whats the difference between dot operator and scope resolution operator
I just wanted to know the difference between . ope开发者_如何学Pythonrator and :: operator?
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;
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;
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"
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