How do I trim leading/trailing whitespace in a standard way?

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

If you can modify the string:

// Note: This function returns a pointer to a substring of the original string.
// If the given string was allocated dynamically, the caller must not overwrite
// that pointer with the returned value, since the original pointer must be
// deallocated using the same allocator with which it was allocated.  The return
// value must NOT be deallocated using free() etc.
char *trimwhitespace(char *str)
{
  char *end;

  // Trim leading space
  while(isspace((unsigned char)*str)) str++;

  if(*str == 0)  // All spaces?
    return str;

  // Trim trailing space
  end = str + strlen(str) - 1;
  while(end > str && isspace((unsigned char)*end)) end--;

  // Write new null terminator character
  end[1] = '\0';

  return str;
}

If you can't modify the string, then you can use basically the same method:

// Stores the trimmed input string into the given output buffer, which must be
// large enough to store the result.  If it is too small, the output is
// truncated.
size_t trimwhitespace(char *out, size_t len, const char *str)
{
  if(len == 0)
    return 0;

  const char *end;
  size_t out_size;

  // Trim leading space
  while(isspace((unsigned char)*str)) str++;

  if(*str == 0)  // All spaces?
  {
    *out = 0;
    return 1;
  }

  // Trim trailing space
  end = str + strlen(str) - 1;
  while(end > str && isspace((unsigned char)*end)) end--;
  end++;

  // Set output size to minimum of trimmed string length and buffer size minus 1
  out_size = (end - str) < len-1 ? (end - str) : len-1;

  // Copy trimmed string and add null terminator
  memcpy(out, str, out_size);
  out[out_size] = 0;

  return out_size;
}

Answer 2

Here's one that shifts the string into the first position of your buffer. You might want this behavior so that if you dynamically allocated the string, you can still free it on the same pointer that trim() returns:

char *trim(char *str)
{
    size_t len = 0;
    char *frontp = str;
    char *endp = NULL;

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

    len = strlen(str);
    endp = str + len;

    /* Move the front and back pointers to address the first non-whitespace
     * characters from each end.
     */
    while( isspace((unsigned char) *frontp) ) { ++frontp; }
    if( endp != frontp )
    {
        while( isspace((unsigned char) *(--endp)) && endp != frontp ) {}
    }

    if( frontp != str && endp == frontp )
            *str = '\0';
    else if( str + len - 1 != endp )
            *(endp + 1) = '\0';

    /* Shift the string so that it starts at str so that if it's dynamically
     * allocated, we can still free it on the returned pointer.  Note the reuse
     * of endp to mean the front of the string buffer now.
     */
    endp = str;
    if( frontp != str )
    {
            while( *frontp ) { *endp++ = *frontp++; }
            *endp = '\0';
    }

    return str;
}

Test for correctness:

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

/* Paste function from above here. */

int main()
{
    /* The test prints the following:
    [nothing to trim] -> [nothing to trim]
    [    trim the front] -> [trim the front]
    [trim the back     ] -> [trim the back]
    [    trim front and back     ] -> [trim front and back]
    [ trim one char front and back ] -> [trim one char front and back]
    [ trim one char front] -> [trim one char front]
    [trim one char back ] -> [trim one char back]
    [                   ] -> []
    [ ] -> []
    [a] -> [a]
    [] -> []
    */

    char *sample_strings[] =
    {
            "nothing to trim",
            "    trim the front",
            "trim the back     ",
            "    trim front and back     ",
            " trim one char front and back ",
            " trim one char front",
            "trim one char back ",
            "                   ",
            " ",
            "a",
            "",
            NULL
    };
    char test_buffer[64];
    char comparison_buffer[64];
    size_t index, compare_pos;

    for( index = 0; sample_strings[index] != NULL; ++index )
    {
        // Fill buffer with known value to verify we do not write past the end of the string.
        memset( test_buffer, 0xCC, sizeof(test_buffer) );
        strcpy( test_buffer, sample_strings[index] );
        memcpy( comparison_buffer, test_buffer, sizeof(comparison_buffer));

        printf("[%s] -> [%s]\n", sample_strings[index],
                                 trim(test_buffer));

        for( compare_pos = strlen(comparison_buffer);
             compare_pos < sizeof(comparison_buffer);
             ++compare_pos )
        {
            if( test_buffer[compare_pos] != comparison_buffer[compare_pos] )
            {
                printf("Unexpected change to buffer @ index %u: %02x (expected %02x)\n",
                    compare_pos, (unsigned char) test_buffer[compare_pos], (unsigned char) comparison_buffer[compare_pos]);
            }
        }
    }

    return 0;
}

Source file was trim.c. Compiled with 'cc -Wall trim.c -o trim'.

