Re: size of pointer variables

junky_fellow@xxxxxxxxxxx wrote:

> Richard Bos wrote:
> > junky_fellow@xxxxxxxxxxx wrote:
> >
> > > node_t * pstr_node;
> > > bzero(((char *)pstr_node, sizeof (*pstr_node));
> > >
> > > Is this legal ? Can we typecast structure pointer to char pointer ?
> >
> > The code is not legal, for a couple of reasons. First, bzero() is not an
> > ISO C function. Second, you don't initialise pstr_node, so it points
> > nowhere when you pass it to bzero(). Third, you have a parenthesis too
> > many (and two superfluous, but harmless ones. BTW, I disagree with
> > bjrnove on using sizeof (node_t); the way you have it is more solid.).
> >
> Can you please specify why sizeof(*pstr_node) is better as compared
> to sizeof(struct node_t) ? I thought they both are equivalent.

To the compiler, yes, they're equivalent. But consider what happens if
you change pstr_node from a node_t to a newnode_t. If you use
sizeof(node_t), you still pass the size of a node_t, which may be larger
or smaller than the size of a newnode_t. This would mean that bzero()
only zeroes part of the struct, or writes over its end. If you use
sizeof *pstr_node, you always pass in the right size: the size of the
object which pstr_node points at.

> > It is probably superfluous, though, since the most usual implementations
> > of bzero() take a void * as their first argument, and you can pass any
> > object pointer as a parameter declared as void *, without a cast.
>
> That means pointer variable to any type may be typecasted to (char *)
> but *not* the viceversa ?

Similarly to a void *, yes, except that void *s don't need casts. You
can assign any object pointer to a void *; you can also assign a void *
to any object pointer; but the latter only makes sense if the void *
contains the value of another pointer of the type assigned to. That is,
this is correct:

int *ip, i=4;
void *vp;

ip=&i;
vp=ip;
ip=NULL;
...
ip=vp;

and now ip again contains the address of i, as long as vp has not been
tampered with in the mean time. _This_, however, is not correct:

int *ip, i=4;
void *vp;
float *fp;

ip=&i;
vp=ip;
...
fp=vp;

Your compiler will probably not spot it, but you're indirectly assigning
an int pointer to a float pointer, which is not required to work.

Richard
.

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