为什么我必须使用寻址运算符来获取指向成员函数的指针？
Why must I use address-of operator to get a pointer to a member function?
struct A
{
void f() {}
};
void f() {}
int main()
{
auto p1 = &f; // ok
auto p2 = f; // ok
auto p3 = &A::f; // ok
//
// error : call to non-static member function
// without an object argument
//
auto p4 = A::f; // Why not ok?
}
为什么必须使用寻址运算符来获取指向成员函数的指针？
auto p1 = &f; // ok
auto p2 = f; // ok
第一个或多或少是正确的。但是因为 non-member 函数有 implicit conversions to pointers, the & isn't necessary. C++ makes that conversion, same applies 到静态成员函数。
引用自cppreference：
An lvalue of function type T can be implicitly converted to a prvalue
pointer to that function. This does not apply to non-static member
functions because lvalues that refer to non-static member functions do
not exist.
struct A
{
void f() {}
};
void f() {}
int main()
{
auto p1 = &f; // ok
auto p2 = f; // ok
auto p3 = &A::f; // ok
//
// error : call to non-static member function
// without an object argument
//
auto p4 = A::f; // Why not ok?
}
为什么必须使用寻址运算符来获取指向成员函数的指针？
auto p1 = &f; // ok
auto p2 = f; // ok
第一个或多或少是正确的。但是因为 non-member 函数有 implicit conversions to pointers, the & isn't necessary. C++ makes that conversion, same applies 到静态成员函数。
引用自cppreference：
An lvalue of function type
Tcan be implicitly converted to a prvalue pointer to that function. This does not apply to non-static member functions because lvalues that refer to non-static member functions do not exist.