confused by operator overloading

[rlwebbnafex]

Hello, I found this example of a class and I'm just trying to
understand what's going on with the operator function, could someone
help? I understand the template part. But i dont get the statment
operator int *(). how can you have a type as part of the function
name? or is it part of the function name? also there is no return
type, is that just because default int is assumed? thanks

class NullClass
{
public:
template<class T>
operator T *() const
{
return 0;
}
};

[Victor Bazarov]

rl*********@gmail.com wrote:

Hello, I found this example of a class and I'm just trying to
understand what's going on with the operator function, could someone
help? I understand the template part. But i dont get the statment
operator int *(). how can you have a type as part of the function
name? or is it part of the function name? also there is no return
type, is that just because default int is assumed? thanks

class NullClass
{
public:
template<class T>
operator T *() const
{
return 0;
}
};

It's called "a type conversion operator" or "a type conversion
function". Whenever an object of your class is going to be used
where a pointer is expected, it would be converted using this
particular operator, i.e. return the null pointer.

V
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