Consider below C++ program:
The output for the above program is 3 since the “a” passed as argument to the “func” shadows the “a” of the class .i.e 1
Then how to output the class’s ‘a’. This is where this pointer comes in handy. A statement like “cout <a” instead of “cout << a" can simply output the value 1 as this pointer points to the object from whom func is called.
How about Scope Resolution Operator? We cannot use Scope resolution operator in above example to print object’s member ‘a’ because scope resolution operator can only be used for a static data member (or class members). If we use scope resolution operator in above program we get compiler error and if we use this pointer in below program, then also we get compiler error.
Conclusion is scope resolution operator is for accessing static or class members and this pointer is for accessing object members when there is a local variable with same name.
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Improved By : Prateek Tamrakar