在PHP中通过引用传递函数 [英] Pass a function by reference in PHP

问题描述

是否可以通过引用传递函数?

Is it possible to pass functions by reference?

类似这样的东西:

function call($func){
    $func();
}

function test(){
    echo "hello world!";
}

call(test);

我知道您可以执行'test'，但是我真的不想要那样，因为我需要通过引用传递函数.

I know that you could do 'test', but I don't really want that, as I need to pass the function by reference.

通过匿名函数这样做是唯一的方法吗?

Is the only way to do so via anonymous functions?

说明:如果您从C ++调用，可以通过指针传递一个函数:

Clarification: If you recall from C++, you could pass a function via pointers:

void call(void (*func)(void)){
    func();
}

或者在Python中:

Or in Python:

def call(func):
    func()

这就是我要完成的工作.

That's what i'm trying to accomplish.

推荐答案

对于它的价值，给这样的镜头打些怎么样? (是的，我知道这是一个匿名函数，在帖子中提到了，但是我对大量的答复没有提及闭包/函数感到不满，根本没有对象，所以这主要是为这篇文章中奔跑的人们提供的注释.)

For what it's worth, how about giving something like this a shot? (Yes, I know it's an anonymous function which was mentioned in the post, but I was disgruntled at the abundance of replies that did not mention closures/function-objects at all so this is mostly a note for people running across this post.)

我不使用PHP ，但是 使用闭包 在 PHP 5.3 (但不是PHP 5.2)中似乎可以作为此处演示. 我不确定有什么限制(如果有的话).(据我所知，关闭会吞噬您的孩子.您已被警告.)

I don't use PHP, but using a closure appears to work in PHP 5.3 (but not PHP 5.2) as demonstrated here. I am not sure what the limitations, if any, there are. (For all I know the closure will eat your children. You have been warned.)

function doIt ($fn) {
  echo "doIt\n";
  return $fn();
}

function doMe () {
  echo "doMe\n";
}

// I am using a closure here.
// There may be a more clever way to "get the function-object" representing a given
// named function, but I do not know what it is. Again, I *don't use PHP* :-)
echo doIt(function () { doMe(); });

快乐的编码.

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