CODEWITHSUNDEEP
c++constants
Jony
Jony

Reputation: 6774

const object in c++

I have a question on constant objects. In the following program:

class const_check{
    int a;
    public:
    const_check(int i);
    void print() const;
    void print2();
};

const_check::const_check(int i):a(i) {}

void const_check::print() const {
int a=19;
    cout<<"The value in a is:"<<a;
}

void const_check::print2() {
    int a=10;
    cout<<"The value in a is:"<<a;
}

int main(){
    const_check b(5);
    const const_check c(6);
    b.print2();
    c.print();
}

void print() is a constant member function of the class const_check, so according to the definition of constants any attempt to change int a should result in an error, but the program works fine for me. I think I am having some confusion here, can anybody tell me why the compiler is not flagging it as an error?

Upvotes: 5

Views: 16286

Answers (4)

Puppy
Puppy

Reputation: 146900

Also, unless I am mistaken, you forgot to actually declare the member variable const to begin with.

Upvotes: -3

Douglas Leeder
Douglas Leeder

Reputation: 53310

You aren't changing the instance variable a you are creating a local variable a in each method.

Upvotes: 4

Jonathan M Davis
Jonathan M Davis

Reputation: 38267

You're not changing the member variable a in either print() or print2(). You're declaring a new local variable a which shadows the member variable a.

Upvotes: 1

Thomas
Thomas

Reputation: 181705

By writing

int a = 19;

inside print(), you are declaring a new local variable a. This has nothing to do with the int a that you declared inside the class const_check. The member variable is said to be shadowed by the local variable. And it's perfectly okay to declare local variables in a const function and modify them; the constness only applies to the fields of the object.

Try writing

a = 19;

instead, and see an error appear.

Upvotes: 22

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