Re: C question

[Kalpesh Rathod <kalpeshrathod@xxxxxxxxx>]

Hi Rick,

gcc can warn about void pointer increment if you use compiler option
-Wpointer-arith

==
Kalpesh

On Thu, Oct 8, 2009 at 8:22 AM, Manish Katiyar <mkatiyar@xxxxxxxxx> wrote:
> On Thu, Oct 8, 2009 at 7:42 AM, Rick Brown <rick.brown.3@xxxxxxxxx> wrote:
>> Hello list,
>>
>> As far as I recall from K&R, isn't pointer arithmetic on a void
>> pointer banned? And any effort to do that results in an error -
>> because the compiler won't know by how much size to increment the
>> pointer for a statement like "ptr++"? But then how about this:
>
> But in the program, you aren't actually trying to dereference the
> value. Just adding means it becomes normal arithmetic and that is why
> you get result as 1. You will see the error if you try to dereference
> it.
>
> /tmp> gcc a.c
> a.c: In function ‘main’:
> a.c:5: warning: format ‘%d’ expects type ‘int’, but argument 2 has type ‘void *’
> a.c:6: warning: dereferencing ‘void *’ pointer
> a.c:6: error: invalid use of void expression
> /tmp> cat a.c
> #include <stdio.h>
> int main()
> {
>   void *ptr = 0;
>   printf("%d \n", ptr+1);
>   printf("%d \n", *(ptr+1));
> }
>
>
>>
>> [rick@linux rick]$ cat t.c
>> #include <stdio.h>
>> int main()
>> {
>>    void *ptr = 0;
>>    printf("%d \n", ptr+1);
>> }
>> [rick@linux rick]$ gcc t.c
>> [rick@linux rick]$ ./a.out
>> 1
>> [rick@linux rick]$
>>
>> It compiles and runs fine ... !
>>
>> TIA,
>>
>> Rick
>>
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>>
>
>
>
> --
> Thanks -
> Manish
> ==================================
> [$\*.^ -- I miss being one of them
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