FreeTDS Development Group

["Lowden, James K" <LowdenJK AT bernstein.com>]

> From: Ian Ashley
> Sent: Friday, July 01, 2005 12:13 PM
> 
> > for (t = istr + ilen; --t > istr && *t == ' '; )
> >   *t = '\0';
> > return t - str + 1;
> > 
> > About char/varchar is the same, server fill needed spaces if
> > necessary.
> > 
> > freddy77
> 
> Shouldn¹t it be
> 
> for (t = istr + ilen; --t >= istr && *t == ' '; )
>    *t = '\0';
> return t - str + 1;
> 
> to cope the string being all spaces. I assume the return value is
> meant to be the number of characters excluding the '\0'

No, Frediano is right.  The loop exits when --t > str is false or when *t == 
' ' is false.  

Whether or not t points to a blank, if t == str  we don't want to return a 
length of  zero.  We want to return a length of 1, else the server will 
interpret '   ' as NULL.  

If the input is 'abc ', the loop exits when *t == 'c'.  At that point, t - 
str == 2, telling the caller to trim the string to 'ab'.  Hence t - str + 1, 
or (as I think is slightly clearer) ++t - str.  

Your loop would exit when t < str, in which case t - str would be a really 
big number, to which adding 1 would (on most architectures, I think) yield 
zero.  Not that it matters in this case, but I don't know what ANSI has to 
say about pointer arithmetic overflow.  

--jkl


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