Why do multiple-inherited functions with same name but different signatures not get treated as overloaded functions?

Member lookup rules are defined in Section 10.2/2

The following steps define the result of name lookup in a class scope, C. First, every declaration for the name in the class and in each of its base class sub-objects is considered. A member name f in one sub-object B hides a member name f in a sub-object A if A is a base class sub-object of B. Any declarations that are so hidden are eliminated from consideration. Each of these declarations that was introduced by a using-declaration is considered to be from each sub-object of C that is of the type containing the declara-tion designated by the using-declaration. If the resulting set of declarations are not all from sub-objects of the same type, or the set has a nonstatic member and includes members from distinct sub-objects, there is an ambiguity and the program is ill-formed. Otherwise that set is the result of the lookup.

class A {
public:
  int f(int);

};
class B {
public:
   int f();

};
class C : public A, public B {};
int main()
{
     C c;
     c.f(); // ambiguous
}

So you can use the using declarations A::f and B::f to resolve that ambiguity

class C : public A, public B {
     using A::f;
     using B::f;

};

int main()
{
     C c;
     c.f(); // fine
}

The second code works flawlessly because void foo(float) is inside C’s scope. Actually d.foo(5); calls void foo(float) and not the int version.