Sum of large numbers in C Revision as of 16:41, 9 May 2014 by Arjun Suresh (talk | contribs) (diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

<syntaxhighlight lang="c" name="largesum">

1. include <stdio.h>
2. include <string.h>
3. include <ctype.h>

void print(char *a, int l); int read(char *a);

int main(void) { char a[1000], b[1000], c[1000], carry; //a and b hold the input numbers and c hold the output number. Each array entry is a digit int al, bl; //for storing the number of digits of the two input numbers printf("Enter the first number "); al = read(a); printf("Enter the second number "); bl = read(b); int l = al > bl? al:bl;//l stores the no.of digit of output which may become l+1 due to carry from MSB int i = l; carry = 0; while(al > 0 && bl > 0) { char val = a[--al] + b[--bl] + carry; c[i--] = val % 10; carry = val / 10; } while(al > 0)//If a has more digits than b { char val = a[--al] + carry;

               c[i--] = val % 10;
               carry =   val / 10;	

} while(bl > 0)//If b has more digits than a { char val = b[--bl] + carry;

               c[i--] = val % 10;
               carry =   val / 10;

} c[0] = carry;//Assigning the final carry printf("sum = "); print(c, l+1); return 0; }

int read(char *a) { char c; int i = 0; do { c = getchar(); } while(isspace(c)); //Reading and discarding any whitespace char typed while(isdigit(c)) { a[i++] = c - 48; //getchar returns the ASCII. So, for 1 it returns 49. Subtracting 48, we get the actual int value entered via keyboard c = getchar(); } return i; } void print(char *a, int l) { int i; if(a[0] != 0) printf("%d", a[0]); for(i = 1; i < l; i++) { printf("%d", a[i]); } printf("\n"); } </syntaxhighlight>

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