Re: Tidying code for PHP5.0.5/PHP4.4.0

Michael Sims wrote:

Jochem Maas wrote:

foo($a = 5);

by definition the expression is evaluated _before_ the function is called - so the expression is not passed to the function, the result of the expression is passed ... I was under the impression that the the expression evaluates to a 'pointer' (I'm sure thats bad terminology) to $a ... which can taken by reference by the function. possibly I am completely misunderstanding what goes on here.

When used as an expression, an assignment evaluates to whatever is on the right side of the assignment operator, not the left. Example:

right so $a evaluates to 5 and a reference to the value of $a is passed. what you say doesn't make sense (to me) take a look at the following: <? $b = 6; $c = "another string"; var_dump(($a = 5), $a, ($a = "some string"), $a, ($a = $b), ($a = $c), $b, $c); ?> as you see dumping the expression ($a = 5) and dumpimg plain old $a result in the same (as far as the output is concerned) its therefore not possible to conclude that

var_dump($a = 5); outputs int(5) var_dump($a = "some string"); outputs string(11) "some string" So, as far as foo() knows: foo($a = 5); and foo(5); are exactly the same...

I don't think they are, and you're examples don't prove it. Anyone care to come up with the proof. or explain to thick-eared old me where I am mistaken.