Answer 3

My solution. String must be changeable. The advantage above some of the other solutions that it moves the non-space part to the beginning so you can keep using the old pointer, in case you have to free() it later.

void trim(char * s) {
    char * p = s;
    int l = strlen(p);

    while(isspace(p[l - 1])) p[--l] = 0;
    while(* p && isspace(* p)) ++p, --l;

    memmove(s, p, l + 1);
}   

This version creates a copy of the string with strndup() instead of editing it in place. strndup() requires _GNU_SOURCE, so maybe you need to make your own strndup() with malloc() and strncpy().

char * trim(char * s) {
    int l = strlen(s);

    while(isspace(s[l - 1])) --l;
    while(* s && isspace(* s)) ++s, --l;

    return strndup(s, l);
}

Answer 4

Here's my C mini library for trimming left, right, both, all, in place and separate, and trimming a set of specified characters (or white space by default).

contents of strlib.h:

#ifndef STRLIB_H_
#define STRLIB_H_ 1
enum strtrim_mode_t {
    STRLIB_MODE_ALL       = 0, 
    STRLIB_MODE_RIGHT     = 0x01, 
    STRLIB_MODE_LEFT      = 0x02, 
    STRLIB_MODE_BOTH      = 0x03
};

char *strcpytrim(char *d, // destination
                 char *s, // source
                 int mode,
                 char *delim
                 );

char *strtriml(char *d, char *s);
char *strtrimr(char *d, char *s);
char *strtrim(char *d, char *s); 
char *strkill(char *d, char *s);

char *triml(char *s);
char *trimr(char *s);
char *trim(char *s);
char *kill(char *s);
#endif

contents of strlib.c:

#include <strlib.h>

char *strcpytrim(char *d, // destination
                 char *s, // source
                 int mode,
                 char *delim
                 ) {
    char *o = d; // save orig
    char *e = 0; // end space ptr.
    char dtab[256] = {0};
    if (!s || !d) return 0;

    if (!delim) delim = " \t\n\f";
    while (*delim) 
        dtab[*delim++] = 1;

    while ( (*d = *s++) != 0 ) { 
        if (!dtab[0xFF & (unsigned int)*d]) { // Not a match char
            e = 0;       // Reset end pointer
        } else {
            if (!e) e = d;  // Found first match.

            if ( mode == STRLIB_MODE_ALL || ((mode != STRLIB_MODE_RIGHT) && (d == o)) ) 
                continue;
        }
        d++;
    }
    if (mode != STRLIB_MODE_LEFT && e) { // for everything but trim_left, delete trailing matches.
        *e = 0;
    }
    return o;
}

// perhaps these could be inlined in strlib.h
char *strtriml(char *d, char *s) { return strcpytrim(d, s, STRLIB_MODE_LEFT, 0); }
char *strtrimr(char *d, char *s) { return strcpytrim(d, s, STRLIB_MODE_RIGHT, 0); }
char *strtrim(char *d, char *s) { return strcpytrim(d, s, STRLIB_MODE_BOTH, 0); }
char *strkill(char *d, char *s) { return strcpytrim(d, s, STRLIB_MODE_ALL, 0); }

char *triml(char *s) { return strcpytrim(s, s, STRLIB_MODE_LEFT, 0); }
char *trimr(char *s) { return strcpytrim(s, s, STRLIB_MODE_RIGHT, 0); }
char *trim(char *s) { return strcpytrim(s, s, STRLIB_MODE_BOTH, 0); }
char *kill(char *s) { return strcpytrim(s, s, STRLIB_MODE_ALL, 0); }

