FreeBSD Port: qmail-1.03_3

[Erik Trulsson]

On Tue, Jan 25, 2005 at 09:34:58PM -0800, Michael Sierchio wrote:
> Joe Marcus Clarke wrote:
> 
> >>-     while ((k > i) && (cmds.s[k - 1] == ' ') || (cmds.s[k - 1] == '\t'))
> >>+     while ((k > i) && ((cmds.s[k - 1] == ' ') || (cmds.s[k - 1] == 
> >>'\t')))
> 
> 
> >Actually, if k is less than or equal to i and cmds.s[k-1] is a tab, then
> >the first bit of code evaluates to true while the second evaluates to
> >false.  I haven't looked at the rest of the code, but those statements
> >are semantically different, and I think the second is more correct.
> 
> You're simply mistaken.
> 
> ( 1 && 0 || 1 ) will ALWAYS evaluate to 1 on any ANSI C compiler.

No, it will evaluate to 0.  && has higher precedence than ||.

(x && y || z) is equivalent to ((x && y) || z) which is different from
(x && (y || z)).


-- 
<Insert your favourite quote here.>
Erik Trulsson
ertr1013 at student.uu.se