How to add two big integers in c++

C/C++: Adding two big numbers

Introduction
Adding two numbers is trivial. In C/C++, the simple mathematical expression such as sum = a + b will add variable 'a' and 'b' an put the result in the variable 'sum'. What is you want to add two number beyond the capacity of any data types in C/C++. Here in this article I have tried to implement addition of two numbers beyond any theoretical limit. Of course, here they are limited by the array size or memory capacity, but can be extended by getting the input numbers from two files and writing the result in another file. Thus the other variation of the same program can be: using character array, using linked list, using file. Here the program uses character array of numbers.

How to find the capacity of primitive data types
If you ask, how to find the capacity of the primitive data types in C/C++? Answer is simple, try writing the following program to find out the integer limits. Same technique applies in finding the capacity of other primitive data types.

#include<stdio.h>
#include<limits.h>
int main(int argc, char **argv)
{
printf("%d", INT_MAX);
printf("%d", INT_MIN);
return 0;
}

The header file “limits.h” contains all the limits defined as constants such as CHAR_MAX, CHAR_MIN, LONG_MAX, LONG_MIN etc. You may check this by typing the following Linux command in the terminal.

$ man limits.h

Also readers should note that I have used GCC compiler, Fedora Linux platform, and Netbeans IDE for programming.

Thinking in code
Logic is pretty simple: we need two character arrays s1[255], s2[255] to get the input stream of numbers as string from standard input(keyboard). For the purpose of calculation we need to convert this input string into integer, so we declare three integer array: num1[255], num2[255] andsum[255] for storing the result. We then convert the character(number) from array s1 to num1 by subtracting ASCII value 48 or '0'. For example, integer 5 in ASCII is 53. To get the integer value we do a simple math: 53 – 48 = 5. Similarly, all the values 0 through 9 can be obtained in much the same way. To make sure the resultant sum and carry a single digit we apply the following logic. For example, 9+9=18, 18 % 10= 8, a single digit sum and 18 / 10 = 1, a single digit carry.

for (; i >= 0 && j >= 0; i--, j--, k++) {
sum[k] = (num1[i] + num2[j] + carry) % 10;
carry = (num1[i] + num2[j] + carry) / 10;
}

Now the number of elements in the arrays as num1 and num2 can have three situations:

• Number of elements in num1 = number of elements in num2. For example, 99 + 11 = {1}10, last carry {1} needs to be included in the sum[k], so
  

  else { 
  if (carry > 0) 
  sum[k++] = carry; 
  }

• Number of elements in num1 > number of elements in num2. For example, 1299 + 11 = {13}10.
  Last carry {13} needs to be included in the sum[k]. To be noted which in turn may have further carry. Following code implements the logic so that carry does not get truncated.
  

  if (l1 > l2) { 
  while (i >= 0) { 
  sum[k++] = (num1[i] + carry) % 10; 
  carry = (num1[i--] + carry) / 10; 
  } 
  }

• Number of elements in num1 < number of elements in num2. For example, 99 + 1234 = {13}33.
  Last carry {13} needs to be included in the sum[k]. Here also to be noted which in turn may have further carry. Following code implements the logic so that carry does not get truncated.
  

  else if (l1 < l2) { 
  while (j >= 0) { 
  sum[k++] = (num2[j] + carry) % 10; 
  carry = (num2[j--] + carry) / 10; 
  } 
  }

And the remaining part of the program is simply printing the resultant array in reverse order

for (k--; k >= 0; k--)
printf("%d", sum[k]);

For the sake of simplicity:

• Only positive numbers are taken into account
• The claim of unlimited addition is limited by array size
• No checking is applied to validate/invalidate non numeric elements. The program may behave erratically in such a case

Putting it all together

#include<stdio.h>
int main() {
int num1[255], num2[255], sum[255];
char s1[255], s2[255];
int l1, l2;

printf("Enter Number1:");
scanf("%s", &s1);
printf("Enter Number2:");
scanf("%s", &s2);

/* convert character to integer*/

for (l1 = 0; s1[l1] != '\0'; l1++)
num1[l1] = s1[l1] - '0';

for (l2 = 0; s2[l2] != '\0'; l2++)
num2[l2] = s2[l2] - '0';

int carry = 0;
int k = 0;
int i = l1 - 1;
int j = l2 - 1;
for (; i >= 0 && j >= 0; i--, j--, k++) {
sum[k] = (num1[i] + num2[j] + carry) % 10;
carry = (num1[i] + num2[j] + carry) / 10;
}
if (l1 > l2) {

while (i >= 0) {
sum[k++] = (num1[i] + carry) % 10;
carry = (num1[i--] + carry) / 10;
}

} else if (l1 < l2) {
while (j >= 0) {
sum[k++] = (num2[j] + carry) % 10;
carry = (num2[j--] + carry) / 10;
}
} else {
if (carry > 0)
sum[k++] = carry;
}


printf("Result:");
for (k--; k >= 0; k--)
printf("%d", sum[k]);

return 0;
}