The one main routine does it all. It trims in place if src == dst, otherwise, it works like the strcpy routines. It trims a set of characters specified in the string delim, or white space if null. It trims left, right, both, and all (like tr). There is not much to it, and it iterates over the string only once. Some folks might complain that trim right starts on the left, however, no strlen is needed which starts on the left anyway. (One way or another you have to get to the end of the string for right trims, so you might as well do the work as you go.) There may be arguments to be made about pipelining and cache sizes and such -- who knows. Since the solution works from left to right and iterates only once, it can be expanded to work on streams as well. Limitations: it does not work on unicode strings.

Answer 5

Here is my attempt at a simple, yet correct in-place trim function.

void trim(char *str)
{
    int i;
    int begin = 0;
    int end = strlen(str) - 1;

    while (isspace((unsigned char) str[begin]))
        begin++;

    while ((end >= begin) && isspace((unsigned char) str[end]))
        end--;

    // Shift all characters back to the start of the string array.
    for (i = begin; i <= end; i++)
        str[i - begin] = str[i];

    str[i - begin] = '\0'; // Null terminate string.
}

Answer 6

Late to the trim party

Features:
1. Trim the beginning quickly, as in a number of other answers.
2. After going to the end, trimming the right with only 1 test per loop. Like @jfm3, but works for an all white-space string)
3. To avoid undefined behavior when char is a signed char, cast *s to unsigned char.

Character handling "In all cases the argument is an int, the value of which shall be representable as an unsigned char or shall equal the value of the macro EOF. If the argument has any other value, the behavior is undefined." C11 §7.4 1

#include <ctype.h>

// Return a pointer to the trimmed string
char *string_trim_inplace(char *s) {
  while (isspace((unsigned char) *s)) s++;
  if (*s) {
    char *p = s;
    while (*p) p++;
    while (isspace((unsigned char) *(--p)));
    p[1] = '\0';
  }

  // If desired, shift the trimmed string

  return s;
}

---

@chqrlie commented the above does not shift the trimmed string. To do so....

// Return a pointer to the (shifted) trimmed string
char *string_trim_inplace(char *s) {
  char *original = s;
  size_t len = 0;

  while (isspace((unsigned char) *s)) {
    s++;
  } 
  if (*s) {
    char *p = s;
    while (*p) p++;
    while (isspace((unsigned char) *(--p)));
    p[1] = '\0';
    // len = (size_t) (p - s);   // older errant code
    len = (size_t) (p - s + 1);  // Thanks to @theriver
  }

  return (s == original) ? s : memmove(original, s, len + 1);
}

Answer 7

Here's a solution similar to @adam-rosenfields in-place modification routine but without needlessly resorting to strlen(). Like @jkramer, the string is left-adjusted within the buffer so you can free the same pointer. Not optimal for large strings since it does not use memmove. Includes the ++/-- operators that @jfm3 mentions. FCTX-based unit tests included.

#include <ctype.h>

void trim(char * const a)
{
    char *p = a, *q = a;
    while (isspace(*q))            ++q;
    while (*q)                     *p++ = *q++;
    *p = '\0';
    while (p > a && isspace(*--p)) *p = '\0';
}

/* See http://fctx.wildbearsoftware.com/ */
#include "fct.h"

FCT_BGN()
{
    FCT_QTEST_BGN(trim)
    {
        { char s[] = "";      trim(s); fct_chk_eq_str("",    s); } // Trivial
        { char s[] = "   ";   trim(s); fct_chk_eq_str("",    s); } // Trivial
        { char s[] = "\t";    trim(s); fct_chk_eq_str("",    s); } // Trivial
        { char s[] = "a";     trim(s); fct_chk_eq_str("a",   s); } // NOP
        { char s[] = "abc";   trim(s); fct_chk_eq_str("abc", s); } // NOP
        { char s[] = "  a";   trim(s); fct_chk_eq_str("a",   s); } // Leading
        { char s[] = "  a c"; trim(s); fct_chk_eq_str("a c", s); } // Leading
        { char s[] = "a  ";   trim(s); fct_chk_eq_str("a",   s); } // Trailing
        { char s[] = "a c  "; trim(s); fct_chk_eq_str("a c", s); } // Trailing
        { char s[] = " a ";   trim(s); fct_chk_eq_str("a",   s); } // Both
        { char s[] = " a c "; trim(s); fct_chk_eq_str("a c", s); } // Both

