How i can utilize S(*)(int)?

user_10737114 0
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For educational reasons, I'm studying the C++ language using clang-12 std=c++17

And I have the following code:

Fullcode

#include <cstdio>
#include <iostream>
#include <type_traits>

using namespace std;

struct S {
  void operator()(int) {}
};

int main()
{
  S(*d)(int); 

  //d = whatValue??

  return 0;
}

I'm studying the types of variables that exist in the C++ language, and I came across this situation above.

What value can I assign to the d variable?

Could it be a language bug? I've already researched several topics on stackoverflow and cppreference and haven't found any way to initialize this variable S(*d)(int) .

c++
2 Answers

The variable d is a pointer to a function taking an int as argument and returning an S. Note that it cannot point to non-static member functions. You could, e.g., use it like this:

struct S {
    void operator()(int) {}
};

S f(int) { return S(); }

int main()
{
    S(*d)(int) = &f;
    S rc = d(17);
}

As functions decay to pointer types, you can actually leave the & out. The same isn't true for non-static member functions, though. If you wanted to get a pointer to a member function of S taking an int as argument and returning void (your operator() returns void not S) you'd need to use the &:

int main() {
    void (S::*d)(int) = &S::operator();
    S obj;
    S rc = (obj.*d)(17); // use d
}

Pointer to member functions are different to pointer to functions because they have an implicit first argument: the this pointer. With C++ it is quite rare that pointer to functions are used for other reasons than interfacing C code: using std::function<void(int)> is more flexible, although also type-erase and most likely involving a virtual function call and not being inlinable. If the function can be a function template it is common to pass function objects which are std::invoke()ed (instead of directly called) which support function pointers, pointer to member functions, pointer to non-function members, lambdas, functions, etc.

user_11202730

What value can I assign to the d variable?

d is a pointer to function, and as such you can assign a function that returns S and accepts int as a parameter. Example:

S some_function(int);
d = some_function;

Could it be a language bug?

No.

user_20793030
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