c permutation code

[TheLionKing]

Hello, I need to have a code to create permutations giving an array of size number of elements and a range...
something like 4 permutation 2, o 5 permutation 3...
unfortunately this code has to be in c and not c++ so things like next permutation cannot be used...
this is the code I was trying to use but it doesnt do it correctly, can anyone propose a better one or correct this one ?

void permutationUtil(int arr[], int store[],int index, int range, int start, int end)
{ //store size == range+1 :: end == sizeof(arr)-1
    // Since index has ==r, current permutation is
    // ready to be printed, print
    if (index == range)
    {
        for (int i = 0; i < range; i++)
		{
            printf("%d ", store[arr[i]]);
		}
        printf("\n");
        return;
    }
 
    // One by one choose all elements (without considering
    // the fact whether element is already chosen or not)
    // and recur
    for (int i = start; i <= end; i++)
    {
        store[index] = i;
        permutationUtil(arr, store, index + 1, range, i, end);
    }
    return;
}

[AvDav]

The following code would generate whole permutation set with the help of held indices' array, obviously the count of the elements in this set should be N!.

#include <stdio.h>

bool permutation(int *ar, int n) {
	int i = n - 2, j, k;
	while (i >= 0 && ar[i] >= ar[i + 1]) --i;
	if (i >= 0) {
		j = i + 1;
		while (ar[i] < ar[j + 1]) ++j;
		ar[i] ^= ar[j] ^= ar[i] ^= ar[j];
	}
	for (j = i + 1, k = n - 1; j < k; ar[j] ^= ar[k] ^= ar[j] ^= ar[k], ++j, --k);
	return i >= 0;
}
int main() {
	int a[] = { 0, 1, 2, 3, 4 };
	int b[] = { 1, 2, 3, 4, 5 };
	const int n = sizeof(a) / sizeof(a[0]);
	do {
		for(int i = 0; i < n; ++i)
			printf("%d\ ", b[a[i]]);
		printf("%c", '\n');
	}
	while (permutation(a, n));
    return 0;
}

Have a nice day.

[superbonzo]

please, note that there's no 'bool' in C, and that the expression ( obfuscation aside ) 'ar[i] ^= ar[j] ^= ar[i] ^= ar[j]' is undefined behaviour because you're reading/modifying the same object twice between two consecutive sequence points... ( this would be true for c++ as well, at least up to c++2003; it's probably valid as for >=c++11, but I should check the wording to be sure ... anyway, there's no point using such experssions unless you're writing in crypto-C )

[AvDav]

Thanks for valuable comments, superbonzo.
The swap expression can be easily substituted with code which uses a temporary variable or just write it to another way.

[2kaud]

For another take on this, consider

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

void perm(const int* const data, int* const column, const int k, const int n, const int r)
{
	for (column[k] = -1; ++column[k] < n;)
		if (k < r - 1)
			perm(data, column, k + 1, n, r);
		else {
			for (int j = 0; j < r; ++j)
				printf("%i ", data[column[j]]);

			printf("\n");
		}
}

int main()
{
	const int data[] = { 11, 12, 13, 14, 15 };
	const int n = sizeof(data) / sizeof(data[0]);
	int *column = (int*)calloc(n, sizeof(int));

	perm(data, column, 0, n, 3);
	free (column);
}

Where this will generate and display nP3 permutations from the array data.