如何更换'调用时间按引用传递关于 PHP

发布于 2025-01-06 13:38:00 字数 454 浏览 1 评论 0原文

考虑到 PHP 已经弃用了一些函数和功能，我想重构我的代码以适应 php 5.3。

现在，我必须消除所有“调用时间传递引用”。所以，我有三个问题：

1 - 如果我替换：

$myclass->myfunc(&$myvar);

$myvar = $myclass->myfunc($myvar);

会起作用吗？

2 - 如果我有类似的情况该怎么办？

$myclass->myfunc(&$myVar, &$ourvar);

3 -

$x = &new myclass();

谢谢您的宝贵时间，我们将非常感谢您的帮助

原文

Considering that PHP has already deprecated a few functions and functionallitys, I would like to refactory my codes to fit it on php 5.3.

now, I have to eliminate all the 'Call-time pass-by-reference' . So, I have three questions:

1 - If I replace:

$myclass->myfunc(&$myvar);

$myvar = $myclass->myfunc($myvar);

will work ?

2 - what do I do if I have something like that?

$myclass->myfunc(&$myVar, &$ourvar);

3 - How about

$x = &new myclass();

Thanks for your time, any help will be very appreciated

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一人独醉 2025-01-13 13:38:00

首先，您熟悉什么是通话时间传递尊重吗？正常的按引用传递是通过在其声明中的参数前面添加 & 来声明某些参数按引用传递的函数来完成的。

调用时按引用传递意味着该函数被声明为按值获取这些参数，并且您在事后将其行为更改为按引用传递。通过引用传递调用时间实际上并不是必要的。每个函数都应该有一个特定的目的，并且为了正确实现其目的，它应该始终通过引用获取参数，或者始终通过值获取参数。让一个函数做一些它没有设计目的的事情是不好的。

回答有关 $myclass->myfunc(&$myvar); 和 $myclass->myfunc(&$myVar, &$ourvar); 的问题> 我想说的是，如果您需要通过引用传递该函数，那么应该将其声明为始终通过引用传递。即

function myfunc(&$x, &$y) { ... }

然后要使用它，您只需调用它而无需 &

$myclass->myfunc($myVar, $ourvar);

$x = &new myclass(); 是完全无关的。你没有传递任何东西。它仍然是有效的语法。

First, are you familiar with what call-time pass by reverence is? Normal pass by reference is accomplished by the function declaring that certain arguments are pass by reference by prepending a & to the parameter in its declaration.

Call-time pass by reference means that the function is declared to take those arguments by value, and you are changing its behavior to pass-by-reference after the fact. Call-time pass by reference shouldn't really ever be necessary. Every function should have a specific purpose, and to correctly accomplish its purpose, it should either always take an argument by reference, or always take it by value. It is bad to make a function do something it was not designed to do.

Responding to your questions about $myclass->myfunc(&$myvar); and $myclass->myfunc(&$myVar, &$ourvar); I would say that, if you need to pass by reference to that function, then it should be declared as always pass by reference. i.e.