I need to 2 strings (10-bit binary numbers). I's easy when i'm adding 1 0 0 0 But how I can modify this, to situation when i'll need to make 11-bit like 1010101010 1111111111=11010101001

i = MAX 1;
    while( i!=0) {
        if ((str1[i - 1] == str2[i - 1]) && (str1[i - 1] == '0' )) {
            str3[i] = '0';
        }
        else if ((str1[i - 1] != str2[i - 1])) {
            str3[i] = '1';
        }
        else if ((str1[i - 1] == str2[i - 1]) && (str1[i - 1] == '1')) {
            str3[i] = '0';
        }
        i--;
    }

MAX=10

CodePudding user response：

#define ADDCHAR(c1,c2)  (((c1) == '1')   ((c2) == '1'))

char *add(char *res, const char *n1, const char *n2)
{
    size_t ln1 = strlen(n1);
    size_t ln2 = strlen(n2);
    char *resend = res   (ln2 > ln1 ? ln2   2 : ln1   2);
    unsigned rem = 0;

    *resend-- = 0;
    while(ln1 || ln2)
    {
        unsigned pres = rem   ADDCHAR(ln1 ? n1[ln1 - 1] : '0', ln2 ? n2[ln2 - 1] : '0');
        *resend-- = '0'   (pres & 1);
        rem = pres >> 1;
        ln1 -= !!ln1;
        ln2 -= !!ln2;
    }
    if(rem) *resend-- = '0'   rem;
    while(resend >= res) *resend-- = ' ';
    return res;
}

CodePudding user response：

   int a = 0 ;
   for(int i = 0 ; i < 10  ; i  ){
   str3[i] = str1[i]   str2[i]   a ;
   if(str3[i] > 1){
   str3[i] %= 2 ;
   a = 1 ;
   }
   else{
   a = 0 ;
   }

} `

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}

CodePudding user response：

It's not clear exactly what the requirements are, but I wrote a function that takes three strings, figures out their lengths using strlen, and then sets the third string equal to the sum of the first two. Here is a program that adds the two 10-bit numbers you gave and prints the expected result of "11010101001".

#include <stdio.h>
#include <string.h>
#include <stdlib.h>

void binary_string_add(const char * a, const char * b, char * sum)
{
  size_t alen = strlen(a);
  size_t blen = strlen(b);
  size_t sumlen = strlen(sum);

  int carry = 0;
  for (size_t i = 0; i < sumlen; i  )
  {
    if (i < alen) { carry  = a[alen - 1 - i] == '1'; }
    if (i < blen) { carry  = b[blen - 1 - i] == '1'; }
    sum[sumlen - 1 - i] = '0'   (carry & 1);
    carry >>= 1;
  }
}

int main()
{
  char str1[] = "1010101010";
  char str2[] = "1111111111";
  char str3[] = "xxxxxxxxxxx";

  binary_string_add(str1, str2, str3);
  printf("Sum: %s\n", str3);
}

By the way, each iteration of the loop actually implements a 1-bit adder.

Update: If you want an easier interface that allocates an output string for you of the appropriate length and returns it, this works:

char * binary_string_add2(const char * a, const char * b)
{
  size_t alen = strlen(a);
  size_t blen = strlen(b);
  size_y sumlen = (alen > blen ? alen : blen)   1;
  char * sum = malloc(sumlen   1);
  if (sum == NULL) { /** Add error handling here. **/ }
  memset(sum, 'x', sumlen);
  sum[sumlen] = 0;
  binary_string_add(a, b, sum);
  return sum;
}