2009/4/27 9el le...@phpxperts.net:
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?
$Count1 =
Thats why I used a (?) after exactly. PJ didn't have a need for the value.
;)
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
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
$Count = $Count + 1; is *exactly(?)* same as $Count++; Â or ++$Count
But not exactly same. Â PostFix notation adds the value after assigning
.
PreFix notation adds the value right away.
But optimized programming argues about how machine is coded nowadays.
Thanks Mike for clarifying this
Richard Quadling wrote:
2009/4/27 9el le...@phpxperts.net:
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 +
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
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
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?
$Count = $Count + 1; is exactly(?) same as $Count++;
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
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