[Resolved] the "&"

[felipe_lopes]

Hi there!

I went thru the following code:

$obj = &new myclass;

It made me wonder what is the "&" for in this case?

Thanx

[LordShryku]

Found this in user comments in the manual

It seems that when a class constructor is called, PHP creates a new instance of the class and then either hands back a reference to this instance if you use the =& assignment operator, or creates and hands back a whole new copy if you use the plain = assignment operator.

[felipe_lopes]

wow...It's kind of hard to understand...

what is the advantage of using =& then if there is one?

[ahundiak]

http://www.php.net/manual/en/language.references.php

[LordShryku]

Gonna point you to the manual again...

References Explained

[felipe_lopes]

Thank you guys!

I read the manual about classes but didn't find anything...

I'll check these links out

[felipe_lopes]

Is it supposed to work the same??

$obj = &new class;

And:

$obj =& new class;

Thanx in advance

[swr]

Yes, "=&" and "= &" are the same. There is also "&=" but that is completely different (like "+=" but with a bitwise AND).

[Weedpacket]

Originally posted by swr
Yes, "=&" and "= &" are the same.

Reading it it's like the difference between "make 'a reference to' something" and "make a 'reference to something'" - just a different way of picturing the same thing 🙂

[swr]

I prefer "$foo =& $bar" because it is a special type of assignment. If you write "$foo = &$bar" I tend to think of "&$bar" as something that evaluates to the address of $bar, a la C. But it's not. You can't go "$foo = &$bar + 1" like you could in C (without the dollar signs of course). "&$bar" on its own doesn't evaluate to anything - it's neither an lvalue nor an rvalue, so putting it on the right-hand side of what otherwise would look like a regular "=" assignment just doesn't sit right with me. As far as I can tell it is only useful in an assignment, and the variable being &'ed must be the only thing in the assignment (it can't be an expression). So I prefer to stick the & with the assignment operator.

[felipe_lopes]

The manual says that do like this: $obj = &new class; should return an error, but I assume that it doesn't since I've seen just this construction on classes....

[zapa]

<?php
7: $aVariable = 42;
8: $anotherVariable = &$aVariable;
9: // a copy of the contents of $aVariable is placed in $anotherVariable
10: $aVariable= 325;
11: print $anotherVariable; // prints 325
12: ?>

Placing an
ampersand (&) in front of the $aVariable variable ensures that a reference to this
variable, rather than a copy of its contents, is assigned to $anotherVariable.

Now
any changes made to $aVariable are seen when accessing $anotherVariable. In
other words, both $aVariable and $anotherVariable now point to the same value.

not sure if that help? got that out of learn php in 24hr

[Weedpacket]

While a program's running, PHP maintains a big table of all the variables curently defined. Each record in this table contains things like the name of the variable and what type of variable it is. Most significantly is a field that contains a (C-like) pointer to a certain location in memory where the value of the variable is to be found.

When you say "$foo=$bar;", the value of $bar is copied to a new location in memory and $foo points to the new location. When you say "$foo=&$bar;" $foo is set t point to the same memory location that $bar does, so that changing the value of $foo or $bar changes the value of $bar or $foo at the same time.

That's only a first approximation: there are all sorts of memory management optimisation fun and games involving reference counts and copy-on-write and so on, but conceptually that's enough to be going on with.

[felipe_lopes]

Originally posted by zapa
<?php
7: $aVariable = 42;
8: $anotherVariable = &$aVariable;
9: // a copy of the contents of $aVariable is placed in $anotherVariable
10: $aVariable= 325;
11: print $anotherVariable; // prints 325
12: ?>

Placing an
ampersand (&) in front of the $aVariable variable ensures that a reference to this
variable, rather than a copy of its contents, is assigned to $anotherVariable.

Now
any changes made to $aVariable are seen when accessing $anotherVariable. In
other words, both $aVariable and $anotherVariable now point to the same value.

not sure if that help? got that out of learn php in 24hr

Thank you Zapa but it not the issue here...