问题
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?
回答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.
回答2:
The return value of operator T() where T is a type is always T.
回答3:
The name of the conversion operator is its type. If this were not the case, you could define an int conversion operator (for example) that actually returned a double. A somewhat similar line of thinking applies to constructors, which also do not have a return type.
来源：https://stackoverflow.com/questions/2036923/how-does-conversion-operator-return-a-value