PHP 类型提示被忽略，没有抛出 TypeError 异常 [英] PHP type hinting is being ignored, no TypeError exception is thrown

问题描述

我刚刚调试了一些 PHP 7.1 代码，其中我忘记删除对返回值的 (bool) 强制转换，但该函数的声明返回类型为 int:

I just debugged some PHP 7.1 code where I'd forgotten to remove a (bool) cast on the return value, but the function had a declared return type of int:

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

$foo = get_foo();

在此处测试代码.

在bool 和int 之间转换当然很容易，但是为什么这不抛出TypeError?

It's easy to cast between bool and int, of course, but why didn't this throw a TypeError?

在三种情况下可能会抛出 TypeError....第二种情况是从函数返回的值与声明的函数返回类型不匹配.

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.

对于类型化的函数参数表现出相同的行为.

The same behaviour is exhibited for typed function parameters.

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

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

推荐答案

您需要添加 strict_types 指令，类型提示才能正常工作

You need to add the strict_types directive for type hinting to work properly

引用自 PHP 7 新功能

要启用严格模式，必须在文件顶部放置一个声明指令.这意味着标量输入的严格性是在每个文件的基础上配置的.该指令不仅会影响参数的类型声明，还会影响函数的返回类型(请参阅返回类型声明、内置 PHP 函数和来自加载扩展的函数.

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.

With that, you should do this.

<?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，你会收到一个Uncaught TypeError Exception

Try this once again, you will receive an Uncaught TypeError Exception

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