Array to int and Sum of them in C - arrays

Question

im learning C and i am a bit confused with array handling.
I have a task that asks me to sum up 2 arrays of non negative integers.
example array 1 = {1,2,3} , array 2 = {4,5,6} -> 123+456 = 579
I searched a bit for a solution on how to convert those arrays of integers to an integer, but didnt really get helpful information.
I ended up with a code:
#include <stdio.h>
int sum(int A[],int B[], int n){
int i,j,t,k;
for(i=0;i<n;i++){
t= t+A[i];
}
for(j=0;j<n;j++){
k= k+B[j];
}
return t+k;
}
int main()
{
int n = 3;
int a[n] = {1,2,3};
int b[n] = {4,5,6};
printf("%d",sum(a,b,n));
return 0;
}
But my result is 1225283 which of course is wrong.
I found a solution where people write something like "t= 10* t+A[i]" , i dont get where that "10* " comes from, but i tested it and then "t" gets "123" but if i try the same for "k" it doesnt work, returning "k" doesnt give me "456". I am a bit confused, whats the proper way of handling this problem?
Thanks for any help.

Answer 1

You're basically adding digits 1+2+3 instead of creating the number 123. Your code also has various other flaws, like uninitialized variables. Here is a working example:
int array2int(int A[], int n) {
int ret = 0;
for(int i=0, k=1; i<n; i++){
ret = ret + k * A[i];
k *= 10;
}
return ret;
}
int sum(int A[],int B[], int n){
return array2int(A, n) + array2int(B, n);
}

Answer 2

First of all, in sum function, you haven't initialized neighter t nor k but you keep summing them and use later, so every time your code is executed, you chould get different result.
On the other hand, in something like "t= 10 t+A[i]", 10 comes from basic math, where a number could be resolved as a10^0 + b10^1 +c*10^2 + .... + m * 10^n. As a result, starting from least significant digit, everytime you try to add new digit (from least to most significant), you need your multipliciant to be 10 times greater.
int sum(int A[],int B[], int n){
int i,j,t=0,k=0,ten=1;
for(i=n-1;i>=0;i--){
t += ten*A[i];
ten *= 10;
}
ten = 1; /* initialize again*/
for(j=n-1;j>=0;j--){
k += ten*B[j];
ten *= 10;
}
return t+k;
}
Something like that should work.