Ford, Mike wrote: > On 27 April 2009 14:21, PJ advised: > > >> Ford, Mike wrote: >> >>> On 26 April 2009 22:59, PJ advised: >>> >>> >>> >>>> kranthi wrote: >>>> >>>> >>>>> if $Count1 is never referenced after this, then certainly this >>>>> assignment operation is redundent. but assignment is not the ONLY >>>>> operation of this statement. if u hav not noticed a post increment >>>>> operator has been used which will affect the value of $Count as >>>>> > well, > >>>>> and this operation is required for the script to work. >>>>> >>>>> the script should work even if u replace >>>>> $Count1 = $Count++; >>>>> with >>>>> $Count++; >>>>> >>>>> Kranthi. >>>>> >>>>> >>>>> >>>>> >>>> Not quite, since that would change the $Count variable and that is >>>> > used > >>>> leater in the code. >>>> >>>> >>> Um -- I must be missing something here, because those two statements >>> have exactly the same effect on $Count, incrementing it by one (and >>> > you > >>> said the first statement fixed your problem, so logically the second >>> > one > >>> must too). >>> >>> In fact, because you've used a post-increment, the statement >>> >>> $Count1 = $Count++; >>> >>> ends up with $Count == $Count1+1, and $Count1 being the original >>> > value > >>> of $Count!! >>> >>> This whole scenario smacks to me of a classic "off-by-one" error -- >>> either $Count actually *needs* to be one greater than the value you >>> first thought of, or some other value you are comparing it to should >>> > be > >>> one smaller than it actually is. >>> >>> Cheers! >>> >>> >>> >> Thanks for the clarification, Mike. In my ignorance, I was under the >> impression that the right side of the equation was only for the use of >> the left part. How stupid of me. So what I should have been doing was >> $Count1 = $Count + 1; right? >> > > No -- because, as you stated, $Count1 was not referred to anywhere else, > so there was and would have been no usefulness in assigning any value to > it. You could equally well have said $Count1 = 2.71828 + 3.14159 * > $Count++; and got the same result -- because this still increments > $Count by 1, and anything else the statement does is irrelevant if > $Count1 is never referred to anywhere else!!! ;) ;) > > What you did worked because $Count++ *always* adds one to the value of > $Count, no matter where it appears -- the crucial part of your original > statement was the $Count++ bit, with the assignment to $Count1 being a > massive red herring. > > For ultimate clarification, the following 4 statements all have > identical effect: > > $Count = $Count + 1; > $Count += 1; > $Count++; > ++$Count; > > and any one of them would have had the same effect on your result. This > is why I say it was an off-by-one error -- your programming logic > somehow *needed* $Count to be 1 greater than you thought it did, so the > inadvertent incrementation of it was doing the trick for you. > > Hope this helps increase the illumination in your head a tiny bit more. > Indeed, it does. Thanks. -- Hervé Kempf: "Pour sauver la planète, sortez du capitalisme." ------------------------------------------------------------- Phil Jourdan --- pj@xxxxxxxxxxxxx http://www.ptahhotep.com http://www.chiccantine.com/andypantry.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
