PHP 7和严格的“资源"类型 [英] PHP 7 and strict "resource" types

问题描述

PHP 7是否支持严格输入资源?如果可以，怎么办?

Does PHP 7 support strict typing for resources? If so, how?

例如:

    declare (strict_types=1);

    $ch = curl_init ();
    test ($ch);

    function test (resource $ch)
    {

    }

上面将给出错误:

致命错误:未捕获的TypeError:传递给test()的参数1必须是资源的实例，资源已给定

Fatal error: Uncaught TypeError: Argument 1 passed to test() must be an instance of resource, resource given

$ch上的var_dump显示它为 resource(4，curl)，并且手册中说curl_init ()返回资源.

A var_dump on $ch reveals it to be resource(4, curl), and the manual says curl_init () returns a resource.

是否完全可以严格键入test()函数以支持$ch variable?

Is it at all possible to strictly type the test() function to support the $ch variable?

推荐答案

PHP没有资源的类型提示因为

PHP does not have a type hint for resources because

不添加资源的类型提示，因为这将阻止从资源移动到现有扩展的对象，某些扩展已经完成了这些操作(例如GMP).

No type hint for resources is added, as this would prevent moving from resources to objects for existing extensions, which some have already done (e.g. GMP).

但是，您可以在函数/方法主体中使用 is_resource() 来验证传递的参数并根据需要进行处理.可重用的版本将是这样的断言:

However, you can use is_resource() within the function/method body to verify the passed argument and handle it as needed. A reusable version would be an assertion like this:

function assert_resource($resource)
{
    if (false === is_resource($resource)) {
        throw new InvalidArgumentException(
            sprintf(
                'Argument must be a valid resource type. %s given.',
                gettype($resource)
            )
        );
    }
}

然后您可以在代码中像这样使用

which you could then use within your code like that:

function test($ch)
{
    assert_resource($ch);
    // do something with resource
}

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