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<?php

$var = "Bob";

$Var = "Joe";

echo "$var, $Var";      // outputs "Bob, Joe"



$4site = 'not yet';     // invalid; starts with a number

$_4site = 'not yet';    // valid; starts with an underscore

$t?yte = 'mansikka';    // valid; 'ä§ is (Extended) ASCII 228.

?>

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<?php

$foo = 'Bob';              // Assign the value 'Bob' to $foo

$bar = &$foo;              // Reference $foo via $bar.

$bar = "My name is $bar";  // Alter $bar...

echo $bar;

echo $foo;                 // $foo is altered too.

?>

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<?php

$foo = 25;

$bar = &$foo;      // This is a valid assignment.

$bar = &(24 * 7);  // Invalid; references an unnamed expression.



function test()

{

   return 25;

}



$bar = &test();    // Invalid.

?>

add a note User Contributed Notes
±äÁ¿

stlawson AT sbcglobal DOT net
06-Jul-2002 09:47

In 'ghent's comment on the 'above example' I think he confuses the
confusion ;)

Here is the example referred
to:

<?php
$foo = 'Bob';              // Assign the value 'Bob'
to $foo
$bar = &$foo;              // Reference $foo via
$bar.
$bar = "My name is $bar";  // Alter $bar...
echo
$bar;
echo $foo;                 // $foo is altered
too.
?>

‘Reference $foo via $bar’/‘Alter $bar…’ IS correct,
but it is a little obscure.  Here’s what it means:

$bar is assigned
a reference to $foo, thus $bar ‘references’ or points to $foo which
contains the string ‘Bob’.  Essentially what is happening here is ‘Bob’ is
stored in memory at a particular address.  &$foo returns that address
[where ‘Bob’ is stored].  That address is assigned to $bar.  So, in the
string “My name is $bar”, $bar ‘uses’ the address it contains to find ‘Bob’
(which is ‘in’ $foo) and thus the string becomes “My name is Bob”.  When
the string is assigned to $bar, because $bar refers to $foo, it gets
assigned to the same address location that ‘Bob’ is stored at, thus ‘Bob’
is overwritten by “My name is Bob”.  The trick here is to realize that $bar
behaves as if it is $foo, so when something is assigned to $bar (the alias
of $foo), it’s as if it was being assigned to $foo!

After the above
script is run, the output will look like this:

  My name is BobMy
name is Bob

e.g. both $foo and $bar print the same thing.

In
‘C++’ the same code snippit would look like this:

foo = “Bob”
;
bar = &foo ;
bar = “My name is “ + *bar ;

Notice that in
C/C++ it is necessary to manually dereference the pointer (*bar) – PHP does
this automagically. 

BTW: You might think that ‘echo $bar;’ would
display the address of $foo – not so!  More PHP automagic ;)

add a note