Why the output of printf(“%d” +1) is d but printf(“%%%d”+1) is %d?

Question

Why is that the output of

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

is d but the output of

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

is %d and not %%d ??

Answer 1

"%d"+1 by pointer arithmetic takes you to the second char in the char array which is d .

In the string literal "%%%d"+1 leaves you with "%%d" which is interpreted as %d by printf . Since %% is escaped to % .

Answer 2

You are doing pointer arithmetic. "%d"+1 is "d" and "%%%d"+1 is "%%d" (it's like you are skipping the first character of the string).

But, as the documentation of printf() explains, the percent sign ( % ) is a special character in the string format argument of printf() . It introduces a "conversion specifications".

Because it is a special character it needs a special sequence (a conversion specification, in fact) in order to be able to print a literal % . And the conversion specification designated to print a literal % is exactly %% , as you can see from the first row of the conversion specifications table in the docs.

Answer 3

You are using +1 in printf which will in turn skip the one character ie % in your case. after skipping % character you will be left with %%d , since %% is used to print % character. Output will be %d .