Michael Sims wrote:
Jochem Maas wrote:
Michael Sims wrote:
So, as far as foo() knows:
foo($a = 5);
and
foo(5);
are exactly the same...
I don't think they are, and you're examples don't prove it.
Anyone care to come up with the proof.
No, I was wrong, Rasmus corrected me. That's my
Lester Caine wrote:
This type of code is used in a few places, so I'd like a little help
converting it to 'good code' under the new rules ;)
They're not new rules. PHP is just warning you where it didn't before.
It was still bad coding practice before.
Get the key from an array ( fails
Jasper Bryant-Greene wrote:
Lester Caine wrote:
This type of code is used in a few places, so I'd like a little help
converting it to 'good code' under the new rules ;)
They're not new rules. PHP is just warning you where it didn't before.
It was still bad coding practice before.
Since
Lester Caine wrote:
I suppose the REAL questions was - Why was using the function in this
way a 'bad practice', is there any way that it could be made a 'good
practice' since the intent is so obvious?
I understand the new checks, but I don't see that the original was
particularly 'bad' - only
Jasper Bryant-Greene wrote:
Lester Caine wrote:
I suppose the REAL questions was - Why was using the function in this
way a 'bad practice', is there any way that it could be made a 'good
practice' since the intent is so obvious?
I understand the new checks, but I don't see that the original
Jochem Maas wrote:
Jasper Bryant-Greene wrote:
From the PHP manual [2]:
| the following examples of passing by reference are invalid:
|
| foo(bar()); // Produces fatal error since PHP 5.1.0
| foo($a = 5); // Expression, not variable
if foo() expects one args by reference then why is doing:
Jasper Bryant-Greene wrote:
Jochem Maas wrote:
Jasper Bryant-Greene wrote:
From the PHP manual [2]:
| the following examples of passing by reference are invalid:
|
| foo(bar()); // Produces fatal error since PHP 5.1.0
| foo($a = 5); // Expression, not variable
if foo() expects one args
Jochem Maas wrote:
foo($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
Jochem Maas wrote:
Basically, in PHP, a reference (such as what key() takes as a
parameter [1]) can only point to an actual variable, not directly to
the result of a function. So you have to assign the output of the
function to a variable first.
wtf, Im now officially confused (before I
thanks everyone for the crash course in better understanding the underlying
mechanisms!
... I'm probably not the only one that learnt something from this ;-)
Dragan Stanojevic - Nevidljivi wrote:
Jochem Maas wrote:
Basically, in PHP, a reference (such as what key() takes as a
parameter [1])
Lester Caine wrote:
This type of code is used in a few places, so I'd like a little help
converting it to 'good code' under the new rules ;)
Get the key from an array ( fails because key(array) )
if( $pId == key( $this-getFunc() ) ) {
In getFunc()
return (
Michael Sims wrote:
Jochem Maas wrote:
foo($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
Rasmus Lerdorf wrote:
Michael Sims wrote:
When used as an expression, an assignment evaluates to whatever is
on the right side of the assignment operator, not the left.
Example:
[...]
foo($a = 5);
and
foo(5);
are exactly the same...
The value passed is the same, but when passed as
Michael Sims wrote:
Jochem Maas wrote:
foo($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
Jochem Maas wrote:
Michael Sims wrote:
So, as far as foo() knows:
foo($a = 5);
and
foo(5);
are exactly the same...
I don't think they are, and you're examples don't prove it.
Anyone care to come up with the proof.
No, I was wrong, Rasmus corrected me. That's my one allowed mistake
Jochem Maas wrote:
Michael Sims wrote:
So, as far as foo() knows:
foo($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.
No, I was wrong, Rasmus corrected me. That's my one
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