Can anyone explain this array declaration in PHP -> $a{0} = "value"

Question

Hi i am using PHP for couple of years ,

these are the ways in PHP that i know to declare an array

$arr    = array();
$arr    = array(1,2);
$arr[0] = 1;
$arr[]  = 1;

In an example I saw this syntax and I ran the code and it was valid:

$a{0} = "value";

but the following code didn't run:

$a{} = "value";

It gave:

Parse error: syntax error, unexpected '}'

How to explain this?

Explain in which sense? Looking into the error message: ***`Parse error: syntax error, unexpected '}`*** - Which part of it in concrete didn't you understood if I may ask? That is a syntax error. — hakre, May 11 '14 at 08:19
Also it is not recommended to use array curly bracket access. — hakre, May 11 '14 at 08:22

score 10 · Accepted Answer · edited Mar 08 '14 at 10:32

10

From the PHP docs:

Both square brackets and curly braces can be used interchangeably for accessing array elements (e.g. $array[42] and $array{42} will both do the same thing ).

{ } is not just for accessing, you can even append elements to the array provided you pass the key !

$arr{34} = 'some data'; // <--- Valid

OUTPUT:

Array
(
    [34] => some data
)

but

$arr{} = 'some data';// <--- This is not a valid and it throws an error.

The only difference between { } and [ ] is that you need to pass the key for the former. Else, it will throw an error: PHP Parse error: syntax error, unexpected '}'.

edited Mar 08 '14 at 10:32

Amal Murali

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answered Mar 05 '14 at 07:04

Shankar Narayana Damodaran

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but $a{} = "value" is invalid , how to explain that – Kanishka Panamaldeniya Mar 05 '14 at 07:06
3

Quoting the snippet above, "interchangeably for accessing array elements". $a{} isn't accessing. – Nathan Dawson Mar 05 '14 at 07:07
There is nothing to explain actually, this is code showing accessing the element `34` and assigning the string "`some data`" to it. (This is also the part where you are wrong - with all due respect.) – hakre May 11 '14 at 08:59

score 1 · Answer 2 · answered Mar 05 '14 at 07:05

1

It's just the same as square brackets but closer to Perl syntax.

http://uk1.php.net/manual/en/language.types.array.php#99015

You're only able to access existing elements this way. As you've already pointed out $a{} = 1; won't work.

answered Mar 05 '14 at 07:05

Nathan Dawson

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Can anyone explain this array declaration in PHP -> $a{0} = "value"

2 Answers2