Re: initial value of reference to non-const must be an lvalue

From:

Sachin <sachinc.biradar@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Thu, 10 Jan 2008 21:58:54 -0800 (PST)

Message-ID:

<61a54c3d-d982-4018-973a-47f7ec2cc2d6@s8g2000prg.googlegroups.com>

On Jan 11, 10:43 am, "Jim Langston" <tazmas...@rocketmail.com> wrote:

asm23 wrote:

Hi,I need some help to clarify the warning "initial value of reference
to non-const must be an lvalue".

I'm searching in this groups to find someone has the same situation
like me. I found in the Post:
http://groups.google.com/group/comp.lang.c++/browse_thread/thread/e81...

//*********************************************************************
// CODE
//*********************************************************************
#include <iostream.h>

class ostream; =

//Need for Overloading <<

class complex {
double re, im; =

//Private members of class

public:
complex() { re=0.0; im=0.0; } //Empty Co=

nstructor

complex(double r, double i=0.0) //Constructor =

from 2

        doubles { re=r; im=i; }
        friend ostream& operator<<(ostream&, complex&);
        friend inline complex operator+(complex, complex);

};

inline complex operator+(complex a1, complex a2) //Add 2 complex
numbers { return complex(a1.re+a2.re, a1.im+a2.im); }
ostream& operator<<(ostream& os, complex& cnum) //Output a complex

operator<< is accepting a non constant ostream& and a non constant complex=

number { os << "(" << cnum.re << "," << cnum.im << ") "; return os;
};
int main(void)
{ complex a(1,2), b(3,4); //Define complex numbe=

cout <<"a=" << a <<",b=" << b <<" a+b = " << a+b << endl; /=

/Print

You are attempting to pass to operator<< a non constant ostream& and a non=

constant *temporary* complex& The temporary is the problem. Any chan=

ges

the function made to the temporary would be lost. A very simple fix. Ch=

ange

your operator<< to:

ostream& operator<<(ostream& os, const complex& cnum)

And your problem should go away.

You need to learn const correctness. A temporary rvalue can be passes a=

s a

const. I'm not sure of all the limitations, but it seems this is one of=

them. Your clue that this is a const correctness error is given in the
error itself:
"initial value of reference to ***non-const*** must be an lvalue".

sum

}

I also get the same warning in my compiler.
but I add some statement like:

//*********************************************************************
// CODE
//*********************************************************************

int a1=2;
int b1=3;
cout<<a1+b1;

this goes very well.

What's the difference between these two code? How can I get rid of
these warnings?

Thanks

--
Jim Langston
tazmas...@rocketmail.com- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -

I am not getting any warning message with above code, using Visual
Studio 6.0 Compiler, You can create temporary object to hold result of
a+b, than pass temporary object to operator<<.

c= a+b;
cout<<c;

Regards,
Sachin