类型声明前的问号 (?) 在 php (?int) 中意味着什么

Question

我在第 40 行的 https://github.com/symfony/symfony/blob/master/src/Symfony/Component/Console/Output/Output.php 中看到了这段代码，他们正在使用 ?int。

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);
    }

Answer 1

叫做Nullable types。

将 ?int 定义为 int 或 null。

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);

function nullOrInt 将接受 null 和 int。

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

类型声明前的问号 (?) 在 php (?int) 中意味着什么

What does question mark (?) before type declaration means in php (?int)

php

php-7.2