PHP 类型提示被忽略，没有抛出 TypeError 异常

Question

我刚刚调试了一些 PHP 7.1 代码，其中我忘记删除 return 值上的 (bool) 强制转换，但该函数声明了 return int 的类型：

function get_foo(): int {
    $val = get_option('foo');
    return (bool) $val;
}

$foo = get_foo();

Test the code here.

当然，在 bool 和 int 之间转换很容易，但是 为什么这没有抛出 TypeError？

There are three scenarios where a TypeError may be thrown. ... The second is where a value being returned from a function does not match the declared function return type.

类型化函数参数表现出相同的行为。

function get_foo(string $foo): int {
    $val = get_option($foo);
    return (bool) $val;
}

// no TypeError!
$foo = get_foo(34);

Answer 1

您需要添加 strict_types 指令以使类型提示正常工作

引用自PHP 7 New Features

To enable strict mode, a single declare directive must be placed at the top of the file. This means that the strictness of typing for scalars is configured on a per-file basis. This directive not only affects the type declarations of parameters, but also a function's return type (see return type declarations, built-in PHP functions, and functions from loaded extensions.

有了这个，你应该这样做。

<?php
declare(strict_types=1);

function get_option_foo() : int {
    $val = get_option('foo'); // Returns a string that can be parsed as an int
    return (bool) $val;
}

$foo = get_option_foo();

再试一次，您将收到 Uncaught TypeError Exception

PHP 类型提示被忽略，没有抛出 TypeError 异常

PHP type hinting is being ignored, no TypeError exception is thrown

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types

casting

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