Asked by Anshya.

Below are the different ways to add base 14 numbers.

**Method 1**

Thanks to Raj for suggesting this method.

1. Convert both i/p base 14 numbers to base 10. 2. Add numbers. 3. Convert the result back to base 14.

**Method 2**

Just add the numbers in base 14 in same way we add in base 10. Add numerals of both numbers one by one from right to left. If there is a carry while adding two numerals, consider the carry for adding next numerals.

Let us consider the presentation of base 14 numbers same as hexadecimal numbers

A --> 10 B --> 11 C --> 12 D --> 13

Example: num1 = 1 2 A num2 = C D 3 1. Add A and 3, we get 13(D). Since 13 is smaller than 14, carry becomes 0 and resultant numeral becomes D 2. Add 2, D and carry(0). we get 15. Since 15 is greater than 13, carry becomes 1 and resultant numeral is 15 - 14 = 1 3. Add 1, C and carry(1). we get 14. Since 14 is greater than 13, carry becomes 1 and resultant numeral is 14 - 14 = 0 Finally, there is a carry, so 1 is added as leftmost numeral and the result becomes 101D

**Implementation of Method 2**

`# include <stdio.h> ` `# include <stdlib.h> ` `# define bool int ` ` ` `int` `getNumeralValue(` `char` `); ` `char` `getNumeral(` `int` `); ` ` ` `/* Function to add two numbers in base 14 */` `char` `*sumBase14(` `char` `*num1, ` `char` `*num2) ` `{ ` ` ` `int` `l1 = ` `strlen` `(num1); ` ` ` `int` `l2 = ` `strlen` `(num2); ` ` ` `char` `*res; ` ` ` `int` `i; ` ` ` `int` `nml1, nml2, res_nml; ` ` ` `bool` `carry = 0; ` ` ` ` ` `if` `(l1 != l2) ` ` ` `{ ` ` ` `printf` `(` `"Function doesn't support numbers of different"` ` ` `" lengths. If you want to add such numbers then"` ` ` `" prefix smaller number with required no. of zeroes"` `); ` ` ` `getchar` `(); ` ` ` `assert` `(0); ` ` ` `} ` ` ` ` ` `/* Note the size of the allocated memory is one ` ` ` `more than i/p lenghts for the cases where we ` ` ` `have carry at the last like adding D1 and A1 */` ` ` `res = (` `char` `*)` `malloc` `(` `sizeof` `(` `char` `)*(l1 + 1)); ` ` ` ` ` `/* Add all numerals from right to left */` ` ` `for` `(i = l1-1; i >= 0; i--) ` ` ` `{ ` ` ` `/* Get decimal values of the numerals of ` ` ` `i/p numbers*/` ` ` `nml1 = getNumeralValue(num1[i]); ` ` ` `nml2 = getNumeralValue(num2[i]); ` ` ` ` ` `/* Add decimal values of numerals and carry */` ` ` `res_nml = carry + nml1 + nml2; ` ` ` ` ` `/* Check if we have carry for next addition ` ` ` `of numerals */` ` ` `if` `(res_nml >= 14) ` ` ` `{ ` ` ` `carry = 1; ` ` ` `res_nml -= 14; ` ` ` `} ` ` ` `else` ` ` `{ ` ` ` `carry = 0; ` ` ` `} ` ` ` `res[i+1] = getNumeral(res_nml); ` ` ` `} ` ` ` ` ` `/* if there is no carry after last iteration ` ` ` `then result should not include 0th character ` ` ` `of the resultant string */` ` ` `if` `(carry == 0) ` ` ` `return` `(res + 1); ` ` ` ` ` `/* if we have carry after last iteration then ` ` ` `result should include 0th character */` ` ` `res[0] = ` `'1'` `; ` ` ` `return` `res; ` `} ` ` ` `/* Function to get value of a numeral ` ` ` `For example it returns 10 for input 'A' ` ` ` `1 for '1', etc */` `int` `getNumeralValue(` `char` `num) ` `{ ` ` ` `if` `( num >= ` `'0'` `&& num <= ` `'9'` `) ` ` ` `return` `(num - ` `'0'` `); ` ` ` `if` `( num >= ` `'A'` `&& num <= ` `'D'` `) ` ` ` `return` `(num - ` `'A'` `+ 10); ` ` ` ` ` `/* If we reach this line caller is giving ` ` ` `invalid character so we assert and fail*/` ` ` `assert` `(0); ` `} ` ` ` `/* Function to get numeral for a value. ` ` ` `For example it returns 'A' for input 10 ` ` ` `'1' for 1, etc */` `char` `getNumeral(` `int` `val) ` `{ ` ` ` `if` `( val >= 0 && val <= 9) ` ` ` `return` `(val + ` `'0'` `); ` ` ` `if` `( val >= 10 && val <= 14) ` ` ` `return` `(val + ` `'A'` `- 10); ` ` ` ` ` `/* If we reach this line caller is giving ` ` ` `invalid no. so we assert and fail*/` ` ` `assert` `(0); ` `} ` ` ` `/*Driver program to test above functions*/` `int` `main() ` `{ ` ` ` `char` `*num1 = ` `"DC2"` `; ` ` ` `char` `*num2 = ` `"0A3"` `; ` ` ` ` ` `printf` `(` `"Result is %s"` `, sumBase14(num1, num2)); ` ` ` `getchar` `(); ` ` ` `return` `0; ` `} ` |

*chevron_right*

*filter_none*

**Notes:**

Above approach can be used to add numbers in any base. We don’t have to do string operations if base is smaller than 10.

You can try extending the above program for numbers of different lengths.

Please comment if you find any bug in the program or a better approach to do the same.

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