Is this correct

[Ravi Uday]

Hi,

Can some-one comment on below snippet ?

static int xxx_list_compar e (void *ptr1, void *ptr2) {
if (ptr1 && ptr2 && (ptr1 == ptr2)) {
return (0);
}
return (1);
}

Are bit operations and comparing operations allowed on void * ?
Any redundancy in the code.

thanks,
- ravi

[Nelu]

Ravi Uday wrote:

Hi,

Can some-one comment on below snippet ?

static int xxx_list_compar e (void *ptr1, void *ptr2) {
if (ptr1 && ptr2 && (ptr1 == ptr2)) {
return (0);
}
return (1);
}

Are bit operations and comparing operations allowed on void * ?
ptr1 and ptr2 are pointers and it's valid to use the operations you are
using.
It doesn't matter what the underlying type for the pointer is.
You do not have bit operations in your code. && is a logical operator
and ptr1 as a condition is true if ptr1 is not NULL. You can always
compare two pointers to see if they point to the same address.
Any redundancy in the code.

Yes. You are trying to do something if the pointers are not null and
they
point to the same address. You can rewrite the condition as:
if(ptr1&&(ptr1= =ptr2)) because ptr2 is not going to be null if it
equals the non-null ptr1.

--
Ioan - Ciprian Tandau
tandau _at_ freeshell _dot_ org (hope it's not too late)
(... and that it still works...)

[Keith Thompson]

Ravi Uday <ra******@gmail .com> writes:

Can some-one comment on below snippet ?

static int xxx_list_compar e (void *ptr1, void *ptr2) {
if (ptr1 && ptr2 && (ptr1 == ptr2)) {
return (0);
}
return (1);
}

Are bit operations and comparing operations allowed on void * ?
Any redundancy in the code.

No, bit operations on void* are not allowed. Fortunately, there are
no bit operations in the code you posted. Equality comparisons are
perfectly ok.

The function returns 0 if both arguments are non-null and are equal to
each other; otherwise it returns 1. If that's what you want, it
looks correct.

Personally, I'd prefer to make the comparisons explicit, and I never
use parentheses on a return statement (they make it look too much like
a function call):

if (ptr1 != NULL && ptr2 != NULL && ptr1 == ptr2) {
return 0;
}
else {
return 1;
}

or perhaps even this:

return ptr1 == NULL ||
ptr1 == NULL ||
ptr1 != ptr2;

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.