title: How to create a program that be able to get the two numbers with a maximum of 1000 digits from the user and then print the result?
body: <p>I'm trying to create a program that be able to get the two numbers with a maximum of 1000 digits from the user and then print the result (with C (programming language))</p>
  
  <p>notes to create this program : </p>
  
  <p>1-to get the two numbers, use strings with a length of 1000 characters.(consider a character for the zero in the end of the sring)</p>
  
  <p>2-to store the numbers , use two array with a length of 1000 block (home).each digit of a large numbers must be store in a separate array block.(for example: 365 ( "3" store in a block(home) and 6 store in another block and ...)</p>
  
  <p>3-To get the numbers as strings and convert it to a numeric array, addition, subtraction and printing large numbers, use separate function. the "main" function must be very small.</p>
  
  <hr>
  
  <p>I wrote this code but can't figure out why it wont work:</p>
  
  <pre><code>/*codes*/
  
  
  #include &lt;conio.h&gt;
  #include &lt;stdio.h&gt;
  #define hang 1
  
  void sinascan1(char sina[2],int i){
       puts("Enter number 1");
       gets(sina);
      while (i != hang){
            sina[i] -= 48;
            printf ("%d\n", sina[i]);
            i++;
            }
       }
  void sinascan2(char sina1[2],int i){
       puts("Enter number 2");
       gets(sina1);
      while (i != hang){
            sina1[i] -= 48;
            printf ("%d\n", sina1[i]);
            i++;
            }
       }
  void sinajam(char sina[2],char sina1[2],char jam[3],int i){
       puts("____");
       int f,g;
       while (i != hang){ 
             jam[i]=sina[i]+sina1[i];
             printf ("%d\n", jam[i]);
             i++;
             }
       }
  int main()
  {   int i=0;
      char sina[2],sina1[2];
      char jam[2];
      sinascan1(sina,i);
      sinascan2(sina1,i);
      sinajam(sina,sina1,jam,i);
      getch();
      return 0;
  }
  </code></pre>
  
  <hr>
  
  <p>how can I do that?</p>