Why buffer overflow doesn't affect to this code?

Question

I have the following code:

int main(int argc, char *argv[])
{
    char ch[10];
    printf(\"String 10 max. :: \"); gets( ch );

    printf(\"String: %s\\n\", ch)         


        

Answer 1

You're not seeing any effect because you don't have any more local variables, change the code to this and you will

int main(int argc, char *argv[])
{
    char ch[10];
    int i=42;

    printf("String 10 max. :: "); gets( ch );

    printf("String: %s\n", ch);
    printf("i: %d\n", i);

    return 0;
}