Re: Reference Member variables

From:
Michael Aaron Safyan <michaelsafyan@aim.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Sat, 29 Mar 2008 05:03:02 CST
Message-ID:
<fsjl01$hpu$1@aioe.org>
smawsk wrote:

Hi,

I am writing a sample program to use a reference member variable. But
I was surprised to know that the below piece of code is not working as
I expected.
Can some one let me know what's wrong with this code?

class X
{
public:
    X(int x) : ref(x)
    {

    }
    void Display() const
    {

        cout<<ref<<endl;
    }

private:
    int& ref;
};

int main()
{
    int c = 10;
    X x(c);
    x.Display(); // This function does not print 10. Instead some junk
value

    return 0;
}

Why So? Am I missing some basic point regarding the working of
References here?

Thanks in advance....

Warm Regards.


A reference must be initialized from a reference or pointer object.
X::X(int) does not take a reference or pointer. What you are doing is
initializing X::ref with the address of "x" in X::X(int), which is
located in the stack frame of the function X::X(int) and which becomes
invalid when X::X(int) has completed execution. Consider the following:

int c = 10;

     ---
c |10 |
     ---

X x(c);

     ---
c |10 |
     ---
x |10 | <= Not the same address as c!
     ---
ref|&x | <= Not pointing to c!
     ---

// After execution of X x(c);

     ---
c |10 |
     ---
ref|? | <= The object "x" no longer exists! This references is invalid.
     ---

// ... You get the point.

What I think you intend to do is:

class X
{
    public:
        X(int& x) : _ref(&x) {}
        ~X(){}
        int get()const{ return _ref: }
        void print(){ std::cout<<get()<<std::endl; }
    private:
        X& operator=(const X& o){ return *this; }
        int& _ref;
};

int main(int argc, char* argv[])
{
    int c = 10;
    X x(c);
    x.print();
    return 0;
}

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