> >> $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 cl
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.
Richard Quadling wrote:
> 2009/4/27 9el :
>
>>> 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; righ
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
Thats why I used a (?) after exactly. PJ didn't have a need for the value.
;)
2009/4/27 9el :
>>
>> 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 = $Count++;
is n
>
> 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 $Coun
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
>>
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 aff
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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