Trim leading spaces on a string (without returning a new string)

I wrote a quick function to trim trailing spaces on a string without generating a new string...

Code:

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void rtrim (char * s ) {
  int end;
  end = strlen(s) - 1;
  if (end >= 0 && s[end] == '\n')
    s[end] = '\0';
}

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Doing it on leading spaces is more complicated. I particularly don't want something that generates a new string. Something I can call like this:

Code:

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ltrim(somestring);

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I can't be the first person to do this - rather than invent the wheel, does someone have code or a pointer to some code?

I'm thinking something that uses memmove() would be the solution...? Or perhaps there's a different technique...?

Using memmove is one way.

Or your own copy, starting from the first non-space char, to the beginning of the string.

Code:

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strcpy(testString,"              now I am learning programming");

char * tmp = testString;
int i;

while (*tmp == ' ') {
        *tmp++;
}
i = 0;
while (*tmp != NULL) {
        testString[i++] = *tmp++;
}
testString[i] = '\0';

printf("Adjusted: %s\n",testString);

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Beginner Computer Programming
teaching programming to beginners using C

john runchey, you might want to remove that dereference of tmp in the first loop, and then you should be comparing *tmp with '\0', not NULL, even though they evaluate to the same integer value.

EDIT:
monk64, there is a slight problem with your rtrim function though. strlen() returns a size_t, which is an unsigned integer type. This means that strlen(s) - 1 on an empty string would not produce a negative value. In fact, it would result in a value that probably is not in the range of an int. Furthermore, it seems that all you are doing is removing a trailing newline character, if it exists, rather than removing trailing spaces.

Good point - too early in the morning for coding :)

Quote:

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Originally Posted by john runchey

Code:

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while (*tmp == ' ') {
        tmp++;
}

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Beginner Computer Programming
teaching programming to beginners using C

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Someone mentioned something about not reinventing the wheel...

sscanf with %s will read a string within a string containing leading/trailing whitespace

Using scansets you can trim other leading/trailing chars like delimeters

Code:

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char *rtrim(char *s)
{
  while( *s && s[strlen(s)-1]==' ' )
    s[strlen(s)-1]=0;
  return s;
}

char *ltrim(char *s)
{
  while( *s==' ' )
    memmove(s,s+1,strlen(s));
  return s;
}

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Code:

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sscanf( str, "%[^ ]", str1 );

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This could be your alternative.

ssharish

Quote:

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Originally Posted by BillyTKid

Code:

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char *rtrim(char *s)
{
  while( *s && s[strlen(s)-1]==' ' )
    s[strlen(s)-1]=0;
  return s;
}

char *ltrim(char *s)
{
  while( *s==' ' )
    memmove(s,s+1,strlen(s));
  return s;
}

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The ltrim function looks good, but why are you moving the whole string multiple times?

Code:

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void ltrim(char *s)
  { char *t = s;
    while( *t ==' ')
        t++;
    memmove(s,t,strlen(t)+1); }

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The same with the rtrim function you only need 1 null terminator.

Code:

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void rtrim(char *s)
  { int n;
    for (n=strlen(s); n>0; n--)
        if (s[n] != ' ')
          { s[n+1]='\0';
            return; } }

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Code:

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void rtrim (char * s ) {
  int end;
  end = strlen(s) - 1;
  while(end >= 0 && s[end] == ' ')
  {
      end--;
  }
  if(end>=0 && s[end]!=' ')
  {
    s[end] = '\0';
  }
}

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@ jesius, in case if you didn't read post#6.

Quote:

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monk64, there is a slight problem with your rtrim function though. strlen() returns a size_t, which is an unsigned integer type. This means that strlen(s) - 1 on an empty string would not produce a negative value. In fact, it would result in a value that probably is not in the range of an int. Furthermore, it seems that all you are doing is removing a trailing newline character, if it exists, rather than removing trailing spaces.

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