Author: Dieter Buerssner
Date: 09:27:36 11/28/01
Go up one level in this thread
On November 28, 2001 at 11:46:45, Miguel A. Ballicora wrote:
>Is it me or an expression like
>
>do_1();
>printf("Print this for goodness sake\n");
>fflush(stdout);
>do_2();
>
>should work no matter what if I see that do_1 and do_2 are executed?
This sounds, like you are overwriting some data in do_1(). The code produced
should be something like this.
char some_string[] = "Print this for goodness sake\n"
...
printf(some_string);
So if you have actually
int array[8];
char some_string[] = "Print this for goodness sake\n"
void do_1(void)
{
array[8] = 0;
}
The problem you described can occure.
Regards, and good luck finding the problem,
Dieter
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