References not working as expected?

MarkR

Consider the following program:

<?php
	// Test program
	foo();

function foo() {
	global $x;
	$y = 42;
	$x =& $y;
	print "x=$x\n";
	print "y=$y\n";
	bar();
}

function bar() {
	global $x;
	print "x=$x\n";
}
?>

I have two questions:

What would you expect the output of this program to be?
Assuming I wanted the output of this program to be
```
x=42
y=42
x=42
```

How do I have to change it?

Note that this is a simple case; in fact I have a massively complex PHP application which is failing because the global reference $x does not continue to refer to $y in the subroutine.

Mark

drew010

it appears to be failing for several reasons. i have commented it and provided 2 ways you can fix it.

your way

<?php
    // Test program
    foo();

function foo() {
    global $x; //doesnt exist
    $y = 42;   //local variable
    $x =& $y; //x is pointer to local y
    print "x=$x\n"; //42
    print "y=$y\n"; //42
    bar();
}

function bar() {
    global $x; //no global x, foo is not global
    print "x=$x\n";  //nothing
}
?>

the first call to global $x in foo doesnt do anything since there is no global var $x. calling global $x is a simpler and cleaner way of doing
function foo(&$x) {
then you create a var local to foo, create a new var x which is a pointer to local y. bar has no idea they exist (x or y)

one fix

<?php
    // Test program

$y = 0;  //create global y
foo();

function foo() {
    global $y;  //bring in y
    $y = 42;    //set it
    $x =& $y;  //local x pointer to glob y
    print "x=$x\n";  //42
    print "y=$y\n";  //42
    bar();
}

function bar() {
    global $y;  //bring in glob y, set as 42
    print "y=$y\n"; //42
}
?>

in this way, you just have to change the variable you use. y is defined in the global scope so it exists to all functions and lasts for the lifetime of the program

other way

<?php
    // Test program
    foo();

function foo() {
    $y = 42;  //set local y = 42
    $x =& $y; //local x pointer to y
    print "x=$x\n";  //42
    print "y=$y\n";  //42
    bar($x);     //pass local x to bar
}

function bar(&$x) { //takes addr of x
    print "x=$x\n";  //42
    //since bar exists in foo, the variable $y in foo isnt destroyed yet so x still points to it.  &$x = &$y
}
?>

no global variables, all exist locally to functions, but we explicitly pass the addr of x into bar before it is destroyed and it will result in what you want.

MarkR

Unfortunately your fix was by making my program do something completely different.

Nevertheless, I fixed it anyway.

By the way, having global $x should work fine even if there is no global $x.

What I wanted to do was have a couple of globals used to store context information so I didn't have to keep passing heaps of references around unnecessarily.

The problem is that

  function something() {
    $y = 42;
    global $x;
    $x = & $y;
  }

Did not do what I expected, because $x = & $y does not set the global $x, it just makes $x a local instead.

I didn't realise this at the time of writing.

The fix is

<?php
	// Test program
	foo();

function foo() {
	$y = 42;
	print "y=$y\n";
	$GLOBALS['x'] =& $y;
	bar();
}

function bar() {
	global $x;
	print "x=$x\n";
}
?>

Mark