k22125208

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

answered Dec 7 '18 at 16:44

NathanOliver

89k15120187

Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.

– Cybran
Dec 7 '18 at 19:08

add a comment |

Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.

– Cybran
Dec 7 '18 at 19:08

Very compact! I can confirm the one-liner also works for template methods, as long as all template arguments specified.

– Cybran
Dec 7 '18 at 19:08

add a comment |

Most simple and concise is probably:

template <typename R, typename... Args>

R return_type_of(R(*)(Args...));



using ReturnTypeOfFoo = decltype(return_type_of(foo));

Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.

But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.

answered Dec 7 '18 at 16:34

Barry

179k18310569

add a comment |

Most simple and concise is probably:

template <typename R, typename... Args>

R return_type_of(R(*)(Args...));



using ReturnTypeOfFoo = decltype(return_type_of(foo));

Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.

But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.

answered Dec 7 '18 at 16:34

Barry

179k18310569

add a comment |

Most simple and concise is probably:

template <typename R, typename... Args>

R return_type_of(R(*)(Args...));



using ReturnTypeOfFoo = decltype(return_type_of(foo));

Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.

But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.

answered Dec 7 '18 at 16:34

Barry

179k18310569

Most simple and concise is probably:

template <typename R, typename... Args>

R return_type_of(R(*)(Args...));



using ReturnTypeOfFoo = decltype(return_type_of(foo));

Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.

But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.

answered Dec 7 '18 at 16:34

Barry

179k18310569

answered Dec 7 '18 at 16:34

Barry

179k18310569

answered Dec 7 '18 at 16:34

Barry

179k18310569

answered Dec 7 '18 at 16:34

Barry

179k18310569

add a comment |

I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:

template <typename R, typename ... Args>

R getRetType (R(*)(Args...));

and using decltype()?

using ReturnTypeOfFoo = decltype( getRetType(&foo) );

Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

answered Dec 7 '18 at 16:34

max66

35.2k73763

add a comment |

I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:

template <typename R, typename ... Args>

R getRetType (R(*)(Args...));

and using decltype()?

using ReturnTypeOfFoo = decltype( getRetType(&foo) );

Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

answered Dec 7 '18 at 16:34

max66

35.2k73763

add a comment |

I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:

template <typename R, typename ... Args>

R getRetType (R(*)(Args...));

and using decltype()?

using ReturnTypeOfFoo = decltype( getRetType(&foo) );

Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

answered Dec 7 '18 at 16:34

max66

35.2k73763

I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:

template <typename R, typename ... Args>

R getRetType (R(*)(Args...));

and using decltype()?

using ReturnTypeOfFoo = decltype( getRetType(&foo) );

Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

answered Dec 7 '18 at 16:34

max66

35.2k73763

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

edited Dec 7 '18 at 17:26

NathanOliver

89k15120187

answered Dec 7 '18 at 16:34

max66

35.2k73763

answered Dec 7 '18 at 16:34

max66

35.2k73763

answered Dec 7 '18 at 16:34

max66

35.2k73763

add a comment |

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