My Code:
$a = [];
isset($a[0]) and unset($a[0]);
it shows "syntax error, unexpected 'unset' (T_UNSET)"
but
$a = [];
isset($a[0]) and exit();
it works!
Both of exit() and unset() are returning no value. Why does one work but not the other?
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Nisse Engström
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yrssoft
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3could you shar your error info? – Ambal Mani Jun 18 '15 at 05:37
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`exit` doesn't return a value because the program _exits_! – Jun 18 '15 at 05:39
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4code is so ugly, what exactly you want to do? – Umair Ayub Jun 18 '15 at 05:51
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@AmbalMani 'syntax error, unexpected 'unset' (T_UNSET)' – yrssoft Jun 18 '15 at 05:53
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@Umair I want delete a value from an array – yrssoft Jun 18 '15 at 05:57
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For Ex : `$val = 1; $a[] = $val; print_r($a); isset($a[0]); unset($a[0]); print_r($a); ` Could you able understood what kind of issue yopu made – Ambal Mani Jun 18 '15 at 06:01
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1You should read [this (*"How does PHP's unset work internally"*)](http://stackoverflow.com/q/24557958/2518525) – Darren Jun 18 '15 at 06:26
2 Answers
3
unset is a language construct, not a real function (this is why you get T_UNSET and not a more generic term), so it doesn't play by the same rules as a normal function would. isset and exit are also language constructs, but they behave more like normal functions.
Austin
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As I asked in comments and you said you want to delete a value from Array,
Why not simply write
$a = [];
if(isset($a[0])){
unset($a[0]);
// And exit() if you want to
}
Umair Ayub
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Because this is simple way to write and read ... else, future programmers working on your code will say bad words about you :P – Umair Ayub Jun 18 '15 at 08:14