Question

can you convert this cod from C++ to C language

#include<bits/stdc++.h>
using namespace std;
// Function for finding sum of two large numbers
string addThese(string a,string b){
  
// Before proceeding further, make sure length
// of a is larger.
if(a.size() < b.size())
swap(a,b);

int j = a.size() - 1;
// Traverse from end of both strings and add (b[i] - '0') to a[j].
for(int i = b.size() - 1; i >= 0; i--, j--)
//(b[i] - '0') is nothing but numerical value of string b char
a[j] += (b[i] - '0'); // if(b[i] ascii value is 49 then we subtract ascii value of '0'== 48 from it.
  
// be store all the values to a string now check ascii value of a[i]
for(int i = a.size()-1; i > 0; i--) {
if (a[i] > '9') { // if a[i] is greater than ascii value of '9' means value must be greater than 9
int d = a[i] - '0'; // calculate value of perticular digit by subtracting ascii value of a[i]-'9'
a[i-1] = ((a[i-1] - '0') + d/10) + '0'; // carry will be added to previous char values
a[i] = (d%10) + '0'; //current a[i] chang to (d%10) + '0'. if d is 11 than a[i] is 1 and carray be 1 added tp previous char.
}
}
  
if(a[0] > '9') // if first char ascii value is grater than ascii value of '9' then ther is a another char has to be added
{
string c; // carry of the first digits
c = a[0];// first store c as a[0]
a[0] = ((a[0]-'0')%10)+'0'; // now change a[0] as modulo 10 of the value
c[0] = ((c[0]-'0')/10)+'0'; // and chnge c to carry of a[0]
a = c + a; // now concatinate string c with string a
}
return a;
  
}

int main(){
   ifstream in ("input.in");
if (in.is_open())
{
string a,b;
int count=0;
while(in>>a>>b)   
cout <<"sample "<<++count<<" output: "<< addThese(a,b) <<"\n";
in.close();
}
else cout << "Unable to open file";

  
return 0;
}

Answer 1

Solution:

C code:

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

char* swap(char *a, char *b)
{
   char temp[100];
    strcpy(temp, a);
   strcpy(a, b);
    strcpy(b, temp);

}
// Function for finding sum of two large numbers
char* addThese( char a[], char b[]){

   // Before proceeding further, make sure length
   // of a is larger.
   if(strlen(a) < strlen(b))
       swap(a,b);

   int j = strlen(a) - 1;
   // Traverse from end of both , char *s and add (b[i] - '0') to a[j].
   for(int i = strlen(b) - 1; i >= 0; i--, j--)
       //(b[i] - '0') is nothing but numerical value of , char * b char
       a[j] += (b[i] - '0'); // if(b[i] ascii value is 49 then we subtract ascii value of '0'== 48 from it.

   // be store all the values to a , char * now check ascii value of a[i]
   for(int i = strlen(a)-1; i > 0; i--) {
       if (a[i] > '9') { // if a[i] is greater than ascii value of '9' means value must be greater than 9
           int d = a[i] - '0'; // calculate value of perticular digit by subtracting ascii value of a[i]-'9'
           a[i-1] = ((a[i-1] - '0') + d/10) + '0'; // carry will be added to previous char values
           a[i] = (d%10) + '0'; //current a[i] chang to (d%10) + '0'. if d is 11 than a[i] is 1 and carray be 1 added tp previous char.
       }
   }

   if(a[0] > '9') // if first char ascii value is grater than ascii value of '9' then ther is a another char has to be added
   {
       char c[100]; // carry of the first digits
       strcpy(c, &a[0]);// first store c as a[0]
       a[0] = ((a[0]-'0')%10)+'0'; // now change a[0] as modulo 10 of the value
       c[0] = ((c[0]-'0')/10)+'0'; // and chnge c to carry of a[0]
       strcat(a, c); // now concatinate , char * c with string a
   }
   return a;

}

int main(){
  
   FILE *fd = fopen("input.in", "rw");
   if (fd != NULL)
   {
       char a[100],b[100];
       int count=0;

       //while(fd>>a>>b)
       while(EOF)
           printf("sample %d, output: %s\n",++count, addThese(a,b));
       fclose(fd);
   }
   else {
       printf("Unable to open file");
       return 1;
   }

   return 0;
}

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Code screenshots:

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Output:

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