一个相同长度的数组，对该数组进行求和

#include <stdio.h>
#include <math.h>

int length(int arr[], int len) {
	int i;
	for(i = len-1; i >= 0; i--) {
		if(arr[i]) {
			return i + 1;
		}
	}
	return len;
}


void main() {
	/*
		加法运算
	*/
	int nums1[10] = {2, 2, 2};
	int nums2[10] = {9, 8, 7};
	int carry = 0;
	int i, n, count = 1, result = 0;
	int len = sizeof nums1 / sizeof nums1[0];

 	for(i = length(nums1, len) - 1; i >= 0; i--) {
		n = nums1[i] + nums2[i] + carry;
		carry = floor( n/10 );
		
		if(count == 1) {
			result = result + n % 10;
			count *= 10;
			continue;
		}

		result = result + n % 10 * count;
		count *= 10;
	}

	if( carry ) result += count;

	printf("%d\n", result);
}