how do i remove the last character by using pointer arithmetic?
I need to call a function which takes a variable as argument,
and removes its last character.

is that possible? how?

Please reply as email.

Thank you.

Maybe I misunderstood the meaning of "pointer arithmetic" but it seem to
do the job.
BTW, this function expect to receive a null terminated string.

void RemoveLastCharacter (char *thisString)
{
    int nLen;

    nLen = strlen (thisString);

    /* Replace the last character in the string by a null */
    if (nLen != 0)
        *(thisString + nLen - 1) = '\0';

Or, if you want to make the 'arithmetic' part clearer:

        *(str + strlen(str)) = '\0';

--
(initiator of the campaign for grumpiness where grumpiness is due in c.l.c)

Attempting to write in a hybrid which can be compiled by either a C compiler
or a C++ compiler produces a compromise language which combines the drawbacks
of both with the advantages of neither.

:>> how do i remove the last character by using pointer arithmetic?
:> str[strlen(str)] = '\0';
:>Or, if you want to make the 'arithmetic' part clearer:
:> *(str + strlen(str)) = '\0';

:How about:
:
:     str[strlen(str)-1] = '\0';
:
:instead.
=============
While wrong, firewind's example has the virtue of having
defined behavior when strlen(str) == 0.
=============
:
:-- glen

--
(initiator of the campaign for grumpiness where grumpiness is due in c.l.c)

*thunk*

     str[strlen(str)-1] = '\0';

instead.

-- glen

#include <string.h>

void removeLastChar(char* string)
{
        if (*string != '\0')
                strchr(string, '\0')[-1] = '\0' ;

On Thu, 28 May 1998 14:12:28 +0500, "Faisal Nasim"

[...]
: >> how do i remove the last character by using pointer arithmetic?

: >  str[strlen(str)] = '\0';

This one is outright wrong.

[...]
:      str[strlen(str)-1] = '\0';

This one is better, but still wrong because it doesn't handle zero-
length strings.  The following should work, assuming that s != NULL.

  if (*s)
  {
    s[strlen(s)-1] = 0;
  }

        char *s;                /* pointer to string, or NULL */

        /* some code skipped */

        if ( s != NULL && *s)   s[strlen(s)-1] = 0;

Now, let's try that with pointer arithmetic only (no array/pointer translation)

        char *s;                /* pointer to string, or NULL */

        /* some code skipped */

        if ( s != NULL && *s)   *(s+(strlen(s)-1)) = 0;

Lew Pitcher
System Consultant, Delivery Systems Architecture
Toronto Dominion Bank

(Opinions expressed are my own, not my employer's.)

--

I believe we can change anything.
I believe in my dream.
    - Joe Satriani

Bruce Wedding
Software Manager

Scientific MicroSystems, Inc.
http://www.scimisys.com
Tomball, Texas

        if (s && *s)                    { s[strlen(s)-1] = 0; }
        if (s != NULL && *s != 0)       { s[strlen(s)-1] = 0; }

Bet whoever posted the original question is sorry he/she asked.  :-)

Cheers,
Richard
--
Richard Stamp
Churchill College, Cambridge

if(*s) s[strlen(s)-1] = 0;

Instead.  Why take four lines for such a simple function?

-- glen

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