#include <stdio.h>
#include <stdlib.h>

/* print a list of ints */
int show(int *x, int len)
{
	int i;
	for (i = 0; i < len; i++)
		printf("%d%c", x[i], i == len - 1 ? '\n' : ' ');
	return 1;
}

/* next lexicographical permutation */
int next_lex_perm(int *a, int n) {
#	define swap(i, j) {t = a[i]; a[i] = a[j]; a[j] = t;}
	int k, l, t;

	/* 1. Find the largest index k such that a[k] < a[k + 1]. If no such
	      index exists, the permutation is the last permutation. */
	for (k = n - 1; k && a[k - 1] >= a[k]; k--);
	if (!k--) return 0;

	/* 2. Find the largest index l such that a[k] < a[l]. Since k + 1 is
	   such an index, l is well defined */
	for (l = n - 1; a[l] <= a[k]; l--);

	/* 3. Swap a[k] with a[l] */
	swap(k, l);

	/* 4. Reverse the sequence from a[k + 1] to the end */
	for (k++, l = n - 1; l > k; l--, k++)
		swap(k, l);
	return 1;
#	undef swap
}

void perm1(int *x, int n, int callback(int *, int))
{
	do {
		if (callback) callback(x, n);
	} while (next_lex_perm(x, n));
}

/* Boothroyd method; exactly N! swaps, about as fast as it gets */
void boothroyd(int *x, int n, int nn, int callback(int *, int))
{
	int c = 0, i, t;
	while (1) {
		if (n > 2) boothroyd(x, n - 1, nn, callback);
		if (c >= n - 1) return;

		i = (n & 1) ? 0 : c;
		c++;
		t = x[n - 1], x[n - 1] = x[i], x[i] = t;
		if (callback) callback(x, nn);
	}
}

/* entry for Boothroyd method */
void perm2(int *x, int n, int callback(int*, int))
{
	if (callback) callback(x, n);
	boothroyd(x, n, n, callback);
}

/* same as perm2, but flattened recursions into iterations */
void perm3(int *x, int n, int callback(int*, int))
{
	/* calloc isn't strictly necessary, int c[32] would suffice
	   for most practical purposes */
	int d, i, t, *c = calloc(n, sizeof(int));

	/* curiously, with GCC 4.6.1 -O3, removing next line makes
	   it ~25% slower */
	if (callback) callback(x, n);
	for (d = 1; ; c[d]++) {
		while (d > 1) c[--d] = 0;
		while (c[d] >= d)
			if (++d >= n) goto done;

		t = x[ i = (d & 1) ? c[d] : 0 ], x[i] = x[d], x[d] = t;
		if (callback) callback(x, n);
	}
done:	free(c);
}

#define N 4

int main()
{
	int i, x[N];
	for (i = 0; i < N; i++) x[i] = i + 1;

	/* three different methods */
	perm1(x, N, show);
	perm2(x, N, show);
	perm3(x, N, show);

	return 0;
}