find() arguments

[crapgarden]

In my book, it says find() must be used like so:
ex:
iter = var.find(scores.begin(), scores.end(), score);

but later it uses this in practice:

ex:
while (used.find(guess) != string::npos)

Why can you simply pass 'guess' var to it here and not need to search a range? Isn't find() an algorithm that requires 3 arguments to work?
*Guess is 'char guess' declared earlier.

[jonsca]

"used" is a std::string so it uses the find() method particular to strings. See http://www.cplusplus.com/reference/string/string/find/

versus the <algorithm> version (your first example)
http://www.cplusplus.com/reference/algorithm/find/

[crapgarden]

Ok I see. so find() operates differently depending which type of variable it's being used on, and what type of arguments are being passed to it right?

[jonsca]

The find() in algorithm is completely independent of find() in string. find() in algorithm should be able to be used on any STL container (including std::string if I am not mistaken). The find() in string can only be used from a string object and takes different kinds of parameters as overloads (see those references I posted).

[crapgarden]

I should have mentioned that used was declared ahead of the code as:

string used = "";

But now I'm curious about this reference you're making. What makes one find() an Algorithm, and another a Member function? I'm looking through the links you sent me, and maybe it's my newb-ness getting in the way, but how do you tell the difference?

[jonsca]

What is the type of var in your above code? The <algorithm> versions are not member functions of an object as normally they take iterators to objects as parameters. That's why I'm confused as to where var is coming from. I misread it initially.

[LevyDee]

Just to add, you can declare multiple functions with the same name in the same namespace as long as the arguments are different. This is called function overloading.