Is c function prototype mismatch merely a warning

Question

please take a look at my codes below

#include 

void printOut()
{
 static int i = 0;
 if (i < 10)
 {
  printOut(i);
 }
}

int main(int argc         


        

Answer 1

In C, a function without any parameters can still take parameters.

That's why it compiles. The way to specify that it doesn't take any parameters is:

void printOut(void)

This is the proper way to do, but is less common especially for those from a C++ background.