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Declaring Variable Types in PHP?#1
I was trying to get my Netbeans to autocomplete with PHP, and I learned that this code is valid in PHP:

function blah(Bur $bur) {}


A couple of questions:


Does this actually impose any limits on what type of variable I can pass to the blah method?
If this is just to help the IDE, that's fine with me. How can I declare the type of a variable in PHP if I'm not in a function?

posted date: 2008-12-23 22:57:00


Re: Declaring Variable Types in PHP?#2
I had made out the solution of this problem. click to view my topic...

hope that hepls.

posted date: 2008-12-23 22:57:01


Re: Declaring Variable Types in PHP?#3
It's called type hinting, added with PHP 5. It isn't quite what you may be expecting if you are coming from a language like Java. It does cause an error to be thrown if you don't pass in the expected type. You can't type-hint primitives, though (no int $bur).

posted date: 2008-12-23 23:03:00


Re: Declaring Variable Types in PHP?#4
Specifying a data type for a function parameter will cause PHP to throw a catchable fatal error if you pass a value which is not of that type. Please note though, you can only specify types for classes, and not primitives such as strings or integers.Most IDE's can infer a data type from a PHPDoc style comment if one is provided. e.g./** * @var string */public $variable = "Blah";

posted date: 2008-12-23 23:04:00


Re: Declaring Variable Types in PHP?#5
Does this actually impose any limits on what type of variable I can pass to the blah method?This is called type hinting. According to the PHP documentation that I just linked to, yes, it does impose limits on the argument type: "Failing to satisfy the type hint results in a catchable fatal error."How can I declare the type of a variable in PHP if I'm not in a function?Read type juggling. You can't explicitly define a variable's type in PHP, its type is decided by the context it is used in.

posted date: 2008-12-23 23:10:00


Re: Declaring Variable Types in PHP?#6
And I'll test the warning/error thing tomorrow morning (it's 5am here)...

posted date: 2008-12-23 23:13:00


Re: Declaring Variable Types in PHP?#7
I'm actually just trying to get my IDE to stop juggling and start hinting...

posted date: 2008-12-23 23:15:00


Re: Declaring Variable Types in PHP?#8
#2 : (...) How can I declare the type of a variable in PHP if I'm not in a function?I recently heard about "settype()" and "gettype()" in PHP4 & 5You can force the variable type anytime easilyFrom PHP.net :bool settype ( mixed &$var , string $type )Parametersvar : The variable being converted.type : Possibles values of type are:"boolean" (or, since PHP 4.2.0, "bool")"integer" (or, since PHP 4.2.0, "int")"float" (only possible since PHP 4.2.0, for older versions use the deprecated variant "double")"string""array""object""null" (since PHP 4.2.0)[ :D First visit, first comment...]

posted date: 2008-12-24 06:46:00


Re: Declaring Variable Types in PHP?#9
Yeah I just saw that. settype and gettype are only for primitives, which doesn't help (although technically it does answer the question, so I'll vote you up. Please provide a link to the PHP doc and remove it from your answer).

posted date: 2008-12-24 07:29:00


Re: Declaring Variable Types in PHP?#10
No need to test. According to the PHP docs "Failing to satisfy the type hint results in a catchable fatal error." ch2.php.net/language.oop5.typehinting

posted date: 2008-12-24 08:26:00


Re: Declaring Variable Types in PHP?#11
This type-hinting only works for validating function arguments; you can't declare that a PHP variable must always be of a certain type. This means that in your example, $bur must be of type Bur when "blah" is called, but $bur could be reassigned to a non-Bur value inside the function.Type-hinting only works for class or interface names; you can't declare that an argument must be an integer, for example.One annoying aspect of PHP's type-hinting, which is different from Java's, is that NULL values aren't allowed. So if you want the option of passing NULL instead of an object, you must remove the type-hint and do something like this at the top of the function:assert('$bur === NULL || $bur instanceof Bur');EDIT: This last paragraph doesn't apply since PHP 5.1; you can now use NULL as a default value, even with a type hint.EDIT: You can also install the SPL Type Handling extension, which gives you wrapper types for strings, ints, floats, booleans, and enums. EDIT: You can also use "array" since PHP 5.1, and "callable" since PHP 5.4.

posted date: 2008-12-24 11:36:00


Re: Declaring Variable Types in PHP?#12
Complete and great answer. I'll admit that my main focus was to get my IDE hinting, which I have already figured out here www.momige.com/390192... personally, I find that many aspects of PHP are depressing hacks, but that's just my preference for typed langs.

posted date: 2008-12-24 12:09:00


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