在 PHP 中声明变量类型？

Declaring Variable Types in PHP?

提问by Dan Rosenstark

I was trying to get my Netbeans to autocomplete with PHP, and I learned that this code is valid in PHP:

我试图让我的 Netbeans 使用 PHP 自动完成，我了解到这段代码在 PHP 中是有效的：

function blah(Bur $bur) {}

A couple of questions:

几个问题：

1. Does this actually impose any limitson 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?

1. 这实际上对我可以传递给 blah 方法的变量类型施加了任何限制吗？
2. 如果这只是为了帮助 IDE，那对我来说没问题。如果我不在函数中，如何在 PHP 中声明变量的类型？

回答by JW.

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.

这种类型提示仅适用于验证函数参数；您不能声明 PHP 变量必须始终是某种类型。这意味着在您的示例中，当“blah”被调用时， $bur 必须是 Bur 类型，但 $bur 可以重新分配给函数内的非 Bur 值。

Type-hinting only works for class or interface names; you can't declare that an argument must be an integer, for example.

类型提示仅适用于类或接口名称；例如，您不能声明参数必须是整数。

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:

PHP 的类型提示与 Java 不同，其中一个令人讨厌的方面是不允许使用 NULL 值。因此，如果您想要传递 NULL 而不是对象的选项，则必须删除类型提示并在函数顶部执行以下操作：

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.

编辑：最后一段自 PHP 5.1 起不再适用；您现在可以使用 NULL 作为默认值，即使有类型提示。

EDIT:You can also install the SPL Type Handlingextension, which gives you wrapper types for strings, ints, floats, booleans, and enums.

编辑：您还可以安装SPL 类型处理扩展，它为您提供字符串、整数、浮点数、布尔值和枚举的包装类型。

EDIT:You can also use "array" since PHP 5.1, and "callable" since PHP 5.4.

编辑：您还可以使用自 PHP 5.1 起的“array”和自 PHP 5.4 起的“callable”。

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

编辑：从 PHP 7.0 开始，您还可以使用“string”、“int”、“float”和“bool”。

回答by Jim OHalloran

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.

1. 如果您传递的值不是该类型的值，则为函数参数指定数据类型将导致 PHP 抛出可捕获的致命错误。但请注意，您只能为类指定类型，而不能为字符串或整数等基元指定类型。
2. 如果提供，大多数 IDE 可以从 PHPDoc 样式注释推断数据类型。例如

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

回答by J Cooper

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).

它被称为类型提示，是在 PHP 5 中添加的。如果您来自 Java 之类的语言，这可能不是您所期望的。如果您没有传入预期的类型，它确实会导致抛出错误。但是，您不能键入提示原语（没有 int $bur）。

回答by Shad

#2 : (...) How can I declare the type of a variable in PHP if I'm not in a function?

#2 : (...) 如果我不在函数中，如何在 PHP 中声明变量的类型？

I recently heard about "settype()" and "gettype()" in PHP4 & 5
You can force the variable type anytime easily

我最近听说PHP4 & 5 中的“ settype()”和“ gettype()”
你可以随时轻松地强制变量类型

---

---

From PHP.net :

从 PHP.net ：

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

bool settype（混合 &$var ，字符串 $type ）

Parameters

参数

var : The variable being converted. type : Possibles values of type are:

var : 被转换的变量。type : type 的可能值是：

• "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)

• “boolean”（或者，自 PHP 4.2.0 起，“bool”）
• “整数”（或者，自 PHP 4.2.0 起，“int”）
• “float”（自 PHP 4.2.0 起才有可能，对于旧版本使用已弃用的变体“double”）
• “细绳”
• “大批”
• “目的”
• “null”（自 PHP 4.2.0 起）

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回答by Jeremy Ruten

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

这实际上对我可以传递给 blah 方法的变量类型施加了任何限制吗？

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."

这称为类型提示。根据我刚刚链接到的 PHP 文档，是的，它确实对参数类型施加了限制：“未能满足类型提示会导致可捕获的致命错误。”

How can I declare the type of a variable in PHP if I'm not in a function?

如果我不在函数中，如何在 PHP 中声明变量的类型？

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.

阅读类型杂耍。你不能在 PHP 中显式定义变量的类型，它的类型由它使用的上下文决定。