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Posted by Ivαn Sαnchez Ortega on 02/13/06 01:59
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leegold wrote:
> Whoa, that's saying that the initial argument to the print function is not
> completely evaluated, but only half-way evaluated and passed and
> then...gosh I'd expect the function to let the args or params - whatever -
So, let's have a look at the operator precedence table* to explain what's
happening behind the scenes while evaluating this:
print '12==12' . print (12==12);
* http://es2.php.net/manual/en/language.operators.php
First of all, the outer "print" wants to be evaluated. In order to do so,
the expression it refers to ("'12==12' . print (12==12)") has to be
evaluated.
The expression "'12==12' . print (12==12)" wants to be evaluated. That
expression can be split into two expressions, "'12==12'" and "print
(12==12)", separated by the left-associative binary operator "." (string
concatenate). So, in order to evaluate this expression, first I'll have to
evaluate "'12==12'", then evaluate "print (12==12)", then apply the binary
operator.
So, the expression "'12==12'" wants to be evaluated. As it is a constant, it
gets evaluated as such.
Next, the expression "print (12==12)" wants to be evaluated. And here comes
the trick. You are not evaluating a numerical expression, but a function
that does I/O (outputs something). But, in order to evaluate this
print(12==12), first we have to evaluate (12==12).
So, (12==12) gets evaluated, as (bool) true.
Then, "print ((bool) true)" is ready to run. This outputs "1" by stdout
(implicit type casting from bool to string), and the expression "print
(12==12)" gets evaluated as (bool) true, as print has been sucessful.
Then, "'12==12' . print(12==12)" is ready to be evaluated. At this time,
that expression equals "'12==12' . (bool) true". So, (bool) true gets
implicitly casted from bool to string, and both strings are concatented.
Now that expression equals "12==121".
Then, the outer print is ready to be run and evaluated. While running, it
outputs "12==121" by stdout. Then, it evaluates to (bool) true.
If that was too much for you, just think about the following code:
<?php
$a = '12==12: ';
$b = print (12==12);
print $a.$b;
?>
$a is evaluated first. Then $b, then "print $a.$b". That's how it happens.
> I'd expect it to print the result of the line and the print fuction itself
> not to be a factor like your saying.
They are not "factors". As in C (and several other imperative languages),
PHP works with expressions.
- --
- ----------------------------------
IvΓ‘n SΓ‘nchez Ortega -i-punto-sanchez--arroba-mirame-punto-net
http://acm.asoc.fi.upm.es/~mr/ ; http://acm.asoc.fi.upm.es/~ivan/
MSN:i_eat_s_p_a_m_for_breakfast@hotmail.com
Jabber:ivansanchez@jabber.org ; ivansanchez@kdetalk.net
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