If you don't have a forward declaration of your function before the place of usage, the compiler will create implicit declaration for you - with the signature int input()
. It will take the name of the function you called, it will assume that the function is returning int
, and it can accept any arguments (as Bartek noted in the comment).
For this function, the implicit declaration matches the real declaration, so you don't have problems. However, you should always be careful about this, and you should always prefer forward declarations instead of implicit ones (no matter if they are same or not). So, instead of just having forward declaration of the void prime()
function (assuming that you will use it somewhere), you should also have a forward declaration of int input()
.
To see how can you pass any number of the arguments, consider this:
#include <stdio.h>
// Takes any number of the arguments
int foo();
// Doesn't takes any arguments
int bar(void)
{
printf("Hello from bar()!\n");
return 0;
}
int main()
{
// Both works
// However, this will print junk as you're not pushing
// Any arguments on the stack - but the compiler will assume you are
foo();
// This will print 1, 2, 3
foo(1, 2, 3);
// Works
bar();
// Doesn't work
// bar(1, 2, 3);
return 0;
}
// Definition
int foo(int i, int j, int k)
{
printf("%d %d %d\n", i, j, k);
return 0;
}
So, inside the definition of the function you're describing function arguments. However, declaration of the function is telling the compiler not to do any checks on the parameters.