Mentor for C self study wanted

[Giorgos Keramidas]

On 2007-10-23 23:24, Harald Schmalzbauer <h.schmalzbauer at omnisec.de> wrote:
> Thanks all,
> here was my example, just for completeness, I found mentors for my
> needs.

> #include <stdio.h>
>
> void main()
> {
>   short nnote;
>
>   // Numerischen Notenwert einlesen
>   printf("Bitte numerischen Schulnotenwert eingeben: ");
>   scanf("%d",&nnote);

You are passing "%d" to scanf() so it expects to find enough 'storage'
in its pointer argument for an 'int'.  If 'short' happens to have a
smaller size (as is commonly the case), scanf() will overwrite random
memory locations after 'nnote'.  On systems where 'nnote' is stored in
the stack (because it's an automatic/local variable of main()), you are
risking stack corruption (and a SEGFAULT *may* happen).

It's also a very good idea to check the return code of scanf():

	int nnote;

	if (scanf("%d", &nnote) != 1) {
		error;
	}

>   switch (nnote)
>   {
>     case 1: printf("Die Note %d entspricht sehr gut.",nnote);
>     break;
>     case 2: printf("Die Note %d entspricht gut.",nnote);
>     break;
>     case 3: printf("Die Note %d entspricht befriedigend.",nnote);
>     break;
>     case 4: printf("Die Note %d entspricht ausreichend.",nnote);
>     break;
>     case 5: printf("Die Note %d entspricht mangelhaft.",nnote);
>     break;
>     case 6: printf("Die Note %d entspricht ungen?gend.",nnote);
>     break;
>     default: printf("%d ist keine zul?ssige Schulnote!");

There's no `int' argument to the printf() call of the default clause.

This will either cause printf() to print random garbage, or try to
access memory regions which are unmapped and SEGFAULT.

> P.S.:
> I found that declaring nnote as int soleves my problem, but I couldn?t
> understand why.
> Another one was the result of default: nnote was -1077942208 instead
> of 9 for example.

It was never assigned to 9 :)