        // Villemoes pointed out an edge case that corrupted memory.  Thank you.
        // http://stackoverflow.com/questions/122616/#comment23332594_4505533
        {
          char s[] = "a     ";       // Buffer with whitespace before s + 2
          trim(s + 2);               // Trim "    " containing only whitespace
          fct_chk_eq_str("", s + 2); // Ensure correct result from the trim
          fct_chk_eq_str("a ", s);   // Ensure preceding buffer not mutated
        }

        // doukremt suggested I investigate this test case but
        // did not indicate the specific behavior that was objectionable.
        // http://stackoverflow.com/posts/comments/33571430
        {
          char s[] = "         foobar";  // Shifted across whitespace
          trim(s);                       // Trim
          fct_chk_eq_str("foobar", s);   // Leading string is correct

          // Here is what the algorithm produces:
          char r[16] = { 'f', 'o', 'o', 'b', 'a', 'r', '\0', ' ',                     
                         ' ', 'f', 'o', 'o', 'b', 'a', 'r', '\0'};
          fct_chk_eq_int(0, memcmp(s, r, sizeof(s)));
        }
    }
    FCT_QTEST_END();
}
FCT_END();

Answer 8

I'm not sure what you consider "painless."

C strings are pretty painful. We can find the first non-whitespace character position trivially:

while (isspace(* p)) p++;

We can find the last non-whitespace character position with two similar trivial moves:

while (* q) q++;
do { q--; } while (isspace(* q));

(I have spared you the pain of using the * and ++ operators at the same time.)

The question now is what do you do with this? The datatype at hand isn't really a big robust abstract String that is easy to think about, but instead really barely any more than an array of storage bytes. Lacking a robust data type, it is impossible to write a function that will do the same as PHperytonby's chomp function. What would such a function in C return?

Answer 9

Another one, with one line doing the real job:

#include <stdio.h>

int main()
{
   const char *target = "   haha   ";
   char buf[256];
   sscanf(target, "%s", buf); // Trimming on both sides occurs here
   printf("<%s>\n", buf);
}

Answer 10

I didn't like most of these answers because they did one or more of the following...

1. Returned a different pointer inside the original pointer's string (kind of a pain to juggle two different pointers to the same thing).
2. Made gratuitous use of things like strlen() that pre-iterate the entire string.
3. Used non-portable OS-specific lib functions.
4. Backscanned.
5. Used comparison to ' ' instead of isspace() so that TAB / CR / LF are preserved.
6. Wasted memory with large static buffers.
7. Wasted cycles with high-cost functions like sscanf/sprintf.

Here is my version:

void fnStrTrimInPlace(char *szWrite) {

    const char *szWriteOrig = szWrite;
    char       *szLastSpace = szWrite, *szRead = szWrite;
    int        bNotSpace;

    // SHIFT STRING, STARTING AT FIRST NON-SPACE CHAR, LEFTMOST
    while( *szRead != '\0' ) {

        bNotSpace = !isspace((unsigned char)(*szRead));

        if( (szWrite != szWriteOrig) || bNotSpace ) {

            *szWrite = *szRead;
            szWrite++;

            // TRACK POINTER TO LAST NON-SPACE
            if( bNotSpace )
                szLastSpace = szWrite;
        }

        szRead++;
    }

    // TERMINATE AFTER LAST NON-SPACE (OR BEGINNING IF THERE WAS NO NON-SPACE)
    *szLastSpace = '\0';
}

Answer 11

If you're using glib, then you can use g_strstrip

Answer 12

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 13

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
    michael.burr@nth-element.com
    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 14

Very late to the party...

Single-pass forward-scanning solution with no backtracking. Every character in the source string is tested exactly once twice. (So it should be faster than most of the other solutions here, especially if the source string has a lot of trailing spaces.)

