Declaring Variable Types in PHP?

Question

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:

1. Does this actually impose any limits on what type of variable I can pass to the blah method?
2. 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?

Villanovan

Answer 1

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.

EDIT: You can also use "string", "int", "float" and "bool" since PHP 7.0.

EDIT: As of PHP 7.4, you can declare member variables of a class/interface/trait as a specific type like public int $a;, and variables that are declared this way cannot be assigned to a value of another type. You can also use union types such as string|int as of PHP 8.0, and you can use classes in the union types as of PHP 8.1.

https://www.php.net/manual/en/language.types.declarations.php

Uttasta

Complete and great answer. I'll admit that my main focus was to get my IDE hinting, which I have already figured out here stackoverflow.com/questions/390192... personally, I find that many aspects of PHP are depressing hacks, but that's just my preference for typed langs. – Villanovan

There there, even those of us who don't prefer strongly-typed languages agree with you :) – Stockbreeder

A good step for PHP would be the option to declare type or leave it to be dynamic. I hope to see that in future PHP – Fiduciary

I slightly disagree or want to make more precise the first EDIT: null is never of valid type $myClass (PHP 5.3). Perhaps it works as default param, but one still can't pass it in (like in Java) if a type hint in function signature are used. – Snapdragon

Actually you can declare myFunc(MyClass $x = NULL), and then call myFunc(NULL) or myFunc() without an error. – Uttasta

Not allowing null for type hinted parameters by default is something that I really like. As you say, Java allows this and they are now screwing around with @Nullable and @NotNull. I think the PHP implementation helps a lot to prevent null reference errors. – Kathie

PHP 7.2 adds object as well – Swindell

Too bad you can't specify an array of objects e.g.: functionName(int[] $someIds) or functionName(SomeObject[] $someObjects), but there is partial workaround: https://mcmap.net/q/203776/-type-hinting-in-php-7-array-of-objects – Piedadpiedmont

For allowing null values, you can also do function blah(?Bur $bur) {} (note the ? in front of Bur). See https://mcmap.net/q/126791/-nullable-return-types-in-php7. – Vivi

Answer 2

1. 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.
2. Most IDE's can infer a data type from a PHPDoc style comment if one is provided. e.g.

/**
 * @var string
 */
public $variable = "Blah";

UPDATE 2021: As of PHP 7 (which is several years old at this point) primitive types can also be declared for function arguments. Nullability can also be indicated with a ? in front of the type from 7.1 onward. You can declare return types now too. So this is valid PHP these days:

public function hasFoo(?int $numFoos) :bool {

phpStorm (my current preferred IDE) is happy to use all of these types for code completion, so I don't need as many phpDoc comments for typing as I used to.

Angloamerican

what if I'm not in a class? The $variable is coming from an included PHP... I think I'm almost where I need to be... – Villanovan

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 – Villanovan

i confirm, it works perfectly with eclipse PDT, thanks for the "Most IDE's can infer... " mention. – Hellhound

Answer 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).

Stockbreeder

Answer 4

#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 & 5
You can force the variable type anytime easily

---

From PHP.net :

bool settype ( mixed &$var , string $type )

Parameters

var : 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...]

Reimburse

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). – Villanovan

$myVar=(string) $variabel; gives a string Variable. Afterwards, assign a value. – Woodall

Answer 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.

Cyclorama

@Jeremy Ruten: actually the link to type hinting is broken. – Vanden