This post is not talking about $b++, but I suddenly recalled a memory... once upon a time, a perl guy told me should avoid to use $a and $b as a var, because we will use this for sorting an numeric array ( i.e. $a<=>$b )... and.. that's all. =)
Smiley Connie =) ----- Original Message ----- From: "Dave Tenen" <[EMAIL PROTECTED]> To: <[EMAIL PROTECTED]> Cc: "Michael Turner" <[EMAIL PROTECTED]> Sent: Sunday, June 30, 2002 11:29 PM Subject: Re: what happens with $b = $b++? Try using a second variable to trace the values as it goes: $a = 0 $b = 1 $a = $b++ print "$a=", $a, " $b=", $b $a = 0 $b = 1 $a = ++$b print "$a=", $a, " $b=", $b -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
