PHP Operators

[Russ.Dilley@gmail.com]

I have an embarrassingly easy PHP question. I have the following line
of code:

$n *= $s = count($data[$i]);

Where $data is an array and $n and $s are scalars.

I'm familiar with the '*=' operator:

$n = $n * $s

But I'm not sure what the second '=' is doing.

Thanks,
R.D.

[yongjin.jiang@gmail.com]

Re: PHP Operators

it is really a confusing expression. i think $n will be set to
count($data[$]).

Russ.Dilley@gma il.com wrote:

I have an embarrassingly easy PHP question. I have the following line
of code:
>
$n *= $s = count($data[$i]);
>
Where $data is an array and $n and $s are scalars.
>
I'm familiar with the '*=' operator:
>
$n = $n * $s
>
But I'm not sure what the second '=' is doing.
>
Thanks,
R.D.

[Tim Hunt]

Re: PHP Operators


Russ.Dilley@gma il.com wrote:

I have an embarrassingly easy PHP question. I have the following line
of code:
>
$n *= $s = count($data[$i]);
>
Where $data is an array and $n and $s are scalars.
>
I'm familiar with the '*=' operator:
>
$n = $n * $s
>
But I'm not sure what the second '=' is doing.
>
Thanks,
R.D.

=, *=, += etc are evaluated from right to left so $s = count(...) is
calculated/assigned before $n = ...

http://www.php.net/manual/en/language.operators.php#language.operators.precedence

Its easier to understand if you split the line into two:

$s = count($data[$i]);
$n *= $s;

[Armando Padilla]

Re: PHP Operators

yongjin.jiang@g mail.com wrote:

it is really a confusing expression. i think $n will be set to
count($data[$]).
>
Russ.Dilley@gma il.com wrote:
>

>>I have an embarrassingly easy PHP question. I have the following line
>>of code:
>>
>>$n *= $s = count($data[$i]);
>>
>>Where $data is an array and $n and $s are scalars.
>>
>>I'm familiar with the '*=' operator:
>>
>>$n = $n * $s
>>
>>But I'm not sure what the second '=' is doing.
>>
>>Thanks,
>>R.D.

>
>

No actually $n will be set to the product of $s*$n where $s is just the
total number of values in the array data[$i]. so if data[$i] contains
array(1, 2, 3, 5) the count will be 4. There for $s will now be 4 and
then $n*= $s is resolved.

Armando Padilla

[Mladen Gogala]

Re: PHP Operators

On Wed, 09 Aug 2006 20:11:15 -0700, Russ.Dilley wrote:

I have an embarrassingly easy PHP question. I have the following line
of code:
>
$n *= $s = count($data[$i]);
>
Where $data is an array and $n and $s are scalars.
>
I'm familiar with the '*=' operator:
>
$n = $n * $s
>
But I'm not sure what the second '=' is doing.
>

Looks fine to me:
$ php -r '$data=array(1, 2,3);$n=2; $n *= $s = count($data); print "N=$n\n";'
N=6
$


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