Question

Say I have three classes:

class X{}; class Y{}; class Both : public X, public Y {}; 

I mean to say I have two classes, and then a third class which extends both (multiple-inheritance).

Now say I have a function defined in another class:

void doIt(X *arg) { } void doIt(Y *arg) { } 

and I call this function with an instance of both:

doIt(new Both()); 

This causes a compile-time error, stating that the function call is ambiguous.

What are the cases, besides this one, where the C++ compiler decides the call is ambiguous and throws an error, if any? How does the compiler determine what these cases are?

Answer 1

Simple: if it’s ambiguous, then the compiler gives you an error, forcing you to choose. In your snippet, you’ll get a different error, because the type of new Both() is a pointer to Both, whereas both overloads of doIt() accept their parameters by value (i.e. they do not accept pointers). If you changed doIt() to take arguments of types X* and Y* respectively, the compiler would give you an error about the ambiguous function call.

If you want to explicitly call one or the other, you cast the arguments appropriately:

void doIt(X *arg) { } void doIt(Y *arg) { } Both *both = new Both; doIt((X*)both);  // calls doIt(X*) doIt((Y*)both);  // calls doIt(Y*) delete both;