printf() with no arguments in C compiles fine. how?

Question

I tried the below c program & I expected to get compile time error, but why compiler isn\'t giving any error?

#include 
int main(void)
{
          


        

Answer 1

This compiles well. Because it matches the printf() prototype which is

printf(const char *,...);

During run-time the call

printf("%d\n");

Tries to fetch the value from second argument and since you have not passed anything it might get some garbage value and print it out so the behavior is undefined here.