Need to add elements of two array together - c++

Question

Adding two integer array elements
if array1 = {0,0,0,0,9,9,9,9}—————> 00009999
and
array2 = {0,0,0,0,0,0,0,1}————————> 00000001
adding the two arrays together should result in 10000 being in array1, since 9999 + 1 = 10000
therefore, the result should be
array1 = {0,0,0,1,0,0,0,0}
Does anyone know how to write a code for this? I was trying to use a while loop which didn't work effectively within another for loop. Fairly new to coding and stack overflow, any help will be appreciated!
CODE THAT I TRIED
Note: both arrays will have the same number of elements, they were initialized with the same size
int length = sizeof(array1)/sizeof(array1[0]);
for(int i = length; i>0; i--){
if(array1[i-1] + array2[i-1] < 10){
array1[i-1] += array2[i-1];
}else{
array1[i-1] = array1[i-1] + array2[i-1] - 10;
if(array1[i-2]!=9){
array1[i-2]++;
} else{
int j = i-2;
while(j == 9){
array1[j] = 0;
j--;
}
array1[j-1]++;
}
}
}

Answer 1

Code below performs base10 arithmetic: you need to iterate the arrays in reverse, do the addition of i-th digit by modulo 10, then carry over any excess digits to the next array element:
#include <iostream>
#include <iterator>
using namespace std;
int main()
{
int a1[] = { 0,0,0,0,9,9,9,9 };
int a2[] = { 0,0,0,0,0,0,0,1 };
const int b = 10;
int s = 0;
for (int i = size(a1) - 1; i >= 0; --i)
{
a1[i] += a2[i] + s;
s = a1[i] / b;
a1[i] %= b;
}
std::copy(a1, a1 + size(a1), ostream_iterator<int>(cout, " "));
cout << endl;
}
Alternative with C arrays and transform algorithm + make_reverse_iterator looks too heavy. Variant with std::arrays looks better:
#include <algorithm>
#include <array>
#include <iterator>
#include <iostream>
using namespace std;
int main()
{
std::array<int, 8> a1 = { 0,0,0,0,9,9,9,9 };
std::array<int, 8> a2 = { 0,0,0,0,0,0,0,1 };
const int b = 10;
int s = 0;
transform(a1.rbegin(), a1.rend(), a2.rbegin(), a1.rbegin(), [b, &s](int &i, int &j)
{
i += j + s;
s = i / b;
return i % b;
});
copy(a1.begin(), a1.end(), ostream_iterator<int>(cout, " "));
cout << endl;
}

Answer 2

It looks like you've overcomplicated the problem a bit. Your task is to perform base 10 addition on two arrays, carrying excess digits. You can simply iterate both arrays in reverse, perform addition on the individual elements, and ensure you carry over a digit when you exceed 10.
Here is an example based on the requirements you've described. You can further abstract this as needed.
I updated this code such that the result is now in array1.
#include <iostream>
int main(int argc, char *argv[], char *argp[]){
/* Initialize arrays and calculate array size */
int array1[] = {0,0,0,0,9,9,9,9};
int array2[] = {0,0,0,0,0,0,0,1};
int array_size = sizeof(array1) / sizeof(int);
/* Iterate the arrays in reverse */
int carry = 0;
for (int i = array_size - 1; i >= 0; i--) {
/* Perform addition on the current elements, remembering to include the carry digit */
int result = array1[i] + array2[i] + carry;
/* Determine if the addition should result in another carry */
if (result >= 10) {
carry = 1;
} else {
carry = 0;
}
/* Store the non-carried addition result */
array1[i] = result % 10;
}
/* View result */
for (int i = 0; i < array_size; i++) {
std::cout << array1[i];
}
std::cout << std::endl;
return 0;
}

Answer 3

uint32_t decode(int array[8]){
uint32_t multiple_of_ten=10000000;
uint32_t result = 0;
for(int i=0;i<8;i++){
result += (array[i] * multiple_of_ten);
multiple_of_ten = multiple_of_ten / 10;
}
return result;
}
int* encode(uint32_t number){
int result[8]={0,0,0,0,0,0,0,0};
uint32_t multiple_of_ten=10000000;
uint32_t recent = number;
for(int i=0;i<8;i++){
if(recent>0 && recent / multiple_of_ten > 0)
{
result[i] = recent / multiple_of_ten;
recent %= multiple_of_ten;
}
multiple_of_ten /= 10;
}
return result;
}
in your case , use encode(decode(array1) + decode(array2))