PHP:将匿名函数作为参数传递 [英] PHP: Pass anonymous function as argument

问题描述

是否可以将匿名函数作为参数传递并使其立即执行，从而传递函数的return 值?

Is it possible to pass an anonymous function as an argument, and have it execute immediately, thus passing the function's return value?

function myFunction(Array $data){
    print_r($data);
}

myFunction(function(){
    $data = array(
        'fruit'     => 'apple',
        'vegetable' => 'broccoli',
        'other'     => 'canned soup');
    return $data;
});

由于 Array 类型提示，这会引发错误，抱怨正在传递对象.好吧，如果我去掉类型提示，它当然会吐出Closure Object，而不是我想要的结果.我知道我在技术上将 Closure 的对象实例传递给 myFunction，但是，我几乎可以肯定我已经在其他地方看到了这一点.这可能吗?如果是这样，我做错了什么?

This throws an error due to the Array type-hint, complaining of an object being passed. Alright, if I remove the type-hint, it of course spits out Closure Object, rather than the results I want. I understand that I am technically passing an object instance of Closure to myFunction, however, I'm near certain that I've seen this accomplished elsewhere. Is this possible? If so, what am I doing wrong?

为了讨论起见，我不能修改传递闭包的函数.

For the sake of this discussion, I cannot modify the function to which I'm passing the closure.

tl;dr:如何将匿名函数声明作为参数传递，从而导致返回值作为参数传递.

PS:如果不清楚，希望的输出是:

PS: If not clear, the desired output is:

Array
(
    [fruit] => apple
    [vegetable] => broccoli
    [other] => canned soup
)

推荐答案

你不能.你必须先打电话给它.而且由于 PHP 尚不支持闭包解除引用，因此您必须先将其存储在变量中:

You can't. You'd have to call it first. And since PHP doesn't support closure de-referencing yet, you'd have to store it in a variable first:

$f = function(){
    $data = array(
        'fruit'     => 'apple',
        'vegetable' => 'broccoli',
        'other'     => 'canned soup');
    return $data;
};
myfunction($f());

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