This includes two solutions, one to copy and trim a source string into another destination string, and the other to trim the source string in place. Both functions use the same code.

The (modifiable) string is moved in-place, so the original pointer to it remains unchanged.

#include <stddef.h>
#include <ctype.h>

char * trim2(char *d, const char *s)
{
    // Sanity checks
    if (s == NULL  ||  d == NULL)
        return NULL;

    // Skip leading spaces        
    const unsigned char * p = (const unsigned char *)s;
    while (isspace(*p))
        p++;

    // Copy the string
    unsigned char * dst = (unsigned char *)d;   // d and s can be the same
    unsigned char * end = dst;
    while (*p != '\0')
    {
        if (!isspace(*dst++ = *p++))
            end = dst;
    }

    // Truncate trailing spaces
    *end = '\0';
    return d;
}

char * trim(char *s)
{
    return trim2(s, s);
}

Answer 15

Just to keep this growing, one more option with a modifiable string:

void trimString(char *string)
{
    size_t i = 0, j = strlen(string);
    while (j > 0 && isspace((unsigned char)string[j - 1])) string[--j] = '\0';
    while (isspace((unsigned char)string[i])) i++;
    if (i > 0) memmove(string, string + i, j - i + 1);
}

Answer 16

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 17

The easiest way to skip leading spaces in a string is, imho,

#include <stdio.h>

int main()
{
char *foo="     teststring      ";
char *bar;
sscanf(foo,"%s",bar);
printf("String is >%s<\n",bar);
    return 0;
}

Answer 18

Ok this is my take on the question. I believe it's the most concise solution that modifies the string in place (free will work) and avoids any UB. For small strings, it's probably faster than a solution involving memmove.

void stripWS_LT(char *str)
{
    char *a = str, *b = str;
    while (isspace((unsigned char)*a)) a++;
    while (*b = *a++)  b++;
    while (b > str && isspace((unsigned char)*--b)) *b = 0;
}

Answer 19

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

char *trim_space(char *in)
{
    char *out = NULL;
    int len;
    if (in) {
        len = strlen(in);
        while(len && isspace(in[len - 1])) --len;
        while(len && *in && isspace(*in)) ++in, --len;
        if (len) {
            out = strndup(in, len);
        }
    }
    return out;
}

isspace helps to trim all white spaces.

• Run a first loop to check from last byte for space character and reduce the length variable
• Run a second loop to check from first byte for space character and reduce the length variable and increment char pointer.
• Finally if length variable is more than 0, then use strndup to create new string buffer by excluding spaces.

Answer 20

This one is short and simple, uses for-loops and doesn't overwrite the string boundaries. You can replace the test with isspace() if needed.

void trim (char *s)         // trim leading and trailing spaces+tabs
{
 int i,j,k, len;

 j=k=0;
 len = strlen(s);
                    // find start of string
 for (i=0; i<len; i++) if ((s[i]!=32) && (s[i]!=9)) { j=i; break; }
                    // find end of string+1
 for (i=len-1; i>=j; i--) if ((s[i]!=32) && (s[i]!=9)) { k=i+1; break;} 

 if (k<=j) {s[0]=0; return;}        // all whitespace (j==k==0)

 len=k-j;
 for (i=0; i<len; i++) s[i] = s[j++];   // shift result to start of string
 s[i]=0;                // end the string

}//_trim

Answer 21

If, and ONLY IF there's only one contiguous block of text between whitespace, you can use a single call to strtok(3), like so:

char *trimmed = strtok(input, "\r\t\n ");

This works for strings like the following:

"   +1.123.456.7890 "
" 01-01-2020\n"
"\t2.523"

This will not work for strings that contain whitespace between blocks of non-whitespace, like " hi there ". It's probably better to avoid this approach, but now it's here in your toolbox if you need it.

Answer 22

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).

