通过const引用传递对象？ [英] Passing an object by const reference?

问题描述

在C ++中，当您希望函数能够从对象读取而不希望对其进行修改时，可以将 const 引用传递给该函数。

In C++ when you want a function to be able to read from an object, but not modify it, you pass a const reference to the function. What is the equivalent way of doing this in php?

我知道php5中的对象默认是通过引用传递的，但是出于可读性考虑，我认为我将继续使用&符在变量名之前，像这样：

I know objects in php5 are passed by reference by default, but for readability I think I will continue to use the ampersand before the variable name, like this:

function foo(&$obj)
{

}

推荐答案

如果要传递对象

<?php
function changeObject($obj) {
  $obj->name = 'Mike';
}

$obj = new StdClass;
$obj->name = 'John';

changeObject(clone $obj);
echo $obj->name; // John

changeObject($obj);
echo $obj->name; // Mike

我知道php5中的对象默认是通过引用传递的，但出于可读性考虑，我想我会继续在变量名之前使用&符号

I know objects in php5 are passed by reference by default, but for readability I think I will continue to use the ampersand before the variable name

这是您的电话，但我发现这样做只会它读起来更像C ++。或者，您可以通过在函数定义中使用类型提示来显示正在传递对象：

That's your call but I find that would simply make it read more like C++. Alternativly, you can show that an object is being passed in by using type-hinting in your function definition:

function foo(StdClass $obj)
{

}

很明显， $ obj 是一个对象，可以假定它已通过引用传递。

Once it's clear that $obj is an object it can be assumed that it's being passed by reference.

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