invalid lvalue in unary `&'

[Suyash Upadhyay]

Hello all,
I was writing a code regarding offset of structure elements.

struct a
{
    struct b
    {
        int i;
        float f;
        char ch;
    }x;
    struct c
    {
        int j;
        float g;
        char ch;
    }y;
};
 
 
main()
{
    int *p;
    struct a z;
    fun(&z.y);
    printf("\n %d %f %c",z.x.f,z.x.i,z.x.ch);
    getch();
}
 
fun(struct c* p)
{
    int offset;
    struct b * address;
    offset=(char *)&((struct c *)(&((struct a *) 0)->y)->j)-(char *)((struct a *)0);    
                address=(struct b*)((char *)&(p->j)-offset);
    address->i=400;
    address->f=5.34;
    address->ch='c';
}

I am getting error "invalid lvalue in unary `&' " on underlined fragment of code. I am not getting what wrong with it.
Please help me.
Thanx


Suyash.Upadhyay

[Ganon11]

The compiler probably thinks you are trying to use '&' as a binary operator with the char* cast to the left and the value to the right. Try enclosing the section you want the address of in parentheses and see if this fixes the issue.

[Banfa]

This line is overly complex

    offset=(char *)&((struct c *)(&((struct a *) 0)->y)->j)-(char *)((struct a *)0);    
 

The main problem is that you have not taken account of the fact that -> has a higher precedence than casting so

(struct c *)(&((struct a *) 0)->y)->j

is equivalent to

(struct c *)((&((struct a *) 0)->y)->j)

that this actually takes an object of type c and tries to treat it as a pointer you need

((struct c *)(&((struct a *) 0)->y))->j

This takes an object of type c, dereferences it to get a pointer to object of type c, unnecessarily casts it to a pointer to an onject of type c (since it already has this type) and the references it to access a member.

This is the same as accessing an integer variable like this

int ix;
 
*(&ix) = 5;
 

i.e. quite unnecessary

So get rid of the cast

(&((struct a *) 0)->y)->j

Get rid of the dereference (&) followed by reference (->) and the extra set of parenthises

((struct a *) 0)->y.j


No I know that you are trying to access x given a pointer to y so what you are looking for here is the address of y so that you can calculate the offest to x. y is an addressable object itself, there is no need to access j in it and if fact this only leaves room for later errors as the calculation would go wrong if someone later added a member to y before j. Since you are going to deference this expression you can drop the .j giving

((struct a *) 0)->y

Putting that back into the line gives

    offset=(char *)&(((struct a *) 0)->y)-(char *)((struct a *)0);    
 

So this is the answer right?


Wrong, this is doesn't address the biggest problem of all which is that since this references the NULL pointer ( ((struct a *) 0)->y ) the code causes undefined behaviour. You have no guarantee it will work and just because it works in 1 place does not mean it will work everywhere.

If you have to do this (it is nasty but sometimes necessary) then all in all you would be better off declaring a variable of type a and using that to perform your calculations on.

[Suyash Upadhyay]

Thanx Banfa,
I got my fault. And I also understood your advise regarding ((struct a*) 0)
thanx very much