Using char arrays to add together large numbers - c++

Question

I have an assignment in which I am supposed to
1.) Read in 2 char arrays of numbers
2.) add together the arrays, performing carry addition, carrying the tens place
3.) print out the newly added array.
I am also supposed to print an error if I need to carry on the last digit of the array (overflow)
so something like 99999999999999999999 +
1 =
________________________
ERROR
That's the part I'm having trouble with.
The above outputs something like "99999999999999999:0" so I have no idea what's going wrong.
I'll post my code, but please be nice :( I know it certainly isn't the most efficient, but I'm just trying to lay things out in a way that is easy for my brain to understand right now.
And yes, I HAVE to use char arrays. I guess it's to help us understand the ascii table.
#include <iostream>
using namespace std;
void InitNumber(char[]);
int AddArrays(char first[], char second[], char combined[]);
void OutputNumber (char[]);
const int LENGTH = 20; // global variable
int main()
{
char set1[LENGTH];
char set2[LENGTH];
char sum[LENGTH];
InitNumber (set1);
InitNumber (set2);
if(AddArrays (set1, set2, sum)) {
cout << "overflow!" << endl;
}
OutputNumber(sum);
}
void InitNumber (char list[])
{
int numberOfDigits;
cout << "Please enter the number of digits in the number: ";
cin >> numberOfDigits;
cout << "Please enter the digits in the number with the LEAST significant first: ";
for (int i = 0; i < numberOfDigits; i++) {
cin >> list [i];
}
for (int l=(numberOfDigits); l < LENGTH; l++) {
list [l] = '0';
}
}
int AddArrays(char first[], char second[], char combined[])
{
for (int h = 0; h < LENGTH; h++)
combined[h]= '0';
int overflow = 0;
for (int i = 0; i < LENGTH; i++) {
int currentSum = (first[i]-'0' + second[i]-'0');
cout << "currentSum = " << currentSum << endl;
if(currentSum / 10 == 0 )
combined[i] += currentSum;
else if (currentSum/10 !=0) {
if (i == LENGTH-1 && currentSum/10 !=0){
overflow = 1;
}
else{
combined [i] += currentSum%10;
cout << "current i: " << combined[i] << endl;
combined [i+1] += currentSum/10;
cout << "current i+1: " << combined[i+1] << endl;
}
}
}
return overflow;
}
void OutputNumber(char arrayOut[])
{
for (int l=LENGTH - 1; l >= 0; l--)
cout << arrayOut[l];
}
working input
input
6
1 2 3 4 5 6
7
1 2 3 4 5 6 7
output
00000000000008308642
Not-working output
input
20
9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9
1
1
output
999999999999999999:0

Answer 1

I'm going to reproduce small parts of your inner loop where you are carrying out the addition, in order to explain why your overflow detection is broken.
for (int i = 0; i < LENGTH; i++) {
int currentSum = (first[i]-'0' + second[i]-'0');
if(currentSum / 10 == 0 )
combined[i] += currentSum;
Here you're adding the corresponding pair of digits from the two numbers you're adding, and checking (in your particular style) if they overflowed. Try to remember this part, it's important. In order to check for overflow, you're only checking the result of adding the pair of digits from the two large numbers you're adding.
else if (currentSum/10 !=0) {
This is a completely useless check. You can only get here if the division is known to produce a non-zero result. This if() can be removed completely. Now, the relevant part of your code is
combined [i] += currentSum%10;
combined [i+1] += currentSum/10;
Do you see the problem yet?
Your approach is, once overflow is detected, is to increment the next higher order digit in the result.
Unfortunately, on the next loop iteration, in order to check for carry over, you're just checking the sum of the next corresponding digit pair, from the two large numbers you're adding. The carry-over you're saving here is going to get completely ignored.
Say your numbers are two digits long max, rather than 20, and you entered the numbers 99 and 1.
On the first iteration, you'll add 9 and 1, save 0 as the first digit, and add 1 to the second digit in the sum.
On the second iteration, you'll add 9 and 0, and, given your logic above, conclude that nothing overflowed.