remove white space

This is a discussion on remove white space within the C Programming forums, part of the General Programming Boards category; Hello I have a char str[100] which contains a sequence of characters with a blank space at the end. I ...

  1. #1
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    remove white space

    Hello

    I have a char str[100] which contains a sequence of characters with a blank space at the end.
    I want to remove the blank space.

    EX:
    str contains: "Pokemons are great "
    I want to change it to
    "Pokemons are great"

    Notice that the string doesn't end with a white space.

    What function can do this?

    Thanks.

  2. #2
    "I Win!" by U. Lose vart's Avatar
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    start with the last char
    while(this is white space) put '\0'
    To be or not to be == true

  3. #3
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    write a loop to copy all the bytes above space symbol to the lower index, sample:
    for (i = spaceIndex; i <= strlen(str); i++)
    str[i] = str[i] + 1;
    str[strlen(str) - 2] = 0; //string end symbol

  4. #4
    Registered User C_ntua's Avatar
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    Don't really get what you mean that it doesn't end with a whitespace since you say it ends with a space.

    Anyhow, the string ends with '\0' always. So to delete a character you can put a '\0' where you want the string to end. Thus to eliminate the space at the end:
    Code:
    str[strlen(str) - 1] = '\0';

  5. #5
    "I Win!" by U. Lose vart's Avatar
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    Quote Originally Posted by baccardi View Post
    write a loop to copy all the bytes above space symbol to the lower index, sample:
    for (i = spaceIndex; i <= strlen(str); i++)
    str[i] = str[i] + 1;
    str[strlen(str) - 2] = 0; //string end symbol
    you need no copy to get rid of trailing spaces...
    To be or not to be == true

  6. #6
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    Code:
    $ cat str.c
    #include <stdio.h>
    #include <stdlib.h>
    #include <ctype.h>
    #include <string.h>
    
    int main ()
    {
     char str[100] = "Pokemons are great ";
     int len;
    
    
     len = (int) strlen(str);
    
     printf("str before [%s], len[%d]\n", str, len);
    
     len--;
     while (isspace(*(str + len) )) {
          len--;
     }
    
     *(str + len + 1) = '\0';
    
     printf("str after [%s], len[%d]\n", str, (int) strlen(str));
    
     exit (0);
    }
    
    $ gcc -Wall str.c
    $ a.out
    str before [Pokemons are great ], len[19]
    str after [Pokemons are great], len[18]

  7. #7
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    thanks

  8. #8
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    I once did that myself. Note that the version above fails for strings consisting only of whitespaces that reside at memory locations with whitespaces just before the start of a string.

    Code:
    /* Remove leading whitespaces */
    char *ltrim(char *const s)
    {
            size_t len;
            char *cur;
    
            if(s && *s) {
                    len = strlen(s);
                    cur = s;
    
                    while(*cur && isspace(*cur))
                            ++cur, --len;
    
                    if(s != cur)
                            memmove(s, cur, len + 1);
            }
    
            return s;
    }
    
    /* Remove trailing whitespaces */
    char *rtrim(char *const s)
    {
            size_t len;
            char *cur;
    
            if(s && *s) {
                    len = strlen(s);
                    cur = s + len - 1;
    
                    while(cur != s && isspace(*cur))
                            --cur, --len;
    
                    cur[isspace(*cur) ? 0 : 1] = '\0';
            }
    
            return s;
    }
    
    /* Remove leading and trailing whitespaces */
    char *trim(char *const s)
    {
            rtrim(s);
            ltrim(s);
    
            return s;
    }
    Greets,
    Philip
    All things begin as source code.
    Source code begins with an empty file.
    -- Tao Te Chip

  9. #9
    C++ Witch laserlight's Avatar
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    From what I see, rtrim() can be simplified to:
    Code:
    /* Remove trailing whitespaces */
    char *rtrim(char *const s)
    {
            if(s && *s) {
                    char *cur = s + strlen(s) - 1;
    
                    while(cur != s && isspace(*cur))
                            --cur;
    
                    cur[isspace(*cur) ? 0 : 1] = '\0';
            }
    
            return s;
    }
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  10. #10
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    Good point! Looks like I did copy and paste from ltrim().

    Thanks,
    Philip
    All things begin as source code.
    Source code begins with an empty file.
    -- Tao Te Chip

  11. #11
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Snafuist
    Looks like I did copy and paste from ltrim().
    Heheh, but a thought just crossed my mind: maybe we could use a similiar idea to simplify ltrim(), e.g.,
    Code:
    /* Remove leading whitespaces */
    char *ltrim(char *const s)
    {
            if(s && *s) {
                    char *cur = s;
    
                    while(*cur && isspace(*cur))
                            ++cur;
    
                    if(s != cur)
                            memmove(s, cur, strlen(cur) + 1);
            }
    
            return s;
    }
    There is also the possibility of replacing this:
    Code:
    cur[isspace(*cur) ? 0 : 1] = '\0';
    with:
    Code:
    cur[!isspace(*cur)] = '\0';
    but it is quite possibly more obfuscation than simplification.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  12. #12
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    Quote Originally Posted by laserlight View Post
    There is also the possibility of replacing this:
    Code:
    cur[isspace(*cur) ? 0 : 1] = '\0';
    with:
    Code:
    cur[!isspace(*cur)] = '\0';
    but it is quite possibly more obfuscation than simplification.
    that only works if isspace returns 0 or 1, and it doesn't have to because any non-zero value counts as true.

  13. #13
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by Meldreth
    that only works if isspace returns 0 or 1, and it doesn't have to because any non-zero value counts as true.
    No, it works if isspace returns zero or non-zero (which is as defined in the C standard), since the logical negation causes zero to become one and non-zero to become zero.

    One more attempt at squeezing some improvement from Snafuist's examples: I am not sure if this is a simplification, an obfuscation, or both, but it is a very micro-optimisation to replace this:
    Code:
    while(*cur && isspace(*cur))
            ++cur;
    with:
    Code:
    while(isspace(*cur) && *++cur);
    EDIT:
    Oh, and we can replace memmove() with:
    Code:
    char *dest = s;
    while ((*dest++ = *cur++));
    Last edited by laserlight; 03-10-2009 at 04:02 PM.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

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