Trimming whitespaces

[john_g83]

have a bit of c code that is ment to take a string (that may or may not
have spaces before or after the string) i.e. " stuff ", and trims off
the whitespace before and after.
Code:

char *trim (char *str, char ch)
{
char *first, *last;
int count;

/* Move first to the first character that isn't the same as ch */
for (first = str; *first == ch; first++);
/* Move last to the null character. Thats the only way to know 100% we
are
* removing items from the end of the string */
for (last = first; *last != '\0'; last++);
/* Ok now we backtrack until we find a character that isn't the same as
ch */
for (last--; *last == ch; last--);

if ( first != str)
{
for (count=0; count< last - first + 1; count++)
str[count] = *(first+count);
str[count] = '\0';
}
else
{
str[last-first] = '\0';
}

return str;
}

the problem is that it always removes the last letter of str as well.
i.e.
" stuff " -> "stuf" any ideas why this is happening.
Cheers
John

[pete]

jo******@hotmai l.com wrote:

have a bit of c code that is ment to take a string
(that may or may not
have spaces before or after the string) i.e. " stuff ", and trims off
the whitespace before and after.
Code:

char *trim (char *str, char ch)
{
char *first, *last;
int count;

/* Move first to the first character that isn't the same as ch */
for (first = str; *first == ch; first++);
That line could over run an array when (ch == '\0').
/* Ok now we backtrack until we find a character that
isn't the same as ch */
for (last--; *last == ch; last--);

When a string consists of a null terminated array of
characters which are all equal to ch, then what happens?

--
pete

[pete]

pete wrote:

jo******@hotmai l.com wrote:

have a bit of c code that is ment to take a string
(that may or may not
have spaces before or after the string)
i.e. " stuff ", and trims off
the whitespace before and after.
Code:

char *trim (char *str, char ch)
{
char *first, *last;
int count;

/* Move first to the first character that isn't the same as ch */
for (first = str; *first == ch; first++);

That line could over run an array when (ch == '\0').

/* Ok now we backtrack until we find a character that
isn't the same as ch */
for (last--; *last == ch; last--);

When a string consists of a null terminated array of
characters which are all equal to ch, then what happens?

#include <string.h>

char *trim(char *str, char ch)
{
char *const p = str;

while (*str != '\0' && *str == ch) {
++str;
}
memmove(p, str, 1 + strlen(str));
str = p + strlen(p);
while (str != p && *--str == ch) {
*str = '\0';
}
return p;
}

--
pete

[Rajan]

John,
Are you passing const char* as an argument to the trim function?
i.e. let's say in main() are you invoking trim(" stuff ", ' ');
If you are passing const char* like this you can't do any changes in
str[] subscript because this is a read-only section which you can't
change.
If you have to pass an argument in trim , it has to be either an array
address or allocated pointer.

[Rajan]

John,
This is a code which will work fine:-

char *trim (char *str, char ch)
{
char *first, *last;

for (first = str; *first == ch; first++);
str = first;
for (last = str; *last != ch; last++) ;
*last = '\0';
return str;
}

[Al Bowers]

Rajan wrote:

John,
This is a code which will work fine:-

char *trim (char *str, char ch)
{
char *first, *last;

for (first = str; *first == ch; first++);
str = first;
for (last = str; *last != ch; last++) ;
*last = '\0';
return str;
}

It would not work if str is a empty string, i.e. "".
Also, it will fail if all the characters in str are value ch, i.e.
char buf[32] = "aaaaaaa";
trim(buf,'a');

--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapi dsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[Ben Bacarisse]

On Thu, 23 Jun 2005 09:24:53 -0400, Al Bowers wrote:

Rajan wrote:

John,
This is a code which will work fine:-

char *trim (char *str, char ch)
{
char *first, *last;

for (first = str; *first == ch; first++);
str = first;
for (last = str; *last != ch; last++) ;
*last = '\0';
return str;
}

It would not work if str is a empty string, i.e. "".
Also, it will fail if all the characters in str are value ch, i.e.
char buf[32] = "aaaaaaa";
trim(buf,'a');

.... and it finds the first "ch" after first "non-ch" not the first of a
conscutive run of "ch" at the end of str as - the original code clearly
intended.

Because this function returns a pointer other than the one it was passed
if the storage is malloced it can't be freed with out holding onto the
original pointer somewhere else making it complicated to use in some
situations.

--
Ben.

[Al Bowers]

Ben Bacarisse wrote:

On Thu, 23 Jun 2005 09:24:53 -0400, Al Bowers wrote:

Rajan wrote:

John,
This is a code which will work fine:-

char *trim (char *str, char ch)
{
char *first, *last;

for (first = str; *first == ch; first++);
str = first;
for (last = str; *last != ch; last++) ;
*last = '\0';
return str;
}

It would not work if str is a empty string, i.e. "".
Also, it will fail if all the characters in str are value ch, i.e.
char buf[32] = "aaaaaaa";
trim(buf,'a') ;

... and it finds the first "ch" after first "non-ch" not the first of a
conscutive run of "ch" at the end of str as - the original code clearly
intended.

Because this function returns a pointer other than the one it was passed
if the storage is malloced it can't be freed with out holding onto the
original pointer somewhere else making it complicated to use in some
situations.

Somewhat similiar to the "complicate d" use of function realloc.
Example:
buf = realloc(buf, size)
intead of
char *tmp = realloc(buf,siz e)

More troublesome to me is that function trim as defined above, must
have synopsis saying to not use the function if the str is an
empty string, or if str consists entirely of characters ch.
--
Al Bowers
Tampa, Fl USA
mailto: xa******@myrapi dsys.com (remove the x to send email)
http://www.geocities.com/abowers822/

[john_bode]

jo******@hotmai l.com wrote:

[snip code]

the problem is that it always removes the last letter of str as well.
i.e.
" stuff " -> "stuf" any ideas why this is happening.
Cheers
John

Huh. I'm not getting that result based on the same test data (" stuff
"). You sure that's the code you're actually running?

[CBFalconer]

jo******@hotmai l.com wrote:

have a bit of c code that is ment to take a string (that may or
may not have spaces before or after the string) i.e. " stuff ",
and trims off the whitespace before and after.
Code:

char *trim (char *str, char ch)

Untested:

char *trim(char *s, char ch)
{
char *p;

if (s && *s && ch) { /* avoid evil cases */
while (ch == *s) s++; /* trims leading. */
p = s; /* must be advanced over entry */
while (*p) p++; /* find end of string */
p-- /* last char in string */
while ((p > s) && (ch == *p)) p--;
*p = '\0';
}
return s; /* ok in evil cases */
}

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