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how to skip the value in printf in c

#include<stdio.h>
void main()
{
int i=6;
printf("%d %*d\n",i,i+9);
}
what is the out put?
Sep 5 '14 #1
2 1590
donbock
2,426 Expert 2GB
What happens is undefined behavior. The call to printf doesn't have enough arguments.

Refer to man page for printf. The format string has two format specifies: "%d" and "%*d"; the first expects one argument and the second expects two arguments. Thus, three arguments should follow the format string but you only have two.
Sep 6 '14 #2
It will most likely crash.
Sep 9 '14 #3

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