How does conversion operator return a value?

Question

For a class A, an integer conversion operator would look something like;

operator int() //Here we don't specify any return type {     return intValue; } 

How is the above function able to return a value when its return value type appears not to be specified? It doesn't appear to return "anything", but I know it's not void.

How is this meaningful when a return type is not specified?

Answer 1

The return type of operator T() is always T. It's a special case of C++.

It does not use standard function prototype syntax T foo() because 2 functions with the same name differing only by the return type cannot coexist (e.g. int foo() conflicts with double foo()). If this syntax is used then you can only define 1 conversion operator overload, which is undesirable.