Behaviour of printf when printing a %d without supplying variable name

I've just encountered a weird problem, I'm trying to printf an integer variable, but I forgot to specify the variable name, i.e.

printf("%d");

instead of

printf("%d", integerName);

Surprisingly the program compiles, there is output and it is not random. In fact, it happens to be the very integer I wanted to print in the first place, which happens to be m-1.

The errorneous printf statement will consistently output m-1 for as long as the program keeps running... In other words, it's behaving exactly as if the statement reads

printf("%d", m-1);

Anybody knows the reason behind this behaviour? I'm using g++ without any command line options.

#include <iostream>
#define maxN 100
#define ON 1
#define OFF 0

using namespace std;

void clearArray(int* array, int n);
int fillArray(int* array, int m, int n);

int main()
{
    int n = -1, i, m;
    int array[maxN];
    int found;

    scanf("%d", &n);

    while(n!=0)
    {
        found=0;
        m = 1;
        while(found!=1)
        {
            if(m != 2 && m != 3 && m != 4 && m != 6 && m != 12)
            {
                clearArray(array, n);
                if(fillArray(array, m, n) == 0)
                {
                    found = 1;
                }
            }
            m++;
        }

        printf("%d\n");

        scanf("%d", &n);
    }

    return 0;
}

void clearArray(int* array, int n)
{
    for(int i = 1; i <= n; i++)
        array[i] = ON;
}

int fillArray(int* array, int m, int n)
{
    int i = 1, j, offCounter = 0, incrementCounter;

    while(offCounter != n)
    {
        if(*(array+i)==ON) 
        {
            *(array+i) = OFF;
            offCounter++;       
        }
        else 
        {
            j = 0;
            while((*array+i+j)==OFF)
            {
                j++;
            }