为什么类方法不能使用相同的名称调用全局函数？

Question

The following code shows a function call another function.
Both have the same name, but different signatures.
This works as expected.

//declarations
void foo();
void foo(int);

int main(){
  foo();
}

//definitions
void foo(){
    foo(1);
}
void foo(int){}

The only difference I will make now, is wrap one of the functions into a structure:

//declarations
struct Bar{
    void foo();
};
void foo(int);

int main(){
  Bar bar;
  bar.foo();
}

//definitions
void Bar::foo(){
    foo(1);
}
void foo(int){}

This fails to compile.

In member function ‘void Bar::foo()’:
error: no matching function for call to ‘Bar::foo(int)’
         foo(1);
              ^
note: candidate: void Bar::foo()
     void Bar::foo(){
          ^
note:   candidate expects 0 arguments, 1 provided

I don't understand why it wants to call foo(int) as a method, when the global function exists.
It doesn't mention anything about ambiguity, it just can't find the function.

Why is this happening, and how can I fix it?

side note: I'm wrapping old C code in a C++ wrapper, and most of the C++ methods are calls to the global C functions which pass in the wrapped struct implicitly. It's a similar situation to what is happening above (in terms of compiler errors).

Answer 1

The member function is hiding the global. It finds the name in the class context, therefore it not continue to search it in other contexts.

You need to call it like this:

::foo(1);

Another solution is to use forward declaration inside the function, like this:

void Bar::foo()
{
    void foo(int);
    foo(1);
}

As Praetorian suggests, here is another option:

void Bar::foo()
{
    using ::foo;
    foo(1);
}