Reputation: 588
I have a couple of queries about modifying an array during a foreach() loop. In the code below I loop through three arrays that contain closures/callbacks and invoke each one. I append a closure to the end of each array during iteration, however sometimes foreach() doesn't seem to recognise that the array has changed size and so the appended closure doesn't get called.
class Foo
{
private $a1 = array();
private $a2 = array();
public function f()
{
echo '<pre style="font-size: 20px;">';
echo 'PHP: ' . phpversion() . '<br><br>';
$this->a1[] = function() { echo 'a1 '; };
$this->a1[] = array($this, 'g');
foreach ($this->a1 as &$v)
{
// The callback added in g() never gets called.
call_user_func($v);
//echo 'count(v) = ' . count($v) . ' ';
}
echo '<br>';
// The same thing works fine with a for() loop.
$this->a2[] = function() { echo 'a2 '; };
$this->a2[] = array($this, 'h');
for ($i = 0; $i < count($this->a2); ++$i)
call_user_func($this->a2[$i]);
echo '<br>';
// It also works fine using a local array as long as it
// starts off with more than one element.
$a3[] = function() { echo 'a3 '; };
//$a3[] = function() { echo 'a3 '; };
$i = 0;
foreach ($a3 as &$x)
{
call_user_func($x);
if ($i++ > 1) // prevent infinite loop
break;
// Why does this get called only if $a3 originally starts
// with more than one element?
$a3[] = function() { echo 'callback '; };
}
echo '</pre>';
}
private function g()
{
echo 'g() ';
$this->a1[] = function() { echo 'callback '; };
}
private function h()
{
echo 'h() ';
$this->a2[] = function() { echo 'callback '; };
}
}
$foo = new Foo;
$foo->f();
Output:
PHP: 5.3.14-1~dotdeb.0
a1 g()
a2 h() callback
a3
Expected output:
a1 g() callback
a2 h() callback
a3 callback
Output for $a3
if I uncomment the second element before the loop:
a3 a3 callback
foreach ($this->a1 as &$v)
realise $v
has another element to iterate over?$a3
work during the third loop foreach ($a3 as &$x)
, but only when the array starts off with more than one element?I realise modifying an array during iteration is probably not a good idea, but since PHP seems to allow it I'm curious why the above works the way it does.
Upvotes: 11
Views: 2591
Reputation: 24551
Interesting observation:
echo "foreach: ";
$a = array(1,2,3);
foreach($a as $v) {
echo $v, " ";
if ($v===1) $a[] = 4;
if ($v===4) $a[] = 5;
}
echo "\nforeach&: ";
$a = array(1,2,3);
foreach($a as &$v) {
echo $v, " ";
if ($v===1) $a[] = 4;
if ($v===4) $a[] = 5;
}
echo "\nwhile: ";
$a = array(1,2,3);
while(list(,$v) = each($a)) {
echo $v, " ";
if ($v===1) $a[] = 4;
if ($v===4) $a[] = 5;
}
echo "\nfor: ";
$a = array(1,2,3);
for($v=reset($a); key($a)!==null; $v=next($a)) {
echo $v, " ";
if ($v===1) $a[] = 4;
if ($v===4) $a[] = 5;
}
results in
foreach: 1 2 3
foreach&: 1 2 3 4
while: 1 2 3 4 5
for: 1 2 3 4 5
This means:
foreach
loop operates on a copy of the array, any modifications of the array within the loop do not affect the loopforeach
with referenced value is forced to use the original array but advances the array pointer before each iteration after assigning key and value variables. Also there is some optimization going on that prevents another check as soon as the pointer reaches the end. So at the beginning of the last iteration the loop is told to run once more and then finish - no more interfering possible.while
loop with each()
advances the array pointer just like foreach
does but explicitly checks it again after the last iterationfor
loop where the array pointer is advanced after each iteration obviously has no problems with changing the array at any point.Upvotes: 6
Reputation:
1.Why doesn't the first loop foreach ($this->a1 as &$v) realise $v has another element to iterate over?
The behaviour looks to be due to the internal pointer being advanced on the array on each foreach iteration. Adding an array element to the end of the array on the last iteration of the array, that is when the internal pointer is already null, means that this element will not be iterated over. With some modifications to your code, this can be seen.
class Foo
{
private $a1 = array();
private $a2 = array();
public function f()
{
echo '<pre style="font-size: 20px;">';
echo 'PHP: ' . phpversion() . '<br><br>';
$this->a1[] = function() { echo 'a1 <br/>'; };
$this->a1[] = array($this, 'g');
foreach ($this->a1 as $key => &$v)
{
//lets get the key that the internal pointer is pointing to
// before the call.
$intPtr = (key($this->a1) === null) ? 'null' : key($this->a1);
echo 'array ptr before key ', $key, ' func call is ',
$intPtr, '<br/>' ;
call_user_func($v);
//echo 'count(v) = ' . count($v) . ' ';
}
echo '<br><br>';
// The same thing works fine with a for() loop.
$this->a2[] = function() { echo 'a2 '; };
$this->a2[] = array($this, 'h');
for ($i = 0; $i < count($this->a2); ++$i)
call_user_func($this->a2[$i]);
echo '<br><br>';
// It also works fine using a local array as long as it
// starts off with more than one element.
$a3[] = function() { echo 'a3 '; };
//$a3[] = function() { echo 'a3 '; };
$i = 0;
foreach ($a3 as &$x)
{
call_user_func($x);
if ($i++ > 1) // prevent infinite loop
break;
// Why does this get called only if $a3 originally starts
// with more than one element?
$a3[] = function() { echo 'callback '; };
}
echo '</pre>';
}
private function g()
{
echo 'g() <br>';
$this->a1[] = function() { echo 'callback '; };
}
private function h()
{
echo 'h() <br>';
$this->a2[] = function() { echo 'callback '; };
}
}
$foo = new Foo;
$foo->f();
Output:
array ptr before key 0 func call is 1
a1
array ptr before key 1 func call is null <-will not iterate over any added elements!
g()
a2 h()
callback
a3
2.Why does modifying $a3 work during the third loop foreach ($a3 as &$x), but only when the array starts off with more than one element?
Of course if you add an element to the array before the internal pointer returns null then the element will be iterated over. In your case if the array has one element then on the first iteration the internal pointer has already returned null. However, if there is initially more than one element then the additional element can be added on the first iteration as the internal pointer will be pointing to the second intial element at this time.
Upvotes: 4