CODEWITHSUNDEEP
carraysdigit
harrisonthu
harrisonthu

Reputation: 454

Check whether the input is digit or not in C programming

I am currently reading this book: The C Programming Language - By Kernighan and Ritchie (second Edition) and one of the examples I am having trouble understanding how to check whether the input is digit or not. The example is on Page 22, explaining under the array chapter.

Below is the example.

#include <stdio.h>

 /* count digits, white space, others */

 main()
 {
   int c, i, nwhite, nother;
   int ndigit[10];

   nwhite = nother = 0;

   for (i = 0; i < 10; ++i)
   {
       ndigit[i] = 0;
   }

   while ((c = getchar()) != EOF)
   {
     if (c >= '0' && c <= '9')
     {
         ++ndigit[c-'0'];
     }
     else if (c == ' ' || c == '\n' || c == '\t')
     {
         ++nwhite;
     }
     else
     {
         ++nother;
     }

   printf("digits =");

   for (i = 0; i < 10; ++i)
   {
      printf(" %d", ndigit[i]);
   }

   printf(", white space = %d, other = %d\n",nwhite, nother);
 }

For this example, what confused me is that the author mentioned that the line ++ndigit[c-'0'] checks whether the input character in c is a digit or not. However, I believe that only the if statement ( if (c>= '0' && c<= '9') ) is necessary, and it will check if c is digit or not. Plus, I do not understand why [c-'0'] will check the input(c) is digit or not while the input variable (c) is subtracted from the string-casting ('0').

Any suggestions/explanations would be really appreciated.

Thanks in advance :)

Upvotes: 2

Views: 1674

Answers (2)

aakansha
aakansha

Reputation: 695

I would try to explain with an example

suppose the input is abc12323

So the frequency of 1=1

frequency of 2=2

frequency of 3=2

if (c >= '0' && c <= '9') //checks whether c is a  digit  
      ++ndigit[c-'0'];

Now if you do printf("%d",c) then you will get the ascii value of the

character

for c='0' the ascii value will be 48,c='1' ascii value will be 49 and it goes

till 57 for c='9'.

In your program you are keeping a frequency of the digits in the input so you need to update the index of the digit in the array every time you get it

if you do ndigit[c]++ then it will update ndigit[48] for c='0',ndigit[49] for c='1'

So either you can do ndigit[c-'0']++ as ascii value of '0'=48 in decimal

or you can simply do ndigit[c-48]++ so for c='0' ndigit[0] is updated,c=1'

ndigit[1] is updated

you can check the re factored code here http://ideone.com/nWZxL1

Hope it helps you,Happy Coding

Upvotes: 0

user3386109
user3386109

Reputation: 34829

The if statement checks whether the character is a digit, and the ++ndigit[c-'0'] statement updates the count for that digit. When c is a character between '0' and '9', then c-'0' is a number between 0 and 9. To put it another way, the ASCII value for '0' is 48 decimal, '1' is 49, '2' is 50, etc. So c-'0' is the same as c-48, and converts 48,49,50,... to 0,1,2...

One way to improve your understanding is to add a printf to the code, e.g. replace 

if (c >= '0' && c <= '9')
     ++ndigit[c-'0'];

with

if (c >= '0' && c <= '9')
{
    ++ndigit[c-'0'];
    printf( "is digit '%c'   ASCII=%d   array_index=%d\n", c, c, c-'0' );
}

Upvotes: 3

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