RE: RES: [PHP] inexplicable behaviour SOLVED

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. Cheers! Mike -- Mike Ford, Electronic Information Developer, C507, Leeds Metropolitan University, Civic Quarter Campus, Woodhouse Lane, LEEDS, LS1 3HE, United Kingdom Email: m.ford@leedsmet.ac.uk Tel: +44 113 812 4730 To view the terms under which this email is distributed, please go to http://disclaimer.leedsmet.ac.uk/email.htm