How to implement std::result_of with only c++98 features?

Question

I wonder how can we implement a result_of template class to get the return type of a function.

I know C++11 has std::result_of or decltype. But how the boost implement this feature in C++98 standard?

I have tried to learn from source code, but I didn't get the point.

It should be like this:

int fint() { return 0;}
double fdouble() { return double(0);}
cout << sizeof(result_of<fint>::type) << endl; // should be 4
cout << sizeof(result_of<fdouble>::type) << endl; // should be 8
result_of<fint>::type x; // same as `int x;`
result_of<fdouble>::type y; // same as `double y;`

its an interesting question, but you already know the answer: boost implements it. Imho it would be better if you say what you dont understand on the boost code, because I doubt that someone will come up with something much simpler than what boost did — 463035818_is_not_an_ai, Jul 13 '17 at 08:43
I can't tell exactly where is wrong, but code is just unable to compile.... — Liu Weibo, Jul 13 '17 at 10:44

score 4 · Answer 1 · answered Jul 13 '17 at 08:43

4

Specialize for Ret (*) (T1, ...,TN) up to some N:

template <typename T> struct result_of;

template <typename Ret> struct result_of<Ret (*)()> {
    typedef Ret type;
};

template <typename Ret, typename T1> struct result_of<Ret (*)(T1)> {
    typedef Ret type;
};

template <typename Ret, typename T1>
struct result_of<Ret (*)(T1, ...)> { /* ellipsis as for printf */
    typedef Ret type;
};

template <typename Ret, typename T1, typename T2> struct result_of<Ret (*)(T1, T2)> {
    typedef Ret type;
};

// ...

answered Jul 13 '17 at 08:43

Jarod42

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result_of.cpp:71:15: error: ‘float f_f()’ cannot appear in a constant-expression result_of::type a = 3; ^ result_of.cpp:71:19: error: a function call cannot appear in a constant-expression result_of::type a = 3; – Liu Weibo Jul 13 '17 at 08:50
I cann't pass a function call to template parameter. It's not defined at compiling time. – Liu Weibo Jul 13 '17 at 08:52
1

@LiuWeibo `f_f()` isnt a function, but `f_f` is – 463035818_is_not_an_ai Jul 13 '17 at 08:53
result_of.cpp: In function ‘int main()’: result_of.cpp:71:18: error: type/value mismatch at argument 1 in template parameter list for ‘template struct result_of’ result_of::type a = 3; – Liu Weibo Jul 13 '17 at 08:58
It seems like that compiler want a type rather than a variable of that type – Liu Weibo Jul 13 '17 at 08:59
vector vec; // is correct. but now I do this like: vector vec; – Liu Weibo Jul 13 '17 at 09:04
@LiuWeibo: You may use this that [way](http://coliru.stacked-crooked.com/a/af1dc0b6c93a2a26). – Jarod42 Jul 13 '17 at 09:05
@Jarod42 how to use these types as type identifier to define some variable? – Liu Weibo Jul 13 '17 at 09:10
`typeof` might help if your compiler supports it. – Jarod42 Jul 13 '17 at 09:18
This will break if the function is overloaded – Passer By Jul 13 '17 at 09:53
1

@PasserBy: If the function is overloaded, getting the result type might be ambiguous anyway ;-) (still possibility to cast to correct type, but then, we don't have to type the name of the function ;-) ) – Jarod42 Jul 13 '17 at 09:56
I'm looking into the possibility of grabbing the right overload if arguments are provided, not too optimistic though :P – Passer By Jul 13 '17 at 09:58
Found [this](https://stackoverflow.com/a/4533801/4832499), not going into that craziness. – Passer By Jul 13 '17 at 10:01
@Jarod42 How can I define variable by typeof? This is mostly confusing part of all – Liu Weibo Jul 13 '17 at 10:42
@LiuWeibo: Something like [that](http://coliru.stacked-crooked.com/a/ba4e116b01be3ac9) – Jarod42 Jul 13 '17 at 12:28

How to implement std::result_of with only c++98 features?

1 Answers1