RE: [SLUG] GCC question

[Amir Binyamini]

Hi,
1) Rob said:
*test is not a string. test is a string
Youre absolteuly right in this. Sorry, I was in a haste.
2)
Rob said:
*test is of type char and value 'a' and should work fine
Well , what I said , and I repeating it now , is :
Take a little program like this:
int main()
   {
    char* test = "ab";
   (*test)++;
   printf("finished\n");
   }
Compiling it with gcc on RefHat Linux an running it gives a segmentation 
fault.

On the other hand,if instead (*test)++ you'll write
  char c = *test;
  c++;
It will not crash.
Regards,
Amir
From: Robert Collins <[EMAIL PROTECTED]>
To: Amir Binyamini <[EMAIL PROTECTED]>
CC: [EMAIL PROTECTED], slug@slug.org.au
Subject: RE: [SLUG] GCC question
Date: Tue, 19 Apr 2005 15:39:47 +1000
On Mon, 2005-04-18 at 14:53 +0300, Amir Binyamini wrote:
> Hi,
>
> The reason is that you are trying to incremnt a string
> and not a char; and the incremetn operation is not allowed.
>
> If you will try to run a simple 2 line program like the following,
> you will get a segmentation fault:
>
>       char* test = "ab";
>       (*test)++;
>
> because *test is a string;
huh? *test is not a string. test is a string, *test is of type char and
value 'a'. 'a'++ is valid here and should work fine = test will be on
the stack.
...
> I assume what you meant to do , which works OK with gcc on linux , is:
>
> void    code(char *);
> main()
>       {
>       code("This is a test string");
>       putchar('\n');
>       return 0;
>       }
>
>
> void code (char * strptr)
>     {
>     while( *strptr)
>         {
>         printf("%c", *strptr);
>         ++strptr;
>         }
>     }
This would be somewhat shorter and also more correct as
void
code (char const * strptr)
{
  while (*strptr)
      printf ("%c", *strptr++);
}
Rob
--
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