Printf Multiple Choice Questions : printf(1+”%d”);

Question : Guess the output of Program

#include<stdio.h>
void main()
{
printf(1+"%d");
printf(1+"%c");
}

Options :

  1. Garbage
  2. dc
  3. cd
  4. %d%c
[toggle title="Output"]dc[/toggle]

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How and Why ?

  1. Printf Returns the Length of String specified in Double Quotes
  2. Consider first printf , Length of String specified in first printf i.e “%d” is 2
  3. It starts printing the String Skipping the number of Characters specified before ‘+’ sign
  4. printf(1+”%d”) skips “%” character and only prints ‘d’ as output.
  5. Similarly printf(1+”PRITESH”) prints “RITESH” skipping first “P”.
  6. Similarly printf(2+”PRITESH”) prints “ITESH” skipping 2 characters “PR”.

Note :

  1. Operation is similar as Substring Operation
  2. While writing such Statement , Variable is not taken into Consideration .
  3. Confused ? 
#include<stdio.h>
void main()
{
int var=10;
printf(1+"%d",var);
}
  1. Above statement also prints “d” as output. i.e Variable is Simply Neglected by Compiler.