OverLoading and Scope

 

     void print(const string &);     void print(double);   // overloads the print function     void fooBar(int ival)     {         void print(int);   // new scope: hides previous instances of print         print("Value: ");  // error: print(const string &) is hidden         print(ival); // ok: print(int) is visible         print(3.14); // ok: calls print(int); print(double) is hidden     } 

When we call print, the compiler first looks for a declaration of that name. It finds the local declaration for print that takes an int. Once the name is found, the compiler does no further checks to see if the name exists in an outer scope. Instead, the compiler assumes that this declaration is the one for the name we are using. What remains is to see if the use of the name is valid.

The first call passes a string literal but the function parameter is an int. A string literal cannot be implicitly converted to an int, so the call is an error. The print(const string&) function, which would have matched this call, is hidden and is not considered when resolving this call.

When we call print passing a double, the process is repeated. The compiler finds the local definition of print(int). The double argument can be converted to an int, so the call is legal.