Non constant member function being called inside a constant member function

#include <iostream>

class Hello {
public:
void Test() {
std::cout << "Testing" << std::endl;
}
};

class Hi {
public:
Hi()
:hello(new Hello())
{}

~Hi()
{
delete hello;
}

void Testing() const {
hello->Test();
}

private:
Hello * hello;
};

int main(int argc, char ** argv) {
Hi hi; ;
hi.Testing();
return 0;
}
As i know a non-constant member function cant be called inside a constant member function but how the above code has been compiled successfully and giving the expected result .

Quote:

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Originally Posted by vikuseth

Code:

---

#include <iostream>

class Hello {
public:
    void Test() {
        std::cout << "Testing" << std::endl;
    }
};

class Hi {
public:
    Hi()
    :hello(new Hello())
    {}

    ~Hi()
    {
        delete hello;
    }

    void Testing() const {
        hello->Test();
    }

private:
    Hello  * hello;
};

int main(int argc, char ** argv) {
    Hi hi; ;
    hi.Testing();
    return 0;
}

---

As i know a non-constant member function cant be called inside a constant member function but how the above code has been compiled successfully and giving the expected result .

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Please use code tags when posting code. I added them in the quote; see the difference?

In the Testing member function, all member variables of the Hi class are const, because the member function is const. That means you have a

Code:

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Hello* const hello;

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In other words, a const pointer to a (non-const) Hello. Hence, you can call non-const functions on the Hello instance, but you cannot modify the pointer.

Thanks for editing the code .Its really looking good now . I understood something from your post .But one thing i want to know is we have not declared hello as constant (Hello *const hello ) inside Hi class , then how Testing member function will take it as constant pointer . OR is there any c++ concept on this topic which i am not aware of since i am very new to it .

Quote:

---

Originally Posted by vikuseth

But one thing i want to know is we have not declared hello as constant (Hello *const hello ) inside Hi class , then how Testing member function will take it as constant pointer .

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Because the object that owns the pointer is const in that context.

Thanks . Now i got it .
whenever we are declaring some member function as constant all the members present inside that function scope will be taken as constant . Like here hello is a constant pointer . so if we do something like hello = new Hello() inside Testing() , then it will give one error (since we are changing the address). But here we are trying to access what hello points to (that is Test()) , So its not showing any error . Is my understanding correct ? Please correct if i am wrong .

Quote:

---

Originally Posted by vikuseth

Thanks . Now i got it .
whenever we are declaring some member function as constant all the members present inside that function scope will be taken as constant . Like here hello is a constant pointer .

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I don't want to take away from your efforts to understand the const keyword but note there's a design alternative called immutability. You simply design your classes not to allow them being changed once created.