Trouble with the carry over when adding 2 arrays together - c++

Question

I'm writing a program that adds two large integers (up to 20 digits) together. I've had no problems so far in storing the two numbers as strings then sorting them into two arrays.
So far, I have half of the addition part working. When the sum of the two digits does not exceed double digits, it works fine.
The issue arises when the sum of the arrays hits double digits. I'm trying to work in the carry over, but it messes with the digits (adding in where it shouldn't.) In addition to that, I'm not sure how to get the carry to appear ahead of the final digits. For example: 9+9 outputs to 8.
Here's my code (please excuse all the letter variables in the for loops.)
#include <iostream>
#include <string>
using namespace std;
int main()
{
string str1;
string str2;
int array1[20];
int array2[20];
int array3[20];
string num3[20];
int i;
int j = 0;
int k;
int l;
int m = 0;
int n;
int o;
int carry = 0;
cout<<"Please enter the first number: "<<endl;
cin>>str1;
for (int i = str1.length() - 1; i >= 0; i--)
{
array1[j] = str1[i];
j++;
}
for (int k = str1.length()-1; k >=0; k--)
{
array1[k] = static_cast<int>(str1[k]) - static_cast<int>('0');
}
cout<<"Please enter the second number: "<<endl;
cin>>str2;
for (int l = str2.length() - 1; l >= 0; l--)
{
array2[m] = str2[l];
m++;
}
for (int n = str2.length()-1; n >=0; n--)
{
array2[n] = static_cast<int>(str2[n]) - static_cast<int>('0');
}
//Where the addition begins
for (int o = 0; o < str1.length(); o++)
{
if (array1[o] + array2[o] > 9)
{
array3[o] = array1[o] + array2[o] + carry;
array3[o] = array3[o] % 10;
carry = 1;
}
else
{
array3[o] = array1[o] + array2[o] + carry;
carry = 0;
}
cout<<array3[o];
}
return 0;
}
I think one thing I have to fix is how this line of code works:
array3[o] = array3[o] % 10;
Which keeps a second digit from appearing in the output. I would imagine if I disabled it once we reach the final numbers in the arrays, it would allow the final carry to show up. Unfortunately, everything I've tried hasn't worked.
Again, thank you!

Answer 1

Try this:
for (int o = 0; o < str1.length(); o++)
{
if (array1[o] + array2[o] + carry > 9)
{
array3[o] = array1[o] + array2[o] + carry;
array3[o] = array3[o] % 10;
carry = 1;
}
else
{
array3[o] = array1[o] + array2[o] + carry;
carry = 0;
}
cout<<array3[o];
}

Answer 2

Modify your for loop for Addition. In condition you need to add carry also
if (array1[o] + array2[o] + carry > 9)
The final for loop will be as below:
for (int o = 0; o < str1.length(); o++)
{
if (array1[o] + array2[o] + carry > 9)
{
array3[o] = array1[o] + array2[o] + carry;
array3[o] = array3[o] % 10;
carry = 1;
}
else
{
array3[o] = array1[o] + array2[o] + carry;
carry = 0;
}
cout<<array3[o];
}

Answer 3

My suggestions:
You can fill up the numbers from the input string in one loop. No need to use two loops.
for (int i = str1.length() - 1; i >= 0; i--)
{
array1[j] = str1[i] - '0';
j++;
}
Similarly for the other loop.
When computing the total you have iterate until the length of the longest string. If the first input is 12 and the second input is 4567, you have to make sure that your iteration stops at 4, not at 2.
The algorithm for computing the sum can be simplified to:
for (int o = 0; o < len+1; o++)
{
array3[o] = array1[o] + array2[o] + carry;
carry = array3[o]/10;
array3[o] %= 10;
}
where len is the maximum of the lengths.
Here's the final code I came up with:
#include <iostream>
#include <string>
using namespace std;
void printNumber(int array[])
{
// Skip the leading zeros.
int i = 19;
for ( ; i >= 0; i-- )
{
if ( array[i] > 0 )
{
break;
}
}
for ( ; i >= 0; i--)
{
cout << array[i];
}
}
int main()
{
string str1;
string str2;
int array1[20] = {0};
int array2[20] = {0};
int array3[20] = {0};
int i;
int j = 0;
int k;
int l;
int m = 0;
int n;
int o;
int carry = 0;
int len = 0;
cout<<"Please enter the first number: "<<endl;
cin>>str1;
len = str1.length();
for (int i = str1.length() - 1; i >= 0; i--)
{
array1[j] = str1[i] - '0';
j++;
}
cout<<"Please enter the second number: "<<endl;
cin>>str2;
if ( len < str2.length() )
{
len = str2.length();
}
for (int l = str2.length() - 1; l >= 0; l--)
{
array2[m] = str2[l] - '0';
m++;
}
//Where the addition begins
for (int o = 0; o < len+1; o++)
{
array3[o] = array1[o] + array2[o] + carry;
carry = array3[o]/10;
array3[o] %= 10;
}
// Print the result.
printNumber(array3);
cout << endl;
return 0;
}

Answer 4

int main()
{
char A[20],B[20],C[22]={0};
int carry,len_a,len_b,x=20,i,j,a,b;
printf("First Number");
gets(A);
printf("Second Number");
gets(B);
len_a=strlen(A);
len_b=strlen(B);
for(i=len_a-1;i>=0;i--)
{
carry=0;
b=(int)B[i]-48;
a=(int)A[len_b-1]-48;
C[x]=C[x]+a+b;
if(C[x]>9)
{
C[x]=C[x]%10;
C[x-1]+=1;
}
x--;
len_b--;
}
int flag=0;
printf("Result :");
for(j=0;j<=20;j++)
{
if(C[j]!=0)
{
printf("%d",C[j]);
flag=1;
}
else if(C[j]==0 && flag==1)
printf("%d",C[j]);
}
if(flag==0)
printf("0");
getch();
return 0;
}

Answer 5

If I were you, would do exactly what I have done here:
inline bigint &bigint::operator+( const bigint & _expr )
{
vector<uint8_t> left = this->_digits;
vector<uint8_t> right = _expr._digits;
vector<uint8_t> sum;
uint8_t carry = 0;
process_operands( left, right ); // makes the two operands have the same length and fills them with leading zeros
for( auto lit = left.cbegin(), rit = right.cbegin(); lit != left.cend(), rit != right.cend(); ++lit, ++rit )
{
uint8_t temp_sum = ( *lit + *rit + carry ) % 10;
carry = ( *lit + *rit + carry ) / 10;
sum.push_back( temp_sum );
}
if( carry ) sum.push_back( carry );
this->_digits = sum;
return *this;
}
To make things look a little bit more clear:
bigint is my class for big integers, and looks something like this:
class bigint
{
private:
vector<uint8_t> _digits;
typedef vector<uint8_t>::size_type size_type;
bigint( vector<uint8_t> & in );
public:
bigint() : _digits() {}
bigint( const string &number );
// ...
};
So you should actually stop using the built-in arrays, since they are error-prone, and because we have better things offered by STL, like std::vector. I am using std::vector<uint8_t> to store the digits of my number, and so, it becomes easier to cycle through the digits: we can use either the range for (for(uint8_t & c : _digits) { }) or the iterators.
Attaching the leading zeros will become easier, since you ony have to do:
_digits.push_back( 0 );
in a for loop.