Printable View
Howdy yall,
I only recall a half semester of c++ before going on to Java for two years so if yall could help me with this simple question, it would be greatly appreciated...
when do you know to use...
void Class::methodName(){
...
}
instead of...
void methodName() {
...
}
when defining a method?
thanks -Cowboy
Err... The first one when the method is part of a class, and the second when it is not part of a class. For more on classes, you can go to the tutorials on this site, or search this forum (I know Stack Overflow made a nice post on classes recently -- I'll see if I can find it).
*edit*
The Brain has a nice post on them here: http://cboard.cprogramming.com/showthread.php?t=59027
Elad makes some good points on them here:
http://cboard.cprogramming.com/showthread.php?t=58661
Hmm... I can't find Stack Overflow's so I may be lying about that one (although I could've sworn there was one recently). :shrugs:
Those others should help, though.
I understand the class idiology, Java is OO. But let me link you to where i got my confusion...
http://www.cprogramming.com/tutorial/lesson18.html
the default constructor uses :: but the destroy_tree does not. They are both declared in the same class (btree) tho...
:: operator is called scope resolution operator.
so a declaration liek this classA::somefunction()
would mean that you are telling compiler that the function somefunction belongs to class calssA.
So whenevr u use the dot( . ) operator with an object to classA you will ahve access to somefunction for that class (ofcourse if it is within the scope of caller) and not somefunction of class B.
:: is also used to call methods of classes direclty using the clas sname without declaring the objject first.
so i can call someFunction in two ways
classA objA;
objA.someFunction();
or
classA::someFunction();
declaring a function without :: operator and class name gives it global scope so you can jsut call it like this
someFunction();
as long as you include the header file.
Hope It helps
venAdder
Appears to be a typo in the tutorial.
It should be 'void btree::destroy_tree(node* leaf)'.
The only time you would not is when you define the function inside the declaration of the class. E.g.
Code:class foo
{
void bar1( )
{
// Stuff here...
}
void bar2( );
};
void foo::bar2( )
{
// Stuff...
}
so a method declared with :: can be a static method or it can be inheirited when an object is declared?
Static methods belong to the class and not to any instance of the class. That is to say, when you call them, there is no 'this' pointer sent to them. Consequently, they cannot modify non-static data of the class. They can be called without an object, though, as in 'SomeClass::SomeFunc( )'. Normal, non-static functions cannot be called this way.
*edit*
You can also use the scope resolution operator (::) when calling a base-class method in a derived class, which somewhat breaks the rule above, but is somewhat of a special case.
Hello,
Member functions maybe defined outside the structure, by using the name of the structure, followed by a scope resolution operator, followed by the name of the member function. The scope resolution operator in C++ is, ::
The scope resolution operator is also used to qualify hidden names so that you can still use them. You can use the unary scope operator if a file scope name is hidden by an explicit declaration of the same name in a block or class. For example:You can also use the class scope operator to qualify class names or class member names. If a class member name is hidden, you can use it by qualifying it with its class name and the class scope operator. Whenever a name is followed by a :: operator, the name is interpreted as a class name.Code:int i = 10;
int f(int i) {
return i ? i : :: i; // return global i if local i is zero
}
In the following example, the declaration of the variable X hides the class type X, but you can still use the static class member count by qualifying it with the class type X and the scope resolution operator.The Binary and Unary scope resolution operator's associativities are read left-to-right. Keep in mind the scope resolution and global scope operators are different. For example:Code:#include <iostream>
using namespace std;
class X {
public:
static int count;
};
int X::count = 10; // define static data member
int main() {
int X = 0; // hides class type X
cout << X::count << endl; // use static member of class X
return 0;
}
All in all, you can use the scope resolution operator to reference global variables, to reference hidden members in base classes, to explicitly reference a member that is inherited, or to otherwise name a member hidden by the current context. When you are in the context of a nested class, you must use the scope resolution operator to access members of the enclosing class.Code:Operator Description Usage and Notes
:: scope resolution class_name :: member
:: global :: name
- Stack Overflow
Okay... Now there is a nice post on the topic by Stack Overflow. ;)
i ran out of ram reading that post...let me try it again lol
Heh. :) I never recalled making a post about this before, so I thought I'd try!Quote:
Originally Posted by Zach L.
Ah, don't worry about it. It was right under your nose. ;)Quote:
Originally Posted by Zach L.
Ah. A 256MB stick can do wonders, sometimes. :pQuote:
Originally Posted by Cowboy
- Stack Overflow
oh my dimm slots are full, its too bad they dont make ram for my brain...
ok so...methods declared outside main have the scope resolution...methods inside main dont have the resolution...and scope resolution is also used when main and class methods or variables have the same name...
Alright,
My examples may have been convincing, though the truth underlying the scope resolution operator is this:
- Reference global variables
- Reference hidden members in base classes
- Explicitly reference a member that is inherited [or to otherwise name a member hidden by the current context]
- Access members of the enclosing class when in the context of a nested class
It is quite simple. My examples showed the use of these somewhat and are not the only ways the scope resolution operator can be used. If you keep the 4 key points of the operator, everything else should become clear.
- Stack Overflow
man you read like my text book lol...
Its pieceing together now, thank you...have you ever considered writing a c++ book (even though there are billions out there)? lol
Glad to be of assistance. I apologize if I came across with too much detail and very little example. Though I have not considered writing a C or C++ book. I may sometime in the future, but I'm still in the learning process. I teach what I know to my best ability.Quote:
Originally Posted by Cowboy
- Stack Overflow