Question

I saw this code in C and run it:

int i,j;
scanf("%d %d"+scanf("%d %d",&i,&j));
printf("%d %d",i,j);

Input:

1 2 3

Output:

3 2

This is quite unexpected (reverse order and three inputs).

Please explain this.

Answer 1

This:

scanf("%d %d"+scanf("%d %d",&i,&j));

Is horrible code! But what’s happening?

First, the inner scanf is called. It stores 1 and 2 into i and j respectively (it should be obvious why this is). It then returns 2, because that is the number of things it stored. That then “skips” the first two characters of the outer scanf’s format statement, making it " %d". Then scanf wants to store 3 from the input somewhere, but no pointer to a variable was given in the outer scanf call. So what happens next? It’s undefined behavior, but the actual fact is that the second (outer) scanf call is just reusing (or stomping on, if you prefer) the arguments passed to the first (inner) scanf. So 3 is stored in i, and that’s it.