Cannot call a method of const reference parameter in C++

Question

class A
{
public:
    A(){};
    ~A(){};
    void method(){};

};

void call(const A &a)
{
    a.method();   // I cannot call this method here if I use "const" but I can call it if not using "const"
}

int main()
{
    A a;
    call(a);
    return 0;
}

In this case, the error is: "passing const A as this argument of void A::method() discards qualifiers [-fpermissive]|"

In function call, if I use const, I get the error, but if I get rid of it, it works.

Can anyone explain it for me?

Solution

You can't call non-const member functions via const references. You can fix this by making the member function const:

void method() const {};
              ^^^^^

This indicates that calling the member does not mutate the object it is called on^*

^{* Conceptually. In practice it can mutate members marked mutable}

Comments

[Lightness Races in Orbit]

What do you think the const does?

[Lightness Races in Orbit]

@MrCold: The only reason that code doesn't work is because you have a bug in the operator[] that you randomly added and that has nothing to do with this question. Try to change just one thing at a time when you are trying to solve problems. If you'd only added the const to method(), you'd have seen that it works fine.

[Mr Cold]

Thank you for your answer but this as I see this rule does not work with overloading operator [] as follow: pastebin.com/1mKek7pM Can you explain more?

[juanchopanza]

@MrCold What do you mean "this rule doesn't work"? Do you mean it doesn't compile? That is because you are returning a non-const reference to a data member via a const reference to the object.

[juanchopanza]

@MrCold In other words, the const overload should return a const reference (or a value): const int& operator [] (int i) const {return num;}. You can also add a non-const overload for when you want to be able to modify the data member: int& operator [] (int i) {return num;}

[Mr Cold]

Oh, i got it. Thank you very much for your explanation :)