PHP unset Behavior

Question

I came across some PHP behavior that I think is subtle, but pretty cool. But I don't understand how...

$test=array('a'=>'c', 'b'=>'c');
unset($test['a']);
var_dump($test);

This prints

array(1) { ["b"]=> string(1) "c" }

I would have expected the array to be emptied out. After all, $test['a'] evaluates to 'c' so the unset function only sees 'c' but knows it was just the first 'c' value I wanted removing?

My guess is the interpeter is super smart and looks inside the array inside the parameter given to it - but that's purely conjecture ...

Possible duplicate of stackoverflow.com/questions/6376702/php-unset-array-value — rink.attendant.6
– rink.attendant.6, Commented Jan 17, 2013 at 4:44
I don't follow your logic in what you were expecting. You unset a specific array key. Why would b=>c be gone? — Brad
– Brad, Commented Jan 17, 2013 at 4:45
@Brad OP expected that PHP first interprets $test['a'] which hold the value of 'c' and then performs a removal of ALL identical values. — Ja͢ck
– Ja͢ck, Commented Jan 17, 2013 at 4:46
@Jack, Oh, that makes sense! I understand now. I think I've been looking at PHP too long when stuff like that makes sense. — Brad
– Brad, Commented Jan 17, 2013 at 4:46
@rink.attendant.6, I did actually look at that question before posting (thanks to SO's helpful feature of showing related questing when you start typing) but no-one explained the WHY - but Kolink below's made it clear now - of course, just use pass by ref. — nchaud
– nchaud, Commented Jan 19, 2013 at 11:33

Niet the Dark Absol · Accepted Answer · 2013-01-17 04:45:20Z

1

Erm, no. unset is not a function, it is a language construct. Therefore it doesn't necessarily follow the same rules.

In this case, however, it actually works similarly to a pass-by-reference. It takes the reference to the variable, and destroys it.

answered Jan 17, 2013 at 4:45

Niet the Dark Absol

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1 Comment

thevikas Over a year ago

it does take by ref and more cause the data structure of php vars within its source may be more pointers than whats seen.

rink.attendant.6 · Accepted Answer · 2013-01-17 04:46:36Z

0

What you're actually doing is destroying (unsetting) the key "a" from the array $test. So after that key is destroyed, only "b" exists in the array.

If you wanted to remove all values of 'c' from the array, use array_diff().

answered Jan 17, 2013 at 4:46

rink.attendant.6

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