How to increase the potential value of an variable in a 16 bit machine using C

Question

I have created a binary to decimal converter but even small numbers written in binary have many digits and thus are too large to be held by an integer variable on a 16 bit machine. Is there any way around this. The program is in C. Here is the code, thanks:

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

void main()
{
 clrscr();
 int a,b,d=0,x=1;
 int check(int y);
 printf("Enter your number in Binary:");
 scanf("%d",&a);
 if(check(a)==0)
 {
  printf("Number tolerable. Conversion Process Initiated.");
 }
 else 
 {
  printf("Number not binary. Try again.");
  exit(1);
 }

 while(a!=0)
 {
   if(a%10==1)
    {
     d=d+x;
    }
   a=a/10;
   x=x*2;
 }
 printf("\nDecimal:%d",d);
 getch();
}

int check(int y)
{
  while(y!=0)
  {
   if(y%10!=0&&y%10!=1)
   {
    return 1;
   }
   else
   {
    y=y/10;
   }
  }
 return 0;
}

either `char array` or some kind of packed bit array like `bitset` — user3528438, Aug 04 '15 at 12:43
@haris Would I be able to conduct mathematical operations on a string? — Sahil Baid, Aug 04 '15 at 12:46
@SahilBaid: Nope, you will have to manually iterate over the string and do the operation — Haris, Aug 04 '15 at 12:48
Note: always check the return value of `scanf`. In case the user entered something that can not be represented as an integer — muXXmit2X, Aug 04 '15 at 12:50
You know there are wider integer types? `long` and `long long`? Note: for binary operations, you should use unsigned integers. — too honest for this site, Aug 04 '15 at 13:04
@LightnessRacesinOrbit OP could be using low-cost chips like PICs or AVRs. Not all embedded platforms are 32/64 bit. — Cloud, Aug 04 '15 at 16:24
@Dogbert: There are many Turbo C cross-compilers for low-cost PIC/AVR chips, are there? — Lightness Races in Orbit, Aug 04 '15 at 16:25
@LightnessRacesinOrbit I'm not sure on TurboC compilers, but for C30 on PIC16 chips, I'm pretty sure 32-bit arithmetic is emulated in s/w, so there's a severe performance penalty. — Cloud, Aug 04 '15 at 16:59
@Dogbert: Um my point is that this question is about Turbo C, not C30 on PIC16 chips. OP should join the 21st century for this particular project, which evidently has nothing to do with PIC16 chips. — Lightness Races in Orbit, Aug 04 '15 at 17:18
@LightnessRacesinOrbit Ah, I misunderstood. I thought you meant he should just use a newer chip due to 16-bit architecture being outdated. NVM. — Cloud, Aug 04 '15 at 17:31
@Dog: No I meant he should use a C++ implementation that is not from 1824 :) — Lightness Races in Orbit, Aug 04 '15 at 17:42

score 2 · Answer 1 · answered Aug 04 '15 at 12:58

A simple way would be to store binary number into a character array.And to convert it into number see following steps-

1.Loop starting from i=n-1 to i>=0 ( array of size n).

2.Check if character at index i is 0 or 1.

Recognize 0 and 1 as follows-

3.If 0 then digit is 0.

4.If 1 then digit will be equal to 2^i (i being the index).

5.Last step would be add them.

Else use an integer array.

score 0 · Answer 2 · answered Aug 04 '15 at 12:47

The %d format specifier to scanf expects a number to be entered in decimal format. There is no binary format specifier, so you have to read in the number as a string and parse through it.

For example, if the user enters "101011", you need to recognize this as 2^5 + 2^3 + 2^1 + 2^0 and add those value to the resulting number (to be clear, I'm using ^ to denote exponentiation instead of bitwise xor).

score 0 · Answer 3 · answered Aug 04 '15 at 13:14

0

there's no real need to store the binary digits.

#include <string.h>

int n=0,c;
printf("Enter your number in Binary:");
while(strchr("01",c=getch()))
    n+=n+c-'0';

answered Aug 04 '15 at 13:14

Jasen

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How to increase the potential value of an variable in a 16 bit machine using C

3 Answers3