ASCII Number

Er.....i want to detect if a character is a number between 0 and 9

i achieve this using ascii rite? er......how do i specify this in the if statement?

Code:

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#include <stdio.h>

int main() {
    char test1 = 'a';
    char test2 = '2';
   
    function(&test1);
    function(&test2);

    return 0;
}

int function(char *test) {
    if(*test < '0' || *test > '9') {
        printf("not a character between 0-9\n");
        return 1;
    }

    printf("is a character between 0-9\n");
    return 0;
}

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The standard library function isdigit will return true if the character is a digit.

Code:

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#include <stdio.h>
#include <ctype.h>

int main(void)
{
  char *ptr, number[] = "*_42-Aa0";
  for ( ptr = number; *ptr != '\0'; ++ptr )
  {
      if ( isdigit(*ptr) )
      {
        printf("'%c' is a digit\n", *ptr);
      }
      else
      {
        printf("'%c' is not a digit\n", *ptr);
      }
  }
  return 0;
}

/* my output
'*' is not a digit
'_' is not a digit
'4' is a digit
'2' is a digit
'-' is not a digit
'A' is not a digit
'a' is not a digit
'0' is a digit
*/

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Quote:

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Originally posted by fkheng
Er.....i want to detect if a character is a number between 0 and 9

i achieve this using ascii rite? er......how do i specify this in the if statement?

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Well, since you are talking about characters, and not numbers- you could just say:

Code:

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void  foo(void)
  {
  char    myLetter;

  myLetter = 'j';              /* stuff a non-digit value in */
  if((myLetter <= '9') && (myLetter >= '0'))
      Success();
  else
      Failure();
  }

---

d

i see, thanx, many alternatives to the code...

about the isdigit, i recall using it, but somehow it could not detect number characters, i really do not know y...anyway, thanx ppl, the codes work!