How can a recursive call in a for loop be optimized? - c

Question

I would like to know how a recursive function called in a loop of its own definition could be optimized like a tail call so as not to suffer from performance and stack size.
Typically, with pseudo code:
fun example(x):
if (something):
return // Stop the recursion
else:
for (/*...*/):
example() // Recursive call
For a concrete example, I would like to know how to apply such an optimization on the following program, found here:
// C program to print all permutations with duplicates allowed
#include <stdio.h>
#include <string.h>
/* Function to swap values at two pointers */
void swap(char *x, char *y)
{
char temp;
temp = *x;
*x = *y;
*y = temp;
}
/* Function to print permutations of string
This function takes three parameters:
1. String
2. Starting index of the string
3. Ending index of the string. */
void permute(char *a, int l, int r)
{
int i;
if (l == r)
printf("%s\n", a);
else
{
for (i = l; i <= r; i++)
{
swap((a+l), (a+i));
permute(a, l+1, r); // Recursive call to be optimized
swap((a+l), (a+i));
}
}
}
/* Driver program to test above functions */
int main()
{
char str[] = "ABC";
int n = strlen(str);
permute(str, 0, n-1);
return 0;
}
If the recursion becomes too deep, there is a risk of stack overflow. So how could we avoid that with this style of recursive functions (if possible, without drastically modifying the algorithm)?

Answer 1

This does not produce the exact same output, but is an iterative way of printing all permutations of a string. Adapted from cppreference.com.
void reverse(char *a, int l, int r)
{
while ((l != r) && (l != --r)) {
swap(a+(l++), a+r);
}
}
bool next_permutation(char *a, int l, int r)
{
if (l == r) return false;
int i = r;
if (l == --i) return false;
while (true) {
int i1 = i;
if (a[--i] < a[i1]) {
int i2 = r;
while (!(a[i] < a[--i2]))
;
swap(a+i, a+i2);
reverse(a, i1, r);
return true;
}
if (i == l) {
reverse(a, l, r);
return false;
}
}
}
void permute(char *a, int l, int r)
{
do {
printf("%s\n", a);
} while(next_permutation(a, l, r+1));
}
Demo