What is the meaning of "Type Qualifier (const)" after function name in function declaration declaration(virtual void printMarksheet() const = 0;)?

Question

I have happened to read a function definition like this:

virtual void printMarksheet() const = 0;

I thought initially that the syntax may be incorrect. So I wrote a dummy code to check. I wrote the following code and compiled. It compiled successfully. But I would like to know the meaning of the line. Is it function definition? or function declaration? What is the significance of const=0.

class Marksheet
        {
        public:
            virtual void printMarksheet() const = 0;
        };
        int main()
        {
            return 0;
        }

possible duplicate of [Why pure virtual function is initialized by 0?](http://stackoverflow.com/questions/2156634/why-pure-virtual-function-is-initialized-by-0) — Levi, Jun 07 '15 at 06:34
That question gives only a partial answer to my question @Levi. — Vigneshkumar Chinnachamy.M, Jun 07 '15 at 06:45

score 0 · Answer 1 · answered Jun 07 '15 at 06:34

0

Its a function declaration. const=0 are two different things. By defining const, the function will not be able to change the data member value. By defining =0, the function declared as pure virtual function.

answered Jun 07 '15 at 06:34

Sumeet

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What is the meaning of "Type Qualifier (const)" after function name in function declaration declaration(virtual void printMarksheet() const = 0;)?

1 Answers1