What could the use of the operator keyword in c++ mean here?

Question

This is for an application that is written in C++. Under what circumstances would this line make sense to any of you, in the context of a struct definition (stream is a member variable of type FILE*):

operator FILE*(){return stream;}

I have been going through with a debugger, trying to make sense of it, but I can't seem to make that line of code activate. I have never encountered the operator overload keyword in such a fashion. What can this line of code do?

Answer 1

This is an implicit conversion operator.

Implicit conversion operators allow a type that would not otherwise implicitly convert to a destination type, the ability to do so. They have the following syntax, where Foo is the class of the object to be implicitly converted, and Bar is the destination class:

class Foo{
public:
    operator Bar(); // allow implicit conversion of Foo objects to Bar
};

The more common instance of this operator is in converting an object to a boolean value as a validity check. This can be seen with the standard library's streams, and smart pointers.

You should note the presence of a variation on the syntax, which prevents the existing conversion, and makes the conversion explicit instead:

class Foo{
public:
    explicit operator Bar(); // allow explicit conversion of Foo objects to Bar
};

This prevents getting bitten by a compiling program when you accidentally feed a type A that can convert into a type B into a function that accepts only B. Sure, that can be what you're going for, but it isn't always, and they decided to add this to the language to help people with a need for an explicit conversion.

With the explicit conversion operator, you can create an object from the origin object either via construction(using it in the construction of an object of the target type) or by explicit casting: B{A}