How do I trim leading/trailing whitespace in a sta

Question

Is there a clean, preferably standard method of trimming leading and trailing whitespace from a string in C? I'd roll my own, but I would think this is a common problem with an equally common solution.

Answer 1

I know there have many answers, but I post my answer here to see if my solution is good enough.

// Trims leading whitespace chars in left `str`, then copy at almost `n - 1` chars
// into the `out` buffer in which copying might stop when the first '\0' occurs, 
// and finally append '\0' to the position of the last non-trailing whitespace char.
// Reture the length the trimed string which '\0' is not count in like strlen().
size_t trim(char *out, size_t n, const char *str)
{
    // do nothing
    if(n == 0) return 0;    

    // ptr stop at the first non-leading space char
    while(isspace(*str)) str++;    

    if(*str == '\0') {
        out[0] = '\0';
        return 0;
    }    

    size_t i = 0;    

    // copy char to out until '\0' or i == n - 1
    for(i = 0; i < n - 1 && *str != '\0'; i++){
        out[i] = *str++;
    }    

    // deal with the trailing space
    while(isspace(out[--i]));    

    out[++i] = '\0';
    return i;
}

Answer 2

Here i use the dynamic memory allocation to trim the input string to the function trimStr. First, we find how many non-empty characters exist in the input string. Then, we allocate a character array with that size and taking care of the null terminated character. When we use this function, we need to free the memory inside of main function.

#include<stdio.h>
#include<stdlib.h>

char *trimStr(char *str){
char *tmp = str;
printf("input string %s\n",str);
int nc = 0;

while(*tmp!='\0'){
  if (*tmp != ' '){
  nc++;
 }
 tmp++;
}
printf("total nonempty characters are %d\n",nc);
char *trim = NULL;

trim = malloc(sizeof(char)*(nc+1));
if (trim == NULL) return NULL;
tmp = str;
int ne = 0;

while(*tmp!='\0'){
  if (*tmp != ' '){
     trim[ne] = *tmp;
   ne++;
 }
 tmp++;
}
trim[nc] = '\0';

printf("trimmed string is %s\n",trim);

return trim; 
 }


int main(void){

char str[] = " s ta ck ove r fl o w  ";

char *trim = trimStr(str);

if (trim != NULL )free(trim);

return 0;
}

Answer 3

Use a string library, for instance:

Ustr *s1 = USTR1(\7, " 12345 ");

ustr_sc_trim_cstr(&s1, " ");
assert(ustr_cmp_cstr_eq(s1, "12345"));

...as you say this is a "common" problem, yes you need to include a #include or so and it's not included in libc but don't go inventing your own hack job storing random pointers and size_t's that way only leads to buffer overflows.

Answer 4

A bit late to the game, but I'll throw my routines into the fray. They're probably not the most absolute efficient, but I believe they're correct and they're simple (with rtrim() pushing the complexity envelope):

#include <ctype.h>
#include <string.h>

/*
    Public domain implementations of in-place string trim functions

    Michael Burr
    [e-mail]
    2010
*/

char* ltrim(char* s) 
{
    char* newstart = s;

    while (isspace( *newstart)) {
        ++newstart;
    }

    // newstart points to first non-whitespace char (which might be '\0')
    memmove( s, newstart, strlen( newstart) + 1); // don't forget to move the '\0' terminator

    return s;
}


char* rtrim( char* s)
{
    char* end = s + strlen( s);

    // find the last non-whitespace character
    while ((end != s) && isspace( *(end-1))) {
            --end;
    }

    // at this point either (end == s) and s is either empty or all whitespace
    //      so it needs to be made empty, or
    //      end points just past the last non-whitespace character (it might point
    //      at the '\0' terminator, in which case there's no problem writing
    //      another there).    
    *end = '\0';

    return s;
}

char*  trim( char* s)
{
    return rtrim( ltrim( s));
}

Answer 5

I'm only including code because the code posted so far seems suboptimal (and I don't have the rep to comment yet.)

void inplace_trim(char* s)
{
    int start, end = strlen(s);
    for (start = 0; isspace(s[start]); ++start) {}
    if (s[start]) {
        while (end > 0 && isspace(s[end-1]))
            --end;
        memmove(s, &s[start], end - start);
    }
    s[end - start] = '\0';
}

char* copy_trim(const char* s)
{
    int start, end;
    for (start = 0; isspace(s[start]); ++start) {}
    for (end = strlen(s); end > 0 && isspace(s[end-1]); --end) {}
    return strndup(s + start, end - start);
}

strndup() is a GNU extension. If you don't have it or something equivalent, roll your own. For example:

r = strdup(s + start);
r[end-start] = '\0';

Answer 6

Personally, I'd roll my own. You can use strtok, but you need to take care with doing so (particularly if you're removing leading characters) that you know what memory is what.

Getting rid of trailing spaces is easy, and pretty safe, as you can just put a 0 in over the top of the last space, counting back from the end. Getting rid of leading spaces means moving things around. If you want to do it in place (probably sensible) you can just keep shifting everything back one character until there's no leading space. Or, to be more efficient, you could find the index of the first non-space character, and shift everything back by that number. Or, you could just use a pointer to the first non-space character (but then you need to be careful in the same way as you do with strtok).