类型声明之前的问号(?)在php(?int)中意味着什么 [英] What does question mark (?) before type declaration means in php (?int)

问题描述

我在中看到了此代码他们正在使用?int的https://github.com/symfony/symfony/blob/master/src/Symfony/Component/Console/Output/Output.php 行号40.

public function __construct(?int $verbosity = self::VERBOSITY_NORMAL, bool $decorated = false, OutputFormatterInterface $formatter = null)
    {
        $this->verbosity = null === $verbosity ? self::VERBOSITY_NORMAL : $verbosity;
        $this->formatter = $formatter ?: new OutputFormatter();
        $this->formatter->setDecorated($decorated);
    }

推荐答案

它称为Nullable types.

将?int定义为int或null.

现在，可以通过在类型名称前加上问号来将

用于参数和返回值的类型声明标记为可为空.这表示与指定的类型一样，NULL可以分别作为参数传递或作为值返回.

Type declarations for parameters and return values can now be marked as nullable by prefixing the type name with a question mark. This signifies that as well as the specified type, NULL can be passed as an argument, or returned as a value, respectively.

示例:

function nullOrInt(?int $arg){
    var_dump($arg);
}

nullOrInt(100);
nullOrInt(null);

函数nullOrInt将同时接受null和int.

function nullOrInt will accept both null and int.

参考: http://php.net/manual/zh/migration71. new-features.php

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