c - Reason for the Output -

#include<stdio.h> int main(void) {  int a=5;  printf("%d"+1,a); } 

output: d. didn't how output coming: d ?

you passed first argument of printf "%d"+1; "%d" seen const char * points memory location %d stored. pointer, if increment one, result point following element, which, in case, d.

a not used, should not problem since in general (i don't know if it's standard-mandated edit: yes is, see bottom) stack cleanup responsibility variadic functions caller (at least, cdecl way, may or may not ub, don't know*).

you can see easier way:

#include<stdio.h> int main(void) {     int a=5;     const char * str="%d";     printf(str + 1, a); } 

 

str ---------+              |              v           +----+----+----+           |  % |  d | \0 |           +----+----+----+  str + 1 ----------+                   |                   v           +----+----+----+           |  % |  d | \0 |           +----+----+----+ 

thus, ("%d"+1) (which "d") interpreted format string, , printf, not finding %, print is. if wanted instead print value of a plus 1, should have done

printf("%d", a+1); 

---

edit: * ok, it's not ub, @ least c99 standard (§7.19.6.1.2) it's ok have unused parameters in fprintf:

if format exhausted while arguments remain, excess arguments evaluated (as always) otherwise ignored.

and printf defined have same behavior @ §7.19.6.3.2

the printf function equivalent fprintf argument stdout interposed before arguments printf.