15

I just wanted to know the difference between . operator and :: operator?

manlio
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defiant
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3 Answers3

26

The former (dot, .) is used to access members of an object, the latter (double colon, ::) is used to access members of a namespace or a class.

Consider the following setup.

namespace ns {
    struct type
    {
        int var;
    };
}

In this case, to refer to the structure, which is a member of a namespace, you use ::. To access the variable in an object of type type, you use ..

ns::type obj;
obj.var = 1;
avakar
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  • so, if we are using object to access something, then we have to use dot operator, and if we are using a class name then we have to use ::, right?, can you give an example also, if u dont mind! – defiant May 24 '10 at 10:59
  • @oDx, that is correct, so if you have a static variable or function of a class, you would use "::" with the name of the class to reference them, whereas if you have a member function or member variable, you would use "." with the name of an instance of the class. – Michael Aaron Safyan May 24 '10 at 11:24
3

Another way to think of the quad-dot '::' is the scope resolution operator. In cases where there are more than one object in scope that have the same name. You explicitly declare which one to use:

 std::min(item, item2);

or

mycustom::min(item, item2);

The dot operator '.' is to call methods and attributes of an object instance

Myobject myobject;
myobject.doWork();
myobject.count = 0;
// etc

It was not asked, but there is another operator to use if an object instance is created dynamically with new, it is the arrow operator '->'

Myobject myobject2 = new Myobject();
myobject2->doWork();
myobject2->count = 1;
mrflash818
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1

If you are using a pointer to an object instance, you'll have to access the members of the object using -> in place of "dot"

Paulo Centeno
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