Answer 23

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 24

This is the shortest possible implementation I can think of:

static const char *WhiteSpace=" \n\r\t";
char* trim(char *t)
{
    char *e=t+(t!=NULL?strlen(t):0);               // *e initially points to end of string
    if (t==NULL) return;
    do --e; while (strchr(WhiteSpace, *e) && e>=t);  // Find last char that is not \r\n\t
    *(++e)=0;                                      // Null-terminate
    e=t+strspn (t,WhiteSpace);                           // Find first char that is not \t
    return e>t?memmove(t,e,strlen(e)+1):t;                  // memmove string contents and terminator
}

Answer 25

These functions will modify the original buffer, so if dynamically allocated, the original pointer can be freed.

#include <string.h>

void rstrip(char *string)
{
  int l;
  if (!string)
    return;
  l = strlen(string) - 1;
  while (isspace(string[l]) && l >= 0)
    string[l--] = 0;
}

void lstrip(char *string)
{
  int i, l;
  if (!string)
    return;
  l = strlen(string);
  while (isspace(string[(i = 0)]))
    while(i++ < l)
      string[i-1] = string[i];
}

void strip(char *string)
{
  lstrip(string);
  rstrip(string);
}

Answer 26

#include "stdafx.h"
#include "malloc.h"
#include "string.h"

int main(int argc, char* argv[])
{

  char *ptr = (char*)malloc(sizeof(char)*30);
  strcpy(ptr,"            Hel  lo    wo           rl   d G    eo rocks!!!    by shahil    sucks b i          g       tim           e");

  int i = 0, j = 0;

  while(ptr[j]!='\0')
  {

      if(ptr[j] == ' ' )
      {
          j++;
          ptr[i] = ptr[j];
      }
      else
      {
          i++;
          j++;
          ptr[i] = ptr[j];
      }
  }


  printf("\noutput-%s\n",ptr);
        return 0;
}

Answer 27

What do you think about using StrTrim function defined in header Shlwapi.h.? It is straight forward rather defining on your own.
Details can be found on:
http://msdn.microsoft.com/en-us/library/windows/desktop/bb773454(v=vs.85).aspx

If you have
char ausCaptain[]="GeorgeBailey ";
StrTrim(ausCaptain," ");
This will give ausCaptain as "GeorgeBailey" not "GeorgeBailey ".

Answer 28

To trim my strings from the both sides I use the oldie but the gooody ;) It can trim anything with ascii less than a space, meaning that the control chars will be trimmed also !

char *trimAll(char *strData)
{
  unsigned int L = strlen(strData);
  if(L > 0){ L--; }else{ return strData; }
  size_t S = 0, E = L;
  while((!(strData[S] > ' ') || !(strData[E] > ' ')) && (S >= 0) && (S <= L) && (E >= 0) && (E <= L))
  {
    if(strData[S] <= ' '){ S++; }
    if(strData[E] <= ' '){ E--; }
  }
  if(S == 0 && E == L){ return strData; } // Nothing to be done
  if((S >= 0) && (S <= L) && (E >= 0) && (E <= L)){
    L = E - S + 1;
    memmove(strData,&strData[S],L); strData[L] = '\0';
  }else{ strData[0] = '\0'; }
  return strData;
}

Answer 29

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 30

Here is how I do it. It trims the string in place, so no worry about deallocating a returned string or losing the pointer to an allocated string. It may not be the shortest answer possible, but it should be clear to most readers.

#include <ctype.h>
#include <string.h>
void trim_str(char *s)
{
    const size_t s_len = strlen(s);

    int i;
    for (i = 0; i < s_len; i++)
    {
        if (!isspace( (unsigned char) s[i] )) break;
    }

    if (i == s_len)
    {
        // s is an empty string or contains only space characters

        s[0] = '\0';
    }
    else
    {
        // s contains non-space characters

        const char *non_space_beginning = s + i;

        char *non_space_ending = s + s_len - 1;
        while ( isspace( (unsigned char) *non_space_ending ) ) non_space_ending--;

        size_t trimmed_s_len = non_space_ending - non_space_beginning + 1;

        if (s != non_space_beginning)
        {
            // Non-space characters exist in the beginning of s

            memmove(s, non_space_beginning, trimmed_s_len);
        }

        s[trimmed_s_len] = '\0';
    }
}

Answer 31

char* strtrim(char* const str)
{
    if (str != nullptr)
    {
        char const* begin{ str };
        while (std::isspace(*begin))
        {
            ++begin;
        }

        auto end{ begin };
        auto scout{ begin };
        while (*scout != '\0')
        {
            if (!std::isspace(*scout++))
            {
                end = scout;
            }
        }

        auto /* std::ptrdiff_t */ const length{ end - begin };
        if (begin != str)
        {
            std::memmove(str, begin, length);
        }

        str[length] = '\0';
    }

    return str;
}

Answer 32

Most of the answers so far do one of the following:

