RFR-8008118

[Martin Buchholz]

Christos, I think you may have overlooked the subtlety that pathv is *not*
a const pointer - it just looks like one.


cat cast.c && echo --- && gcc -Wall -Wcast-qual cast.c
#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]) {
    const char **p = (const char **) malloc(5);
    char *q = (char *)p + 3;
    printf("%p %p\n", p, q);
    return 0;
}



On Wed, Apr 10, 2013 at 1:30 PM, Christos Zoulas <christos at zoulas.com>wrote:

> On Apr 10, 11:54am, martinrb at google.com (Martin Buchholz) wrote:
> -- Subject: Re: RFR-8008118
>
> | > 1. We did we switch from NEW() to xmalloc()? Why is the xmalloc cast
> | > needed?
> |
> | NEW is for allocating homogeneous arrays, but here the memory block is
> | being used for both chars and pointers.
>
> I did not know that. Why the cast though? xmalloc() returns void *, no?
> Extraneous casts are bad because they hide conversion errors. For example,
> if you don't have the xmalloc prototype in scope, without the cast you
> get a warning of casting integer to pointer of different size. With
> the cast you get the wrong data assigned to the pointer.
>
> | > 2. I would not declare pathv "const char **", but "char **", and then
> | >    cast the return if needed. This will make life easier in the future
> | >    if we decide to turn on warnings about const-castaways.
> | >
> | >
> | I believe the current code doesn't cast away const and doesn't write to
> | const.  The only cast is to the return from xmalloc, which is expected.
> |  What might a compiler warn about?
>
> It is casting away const before the memcpy:
>
> +    p = (char *) pathv + pathvsize;
>
> Try to compile with -Wcast-qual.
>
> christos
>