Posted by "Michael Sims" on 09/20/05 17:11

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:

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...

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