Declaring Variable Types in PHP?

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.

#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...]

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).

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