It's actually bug in your program. The scanf reads two characters and
then terminates it with '\0' which overrides the next variable on the
stack which happens to be "i". Classical example of buffer overflow :)
F.
On Thu, May 29, 2008 at 12:09 PM, G. Belllini <[EMAIL PROTECTED]> wrote:
>
>
> Hi guys,
>
> I fell into a problem in using "tcc 0.9.24" if you are interested.
>
> Here is a code showing the error :
> You should notice that variable "i" is not incremented. It looks like
> variable "i" is resetted
> just after the "scanf" instruction.
>
> #include <stdio.h>
>
> main ()
> {
> char i; // used as number that we want on 1 byte only
> char tmp_str[2]; // this is a string
>
> i = 0;
>
> do
> {
>
> printf ("\n");
> printf ("Type 2 characters : ");
> scanf ("%s", &tmp_str);
>
> printf("\n");
> printf("You typed %s\n", tmp_str);
>
> i++;
> printf("DEBUG : i = %d\n", i);
>
> } while (i < 5);
>
> }
>
>
> Ok, now a workaround is to declare variable "tmp_str" as first. Like this :
>
> #include <stdio.h>
>
> main ()
> {
> char tmp_str[2]; // this is a string
> char i; // used as number that we want on 1 byte only
>
> i = 0;
>
> do
> {
>
> printf ("\n");
> printf ("Type 2 characters : ");
> scanf ("%s", &tmp_str);
>
> printf("\n");
> printf("You typed %s\n", tmp_str);
>
> i++;
> printf("DEBUG : i = %d\n", i);
>
> } while (i < 5);
>
> }
>
>
>
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