1. Backtrack at the end of the string (i.e. find the end of the string and then seek backwards until a non-space character is found,) or
2. Call strlen() first, making a second pass through the whole string.

This version makes only one pass and does not backtrack. Hence it may perform better than the others, though only if it is common to have hundreds of trailing spaces (which is not unusual when dealing with the output of a SQL query.)

static char const WHITESPACE[] = " \t\n\r";

static void get_trim_bounds(char  const *s,
                            char const **firstWord,
                            char const **trailingSpace)
{
    char const *lastWord;
    *firstWord = lastWord = s + strspn(s, WHITESPACE);
    do
    {
        *trailingSpace = lastWord + strcspn(lastWord, WHITESPACE);
        lastWord = *trailingSpace + strspn(*trailingSpace, WHITESPACE);
    }
    while (*lastWord != '\0');
}

char *copy_trim(char const *s)
{
    char const *firstWord, *trailingSpace;
    char *result;
    size_t newLength;

    get_trim_bounds(s, &firstWord, &trailingSpace);
    newLength = trailingSpace - firstWord;

    result = malloc(newLength + 1);
    memcpy(result, firstWord, newLength);
    result[newLength] = '\0';
    return result;
}

void inplace_trim(char *s)
{
    char const *firstWord, *trailingSpace;
    size_t newLength;

    get_trim_bounds(s, &firstWord, &trailingSpace);
    newLength = trailingSpace - firstWord;

    memmove(s, firstWord, newLength);
    s[newLength] = '\0';
}

Answer 33

As the other answers don't seem to mutate the string pointer directly, but rather rely on the return value, I thought I would provide this method which additionally does not use any libraries and so is appropriate for operating system style programming:

// only used for printf in main
#include <stdio.h>

// note the char ** means we can modify the address
char *trimws(char **strp) { 
  char *str;
  // check if empty string
  if(!*str)
    return;
  // go to the end of the string
  for (str = *strp; *str; str++) 
    ;
  // back up one from the null terminator
  str--; 
  // set trailing ws to null
  for (; *str == ' '; str--) 
    *str = 0;
  // increment past leading ws
  for (str = *strp; *str == ' '; str++) 
    ;
  // set new begin address of string
  *strp = str; 
}

int main(void) {
  char buf[256] = "   whitespace    ";
  // pointer must be modifiable lvalue so we make bufp
  char **bufp = &buf;
  // pass in the address
  trimws(&bufp);
  // prints : XXXwhitespaceXXX
  printf("XXX%sXXX\n", bufp); 
  return 0;
}

Answer 34

IMO, it can be done without strlen and isspace.

char *
trim (char * s, char c)
{
    unsigned o = 0;
    char * sb = s;

    for (; *s == c; s++)
        o++;

    for (; *s != '\0'; s++)
        continue;
    for (; s - o > sb && *--s == c;)
        continue;

    if (o > 0)
        memmove(sb, sb + o, s + 1 - o - sb);
    if (*s != '\0')
        *(s + 1 - o) = '\0';

    return sb;
}

Answer 35

Here is a function to do what you want. It should take care of degenerate cases where the string is all whitespace. You must pass in an output buffer and the length of the buffer, which means that you have to pass in a buffer that you allocate.

