Re: [PHP] Re: PHP5 and pass by reference bug.
* Thus wrote Daevid Vincent:
Yeah, I get what references are. The point is that when it was on the user
to decide, they could do it. Now that PHP5 makes you put the in the
function declaration instead of the passing parameter, you don't know what
the user is going to send. Therefore it renders the in the function
declaration a useless thing.
I could have this function
Function add ($a, $b)
{
return ($a + $b);
}
And as a user I could use it like so:
$x = 5;
$y = 10;
add($x, $y);
Or I could also use it like this:
add(5,10);
But since the function is now responsible in PHP5 to use the [since
passing add($x, $y); is now invalid], it makes my function add basically
useless.
The only time you should pass by reference is when you plan on
modifiying that variable and the caller is going to get a modified
version back:
function foo($modify_me) {
$modify_me = 'another value';
}
foo($var);
// var contains now 'another value'
Otherwise, always declare your functions as normal variables.
Or for your example, you provided, that would mean that in your
original code you had called your functions as:
add(5, 10);
Which is obviosly wrong, so going that route wont really work.
If you're concerned about memory usage, like why do I need to
duplicate memory if I'm just passing variables that will just
simply be read. Well, PHP takes that pretty much into
consideration. And the only time a duplicate is created is when you
modify the variable.
If for example you have:
$var = 'a really long string';
And you assign that variable to something else:
$var2 = $var;
PHP doesn't allocate all the memory of var for that var2 until
you try to modify the $var2. So not until you do something like:
$var2 .= 'some more text';
PHP then will make a copy of the original $var, and then make
modifications.
The same principle applies to function parameters.
Curt
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[PHP] Re: PHP5 and pass by reference bug.
Maybe you recheck the dokumentation on what exactly referenzes are. Do
you expect the function to alter the string something here and every-
time you later print the string within your script you get the altered one?
ONLY variables can be passed by referenze !
-- red
Daevid Vincent wrote:
So, I'm getting all these errors/warnings in PHP5 now saying that I have to
put the on the function and not in the passing (which sorta makes sense
and puts the burden on the function rather than the user, which I like too).
So I spend the time to go and fix several thousand lines of code.
Then I start to see these other errors...
Maybe I'm missing something, but this seems like a glaring bug in passing by
reference that nobody caught...
say you have
function foo($bar)
{
}
well that works great as long as you use it like
foo($x);
but if you try
foo(something here);
Or
foo( array('a','b','c') );
it shits the bed. :-(
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[PHP] Re: PHP5 and pass by reference bug.
Maybe you recheck the dokumentation on what exactly referenzes are. Do
you expect the function to alter the string something here and every-
time you later print the string within your script you get the altered one?
ONLY variables can be passed by referenze !
-- red
Daevid Vincent wrote:
So, I'm getting all these errors/warnings in PHP5 now saying that I have to
put the on the function and not in the passing (which sorta makes sense
and puts the burden on the function rather than the user, which I like too).
So I spend the time to go and fix several thousand lines of code.
Then I start to see these other errors...
Maybe I'm missing something, but this seems like a glaring bug in passing by
reference that nobody caught...
say you have
function foo($bar)
{
}
well that works great as long as you use it like
foo($x);
but if you try
foo(something here);
Or
foo( array('a','b','c') );
it shits the bed. :-(
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RE: [PHP] Re: PHP5 and pass by reference bug.
Yeah, I get what references are. The point is that when it was on the user
to decide, they could do it. Now that PHP5 makes you put the in the
function declaration instead of the passing parameter, you don't know what
the user is going to send. Therefore it renders the in the function
declaration a useless thing.
I could have this function
Function add ($a, $b)
{
return ($a + $b);
}
And as a user I could use it like so:
$x = 5;
$y = 10;
add($x, $y);
Or I could also use it like this:
add(5,10);
But since the function is now responsible in PHP5 to use the [since
passing add($x, $y); is now invalid], it makes my function add basically
useless.
-Original Message-
From: Red Wingate [mailto:[EMAIL PROTECTED]
Sent: Friday, July 16, 2004 5:35 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Re: PHP5 and pass by reference bug.
Maybe you recheck the dokumentation on what exactly
referenzes are. Do
you expect the function to alter the string something here
and every-
time you later print the string within your script you get
the altered one?
ONLY variables can be passed by referenze !
-- red
Daevid Vincent wrote:
So, I'm getting all these errors/warnings in PHP5 now
saying that I have to
put the on the function and not in the passing (which
sorta makes sense
and puts the burden on the function rather than the user,
which I like too).
So I spend the time to go and fix several thousand lines of code.
Then I start to see these other errors...
Maybe I'm missing something, but this seems like a glaring
bug in passing by
reference that nobody caught...
say you have
function foo($bar)
{
}
well that works great as long as you use it like
foo($x);
but if you try
foo(something here);
Or
foo( array('a','b','c') );
it shits the bed. :-(
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To unsubscribe, visit: http://www.php.net/unsub.php
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Re: [PHP] Re: PHP5 and pass by reference bug.
But that function doesn't need to pass by reference. If you have a
function that has a parameter that is passed by ref, it should always
be a variable that is passed in. You should only be using pass by ref
when the function changes the value and it simply doesn't make sense
to change a constant.
On Fri, 16 Jul 2004 18:03:08 -0700, Daevid Vincent [EMAIL PROTECTED] wrote:
Yeah, I get what references are. The point is that when it was on the user
to decide, they could do it. Now that PHP5 makes you put the in the
function declaration instead of the passing parameter, you don't know what
the user is going to send. Therefore it renders the in the function
declaration a useless thing.
I could have this function
Function add ($a, $b)
{
return ($a + $b);
}
And as a user I could use it like so:
$x = 5;
$y = 10;
add($x, $y);
Or I could also use it like this:
add(5,10);
But since the function is now responsible in PHP5 to use the [since
passing add($x, $y); is now invalid], it makes my function add basically
useless.
-Original Message-
From: Red Wingate [mailto:[EMAIL PROTECTED]
Sent: Friday, July 16, 2004 5:35 PM
To: [EMAIL PROTECTED]
Subject: [PHP] Re: PHP5 and pass by reference bug.
Maybe you recheck the dokumentation on what exactly
referenzes are. Do
you expect the function to alter the string something here
and every-
time you later print the string within your script you get
the altered one?
ONLY variables can be passed by referenze !
-- red
Daevid Vincent wrote:
So, I'm getting all these errors/warnings in PHP5 now
saying that I have to
put the on the function and not in the passing (which
sorta makes sense
and puts the burden on the function rather than the user,
which I like too).
So I spend the time to go and fix several thousand lines of code.
Then I start to see these other errors...
Maybe I'm missing something, but this seems like a glaring
bug in passing by
reference that nobody caught...
say you have
function foo($bar)
{
}
well that works great as long as you use it like
foo($x);
but if you try
foo(something here);
Or
foo( array('a','b','c') );
it shits the bed. :-(
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To unsubscribe, visit: http://www.php.net/unsub.php
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