{"id":25625,"date":"2017-10-15T20:02:16","date_gmt":"2017-10-15T14:32:16","guid":{"rendered":"https:\/\/www.wikitechy.com\/technology\/?p=25625"},"modified":"2017-10-15T20:02:16","modified_gmt":"2017-10-15T14:32:16","slug":"c-programming-program-add-two-numbers-base-14","status":"publish","type":"post","link":"https:\/\/www.wikitechy.com\/technology\/c-programming-program-add-two-numbers-base-14\/","title":{"rendered":"C Programming-program to add two numbers in base 14"},"content":{"rendered":"

Method 1<\/strong>
\nThanks to Raj for suggesting this method.<\/p>\n

  1. Convert both i\/p base 14 numbers to base 10.\r\n  2. Add numbers.\r\n  3. Convert the result back to base 14.\r\n<\/pre>\n
Method 2<\/strong>
\nJust 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.<\/p>\n
Let us consider the presentation of base 14 numbers same as hexadecimal numbers<\/p>\n
   A --> 10\r\n   B --> 11\r\n   C --> 12\r\n   D --> 13\r\n<\/pre>\n
Example:\r\n   num1 =       1  2  A\r\n   num2 =       C  D  3   \r\n\r\n   1. Add A and 3, we get 13(D). Since 13 is smaller than \r\n14, carry becomes 0 and resultant numeral becomes D         \r\n\r\n  2. Add 2, D and carry(0). we get 15. Since 15 is greater \r\nthan 13, carry becomes 1 and resultant numeral is 15 - 14 = 1\r\n\r\n  3. Add 1, C and carry(1). we get 14. Since 14 is greater \r\nthan 13, carry becomes 1 and resultant numeral is 14 - 14 = 0\r\n\r\nFinally, there is a carry, so 1 is added as leftmost numeral and the result becomes \r\n101D<\/pre>\n[ad type=”banner”]\n
Implementation of Method 2:<\/strong><\/p>\n
 
 C<\/span> <\/div> 
# 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;}<\/code><\/pre> <\/div>\n","protected":false},"excerpt":{"rendered":"
C Programming-program to add two numbers in base 14 – Mathematical algorithms – Just add the numbers in base 14 in same way we add in base 10. <\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[69969,69866,1,74058],"tags":[74356,74367,74347,74360,74343,74342,74371,74379,74378,70194,74357,74341,74345,74366,70223,74346,74352,74374,74361,74349,74353,74362,74337,74377,74364,74348,74358],"class_list":["post-25625","post","type-post","status-publish","format-standard","hentry","category-algorithm","category-c-programming","category-coding","category-mathematical-algorithms","tag-binary-language-translator","tag-binary-math","tag-binary-number-system-in-computer","tag-binary-number-system-in-hindi","tag-binary-numbers-definition","tag-binary-numbers-in-computer","tag-binary-numbers-table","tag-binary-subtraction-examples","tag-binary-subtraction-rules","tag-binary-system","tag-binary-system-uses-power-of","tag-computer-binary-numbers","tag-computer-number-system","tag-conversion-of-number-system","tag-dec-to-binary","tag-different-number-systems","tag-number-conversion-system","tag-number-system","tag-number-system-conversion","tag-number-system-in-computer","tag-number-system-in-maths","tag-number-system-math","tag-tonights-lucky-numbers","tag-what-is-a-binary-number","tag-what-is-binary","tag-what-is-binary-system","tag-what-is-number-system-in-maths"],"_links":{"self":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/25625","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/comments?post=25625"}],"version-history":[{"count":0,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/posts\/25625\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/media?parent=25625"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/categories?post=25625"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.wikitechy.com\/technology\/wp-json\/wp\/v2\/tags?post=25625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}