Take the following code :
int *p = malloc(2 * sizeof *p);
p[0] = 10; //Using the two spaces I
p[1] = 20; //allocated with malloc before.
p[2] = 30; //U
As everyone has said, you're writing to memory that isn't actually allocated, meaning that something could happen to overwrite your data. To demonstrate the problem, you could try something like this:
int *p = malloc(2 * sizeof(int));
p[0] = 10; p[1] = 20; p[2] = 30;
int *q = malloc(2 * sizeof(int));
q[0] = 0; // This may or may not get written to p[2], overwriting your 30.
printf("%d", p[0]); // Correctly prints 10
printf("%d", p[1]); // Correctly prints 20
printf("%d", p[2]); // May print 30, or 0, or possibly something else entirely.
There's no way to guarantee your program will allocate space for q at p[2]. It may in fact choose a completely different location. But for a simple program like this, it seems likely, and if it does allocate q at the location where p[2] would be, it will clearly demonstrate the out-of-range error.