c++ template argument match std::function with function pointer

Question

I have a function wrapper that takes a function and a list of arguments and calls the function with those arguments:

template<typename R, typename... Args>
R wrapper(std::function<R(Args...)> f, Args... args) {
    auto ret = f(args...);
    return ret;
}

But the following does nott compile:

int f1(int a, int b) { return a+b; }

wrapper(f1, 1, 2);

It says:

mismatched types ‘std::function<_Res(_ArgTypes ...)>’ and ‘int (*)(int, int)’

However, if I do:

wrapper(std::function<int(int, int)>(f1), 5, 10);

...it will be well-formed, which is contradictory with the compiler error.

Why doesn't the compiler accept the first version?

Demo trace: https://onlinegdb.com/rJWzZIQX4

There is no contradiction here. The compiler's telling you that `std::function<_Res(_ArgTypes ...)>` is not the same type as `int (*)(int, int)` (and that it's not allowed to implicitly convert the latter to the former). In your second example, you explicitly convert it so the compiler is happy. — JMAA, Jan 21 '19 at 14:16
FWIW, in C++17 `std::function(f1)` can be shortened to `std::function{f1}` to save you some typing and repeating. — NathanOliver, Jan 21 '19 at 14:17
@StoryTeller Hm, that doesn't work for me: `candidate template ignored: could not match 'function' against 'int (*)(int, int)'` — lubgr, Jan 21 '19 at 14:22
Also highly related and also tells you why lambdas also have this issue: https://stackoverflow.com/questions/9998402/c11-does-not-deduce-type-when-stdfunction-or-lambda-functions-are-involved — NathanOliver, Jan 21 '19 at 14:23
@lubgr - Interesting. I would have thought that with the types deduced/specified this should be enough. Oh well. — StoryTeller - Unslander Monica, Jan 21 '19 at 14:24

score 1 · Answer 1 · answered Jan 21 '19 at 14:16

When deducing template parameters, the compiler does not perform any conversions. Hence, in

wrapper(f1, 1, 2);

you are requesting the implicit construction of std::function<int(int, int)> from a function pointer of type int (*)(int, int). When explicitly passing std::function<int(int, int>(f1), there is no need for converting the argument, and the argument deduction works out.

c++ template argument match std::function with function pointer

1 Answers1