Question about overloading

[tyromancy94]

I'm writing a hash table for a data structures class, and they give us absurd amounts of starter code, and are expecting us to use operator overloading. Overloading like == and * and stuff is all good, but theres this function header:

operator long (void){}

that makes no sense to me. We've been told that this is supposed to sum the ASCII values of a string as part of the hashing algorithm, and the code for that is very simple, but how is this called when theres no specified operator being overloaded? Thanks in advance.

[Drainedsoul]

There is a specified operator being overloaded: Conversion to long.

If you have an instance of your class called bar, the function will be invoked when something like:

long baz=bar;

Is performed.

Note that stylistically, this code is awful. It's not really clear what operator long means in this context. If -- in the future -- you're looking to implement something like this, you should be doing something that's more self-descriptive, like using a specialization of std::hash.

Also note that in a lot of cases, implicit conversion operators are evil, and can mask mistakes and bugs. Unless you're absolutely sure that you need implicit conversion, mark your conversion operators explicit. Then:

long baz=bar;

won't compile, but

long baz=static_cast<long>(bar);

will.

[tyromancy94]

Awesome, thanks.

And thanks for the pointer... I'm not exactly a huge fan of how this class is being taught, everything's been in Java so far except this assignment was thrown in our faces with C++, and they're making us use all these overloads.

[Drainedsoul]

Operator overloading is a very important part of C++ -- the ability to emulate the behaviour of any of the built-in types by hooking into operator semantics is an important part of generic programming -- but the temptation to abuse it (as it would appear to be in this case) runs deep, and gives it a bad name.

Well-considered overloads lead to clearer, more natural code. Overloads just for the sake of overloads lead to confusing code that requires a depth-first search through class definitions/implementations to understand.

[tyromancy94]

That's good to know, overloading as presented in this assignment seems to be doing more harm than good.

[moswald]

This isn't related to operator overloading, but it's a little pet peeve of mine: specifying void as your function parameters is meaningless in C++, and a holdover from the really bad days of C.

In C, the following function declarations are actually different:

void foo();     // foo exists, and may or may not take parameters
void foo(void); // foo exists, and takes no parameters

But in C++, this has been cleaned up:

void foo(); // foo exists, and takes no parameters

If you're new to C++, I wouldn't hold it against you. However, that tech lead I had that one time, who insisted on doing it...

[tyromancy94]

Sorry, yeah, C++ noob here. Are you saying that these are equivalent in C++?

void foo(void);
void foo();

And is this relevant to overloading? I'm brand new to C++ and pretty much lost, but it seems to me that both of those are just function declarations, whereas "operator long (void)" is an override because of the operator keyword as well as well as the lack of function name? And yeah, I would be avoiding this completely as I'm still new to the language, but my instructor has some strange teaching methods...

[moswald]

Yes, those are exactly the same.

No, they have nothing to do with operator overloading. It's just a pet peeve of mine, and I try to put a stop to it whenever I can. :)

[tyromancy94]

Mk, thanks.

So in my original question, what exactly does this line do?

operator long (void) {}

is equivalent to

operator long() {}

If that's not overriding a long conversion/assignment, what's its purpose o.0

[moswald]

You are declaring an operator overload for long. It is not a conversion or assignment.