Search for a newline charactor or CR/LF characters

Question

I am searching a string and the string could contain a newline or a CRLF. I increment the pointer to search through each character. I have converted the hex to decimal. However, as I am looking for a single char, I am not sure this would be the best way to do this.

Many thanks for any suggestions,

   /*
      0x0a (ASCII newline)
      0x0d (ASCII carriage return)
      CRLF (0x0d0a)
    */

    while(*search != '\0') {
        /* Seach for a newline */
        if(*search == 10) {
            printf("\nnewline Found\n");
        }

        /* Search for a CRLF */
        if(*search == 3338) {
            printf("\nCRLF Found\n");
        }

        search++;
    }

I'm compiling with gcc 4.6.2 in C89-mode.

Here's the Q&A with some details on Japanese encodings https://stackoverflow.com/questions/724247/newline-control-characters-in-multi-byte-character-sets?rq=1 — Eugene Gr. Philippov, Jan 30 '19 at 17:11

score 4 · Accepted Answer · answered Jan 05 '12 at 07:53

CRLF is two characters: CR, and LF. You need to match them both individually, not together as you've done with 3338 (I think: where did you get that number?) Also, use the character constants: '\0', '\n', '\r', it will make your code clearer.

/*
  0x0a (ASCII newline)
  0x0d (ASCII carriage return)
  CRLF (0x0d0a)
*/

while(*search != '\0') {
    /* Seach for a newline */
    if(*search == '\n') {
        printf("\nnewline Found\n");
        search++;
    }

    /* Search for a CR or a CRLF */
    if(*search == '\r') {
        // OK, we found a CR, is it followed by a LF?
        if(*(search + 1) == '\n') {
            // Yes, it is, thus, it is a CRLF
            printf("\nCRLF Found\n");
            search += 2; // Note the 2! CRLF is 2 characters!
        }
        else {
            // No, just a lonely CR, forever alone.
            printf("\nCarriage return found\n");
            search++;
        }
    }
}

As another poster suggests, you can also use the strchr and strstr functions, and take the minimum valid result from them. (It'll be like 3 lines of code.)

The check `*(search + 1) != '\0'` is unnecessary in the inner `if` statement, when you're also checking if `*(search + 1) == '\n'`. — AusCBloke, Jan 05 '12 at 07:56
@AsusCBloke: You're absolutely right — edited. Being too careful with my bounds checking… — Thanatos, Jan 05 '12 at 07:58
`'\n'` isn't just clearer, it's more correct - this code will now work on EBCDIC systems! (...that also happen to use CRLF for their line endings...) — Chris Lutz, Jan 05 '12 at 07:59

score 2 · Answer 2 · answered Jan 05 '12 at 07:53

2

Look up the strchr and strstr functions. They've been made for this.

But CRLF is not one character, that's two characters. To locate that, search for '\n' and check the previous character.

answered Jan 05 '12 at 07:53

Daniel Fischer

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AusCBloke · Answer 3 · 2012-01-05T08:05:16.267

You could simply modify your code to:

for(; *search != '\0'; search++) {
    /* Seach for a newline */
    if(*search == '\n') {
        printf("\nnewline Found\n");
    }
    /* Search for a CRLF */
    else if(*search == '\r' && *(search + 1) == '\n') {
        printf("\nCRLF Found\n");
        search++; /* search will be incremented twice (bit hacky?) */
    }
}

Short-circuiting in the if statement will prevent (search + 1) being dereferenced if *search is not '\r', so that won't cause you to go past the bounds of the string if *search turns out to be '0'.

*search == 3338 in your previous code example isn't going to work if search is a char*/char[], since a single char can only represent 256 different values and CRLF is a sequence of 2 bytes (2 chars).

score 1 · Answer 4 · answered Jan 05 '12 at 08:58

1

I'd recommend to use strstr() and not to use strchr(), as the latter is not multi-byte character save (per definition).

Also the two methods strspn(), strcspn() could help.

answered Jan 05 '12 at 08:58

alk

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255

score 0 · Answer 5 · 2015-08-14T15:29:02.613

personally i'm doing it like this for the time being with the test stub i'm using ~

..mainly because I just get to use the modulo operator :D

int count(char *s) {
    char *p = s; 
    int i = 0;         
    while (*p) { 
        if (*p == '\r' || *p == '\n') { 
            i++;
        }
        p++; 
    }        
    if (i % 2 != 0) { 
            fprintf(stderr, "%s", "\n\x1b[31mgooby pls?\x1b[0m\n");
            dosomething;  
    }
    return i / 2; 
}

Search for a newline charactor or CR/LF characters

5 Answers5