为什么C ++不允许进行static_cast <function_type_of_f> （F）？

Question

#include <functional>

int f(int x)
{
    return 0;
}

int main()
{
    std::function<int(int)> fn1 = f; // ok
    std::function<int(int)> fn2 = static_cast<int(*)(int)>(f); // ok

    //
    // error C2066: cast to function type is illegal
    //
    std::function<int(int)> fn3 = static_cast<int(int)>(f); 
}

My C++ compiler is VS 2015 Update 3.

I just wonder:

Why doesn't the C++ standard allow std::function<int(int)> fn3 = static_cast<int(int)>(f); ?

What's the rationale behind?

Answer 1

This is presumably because you cannot have an object of function type (under the definition of object used in the standard: a region of storage). The cast would need to create an object of the type int(int) , but you can't have objects of that type (functions are not objects).

You can, however, cast to a function pointer because you can have objects of a function pointer type. In fact, the function argument to static_cast already decays to a function pointer before being casted to the now-same type (much like an array readily decays to a pointer). This is because of [expr.static.cast]/8:

The lvalue-to-rvalue (4.1), array-to-pointer (4.2), and function-to-pointer (4.3) conversions are applied to the operand.

Long story short, fn2 is equivalent to fn1 with a redundant cast because f is already converted to a function pointer when intializing fn1 .