Simple if character does not equal

Question

void encode(char* dst, const char* src) {
    while (1) {
       char ch = *src;
       if (!ch || ch != '0','1','2','3') { //pseudo-code
          *dst = 0;
          return;
       }

size_t count = 1;
       while (*(++src) == ch)
          ++count;

       *(dst++) = ch;
       dst += sprintf(dst, "%zu", count);
}

How can I say, ch does not equal a number.. then return. I am trying to get rid of the infinite loop as well.

Do you know about the `isdigit()` function? – Barmar Oct 05 '17 at 03:03 — Barmar, Oct 05 '17 at 03:03
Also, what's wrong with using *`sscanf("%[0-9]")`* – Antti Haapala -- Слава Україні Oct 05 '17 at 07:34 — Antti Haapala -- Слава Україні, Oct 05 '17 at 07:34

YaatSuka · Answer 1 · 2017-10-05T03:30:33.510

2

Try this:

#include <ctype.h>

void encode(char* dst, const char* src) {
    while (1) {
       char ch = *src;
       if (!isdigit(ch)) { //Should works
          *dst = 0;
          return;
   }
   // Rest of the method
}

isdigit(int) return true if the character is an integer (as '1', '2',..., '9') or false in the other case.

(I suppose that the infinity loop is done on purpose)

edited Oct 05 '17 at 03:30

answered Oct 05 '17 at 03:08

YaatSuka

231
1
9

Perfect - thank you! – sgerbhctim Oct 05 '17 at 03:12
1

'42' is not valid – savram Oct 05 '17 at 03:12
Yeah, you're right, sorry :) – YaatSuka Oct 05 '17 at 03:13
1

The `!ch` is redundant, `isdigit` knows about that – M.M Oct 05 '17 at 03:13
@sgerbhctim can you valid the answer ? – YaatSuka Oct 05 '17 at 03:16

score 1 · Answer 2 · answered Oct 05 '17 at 03:14

1

Here is a little code i've made for you, it will parse all the src string, and tells if the string is a number or not. The probleme you may have with your while(1) is infinite loop and you're only looking for the first caractere so if I pass 2toto as argument it would say ok it's a number.

This function return true if src if a number and false if not.

#include <stdbool.h>

bool encode(char* dst, const char* src) {
  int i = 0;
  char ch;

  while (src[i] != '\0')
    {
      ch = src[i];
      if (!isdigit(ch)) { //pseudo-code                                                                                                                                                                                
        *dst = 0;
        return false;
      }
      i++;
    }
  return true;
}

answered Oct 05 '17 at 03:14

Nicolas Guerin

356
2
18

Oh le beau code :o – YaatSuka Oct 05 '17 at 03:19
1

@YaatSuka bah tu sais la piscine ca forge un homme hein.. xD – Nicolas Guerin Oct 05 '17 at 03:20
what exactly is `'\0'` – sgerbhctim Oct 05 '17 at 03:53
This the null terminated caracter @sgerbhctim it marks the end on a string. – Nicolas Guerin Oct 05 '17 at 03:54
also, how would i integrate this with my above edit – sgerbhctim Oct 05 '17 at 03:55
`while (*src || (*(++src) == ch))` means while `src[..]` is not finished (`\0`) and while `src[..]` is the same as `ch` – Nicolas Guerin Oct 05 '17 at 04:00

synchronizer · Answer 3 · 2017-10-05T03:28:13.713

isdigit works, but for completion's sake, this is how you would begin to write your own check for char ranges:

void encode(char* dst, const char* src) {
    while (1) {
        char ch = *src;
        // the char numeric value minus the 
        // numeric value of the character '0'
        // should be greater than or equal to
        // zero if the char is a digit,
        // and it should also be less than or
        // equal to 9
        if (!ch || (ch - '0' >= 0 && ch - '0' <= 9)) { 
            *dst = 0;
            return;
        }
        // ...
    }
}

As Nicolas says, you need to guarantee that you'll exit the loop though.

Simple if character does not equal

3 Answers3