CODEWITHSUNDEEP
cscanf
harsh vardhan
harsh vardhan

Reputation: 25

Related to scanf statements in c

I have come across the following piece of code and i'm not being able to understand the scanf portion.

int main()
{
  int i,j;
  scanf("%d %d"+scanf("%d %d",&i,&j));
  printf("%d %d",i,j);
  return 0;
}

I ran the code on inputs 4 8 9 and it returned 9 8.

Can someone please explain the working?

Upvotes: 2

Views: 72

Answers (1)

chux
chux

Reputation: 154198

The inner scanf("%d %d",&i,&j) returns a count, like 2,1, EOF (or maybe 0).

Adding that count to the format string "%d %d", offsets the format by the count, such as by 2, to form " %d". This is simply pointer addition. @John Bollinger

Then code does the equivalent of scanf(" %d"); which is undefined behavior (UB) as it is missing a matching int * to go with the " %d". @mch

Can someone please explain the working?

Its not "working", it is UB.

A variation that treads on thin ice. It will "work" if the first scanf() returns 2. Yet this all looks like hacker code to me.

int main() {
  int i,j;
  scanf("%d %d"+scanf("%d %d",&i,&j), &i);
  printf("%d %d",i,j);
  return 0;
}

Upvotes: 5

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