void str_trim(char *output, const char *text, int32 max_len)
{
    int32 i, j, length;
    length = strlen(text);

    if (max_len < 0) {
        max_len = length + 1;
    }

    for (i=0; i<length; i++) {
        if ( (text[i] != ' ') && (text[i] != '\t') && (text[i] != '\n') && (text[i] != '\r')) {
            break;
        }
    }

    if (i == length) {
        // handle lines that are all whitespace
        output[0] = 0;
        return;
    }

    for (j=length-1; j>=0; j--) {
        if ( (text[j] != ' ') && (text[j] != '\t') && (text[j] != '\n') && (text[j] != '\r')) {
            break;
        }
    }

    length = j + 1 - i;
    strncpy(output, text + i, length);
    output[length] = 0;
}

The if statements in the loops can probably be replaced with isspace(text[i]) or isspace(text[j]) to make the lines a little easier to read. I think that I had them set this way because there were some characters that I didn't want to test for, but it looks like I'm covering all whitespace now :-)

Answer 36

Here is what I disclosed regarding the question in Linux kernel code:

/**
 * skip_spaces - Removes leading whitespace from @s.
 * @s: The string to be stripped.
 *
 * Returns a pointer to the first non-whitespace character in @s.
 */
char *skip_spaces(const char *str)
{
    while (isspace(*str))
            ++str;
    return (char *)str;
}

/**
 * strim - Removes leading and trailing whitespace from @s.
 * @s: The string to be stripped.
 *
 * Note that the first trailing whitespace is replaced with a %NUL-terminator
 * in the given string @s. Returns a pointer to the first non-whitespace
 * character in @s.
 */
char *strim(char *s)
{
    size_t size;
    char *end;

    size = strlen(s);

    if (!size)
            return s;

    end = s + size - 1;
    while (end >= s && isspace(*end))
            end--;
    *(end + 1) = '\0';

    return skip_spaces(s);
}

It is supposed to be bug free due to the origin ;-)

Mine one piece is closer to KISS principle I guess:

/**
 * trim spaces
 **/
char * trim_inplace(char * s, int len)
{
    // trim leading
    while (len && isspace(s[0]))
    {
        s++; len--;
    }

    // trim trailing
    while (len && isspace(s[len - 1]))
    {
        s[len - 1] = 0; len--;
    }

    return s;
}

Answer 37

void trim(char* const str)
{
    char* begin = str;
    char* end = str;
    while (isspace(*begin))
    {
        ++begin;
    }
    char* s = begin;
    while (*s != '\0')
    {
        if (!isspace(*s++))
        {
            end = s;
        }
    }
    *end = '\0';
    const int dist = end - begin;
    if (begin > str && dist > 0)
    {
        memmove(str, begin, dist + 1);
    }
}

Modifies string in place, so you can still delete it.

Doesn't use fancy pants library functions (unless you consider memmove fancy).

Handles string overlap.

Trims front and back (not middle, sorry).

Fast if string is large (memmove often written in assembly).

Only moves characters if required (I find this true in most use cases because strings rarely have leading spaces and often don't have tailing spaces)

I would like to test this but I'm running late. Enjoy finding bugs... :-)

Answer 38

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

main()
{
    char sent[10]={' ',' ',' ','s','t','a','r','s',' ',' '};
    int i,j=0;
    char rec[10];

    for(i=0;i<=10;i++)
    {
        if(!isspace(sent[i]))
        {

            rec[j]=sent[i];
            j++;
        }
    }

printf("\n%s\n",rec);

}

Answer 39

C++ STL style

std::string Trimed(const std::string& s)
{
    std::string::const_iterator begin = std::find_if(s.begin(),
                                                 s.end(),
                                                 [](char ch) { return !std::isspace(ch); });

    std::string::const_iterator   end = std::find_if(s.rbegin(),
                                                 s.rend(),
                                                 [](char ch) { return !std::isspace(ch); }).base();
    return std::string(begin, end);
}

http://ideone.com/VwJaq9

Answer 40

void trim(char* string) {
    int lenght = strlen(string);
    int i=0;

    while(string[0] ==' ') {
        for(i=0; i<lenght; i++) {
            string[i] = string[i+1];
        }
        lenght--;
    }


    for(i=lenght-1; i>0; i--) {
        if(string[i] == ' ') {
            string[i] = '\0';
        } else {
            break;
        }
    }
}