[求助]怎么用递归调用编x的n次方

请教大家：怎么用递归调用来编写x的n次方？谢谢！
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#include <stdlib.h>
#include <stdio.h>

void square(int x, int n, int *result);

int main(void)
{
int x, n, result = 1;

printf("Enter x and n: \n");
scanf("%d %d", &x, &n);

square(x, n, &result);
printf("result = %d\n", result);

exit(0);
}

void square(int x, int n, int *result)
{
if (n > 0) {
(*result) *= x;
square(x, --n, &(*result));
}
}





----------------解决方案--------------------------------------------------------
//这个递归函数计算x的n次方
void f(x, n) {
if (n == 1) return x;
return f(x,n-1)*x;
}
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谢谢两位.

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