澄清PHP手册;通过引用传递的默认值 [英] Clarification of PHP manual; default values passed by reference

问题描述

由于PHP手册中的不一致（我已经在之前发布过），我只是询问一些澄清。

Due to inconsistencies in the PHP manual (as I've posted about before) I'm just inquiring about some clarification.

功能参数页面（ http：// ca2.php.net/manual/en/functions.arguments.php ）具有以下注意事项：

The Function Arguments page (http://ca2.php.net/manual/en/functions.arguments.php) has the following note:

注意：从PHP 5起，默认值可以通过引用传递。

Note: As of PHP 5, default values may be passed by reference.

现在，我假设这只是意味着以下语法是可以接受的：

Now, I assume this simply means that the following syntax is acceptable:

function foo(&$bar = null){
    // ...
}

然而，再次由于其他不一致，我想知道也许这与其他东西有关。

However, again due to other inconsistencies, I was wondering if perhaps this pertains to something else.

推荐答案

这意味着在PHP 4中，对引用传递的参数使用默认值会导致在解析错误中：

It means that in PHP 4, using a default value for arguments passed by reference would result in a parse error:

Parse error: syntax error, unexpected '=', expecting ')' in ...

演示

在PHP5中，当没有参数被传递时，你的函数将有一个普通的局部变量叫做$ code> $ bar 初始化为 null 。

In PHP5, when no argument is passed, your function will have a normal local variable called $bar initialized to null.

应该重写：

注意：从PHP 5开始，函数声明可以定义引用传递参数的默认值。

Note: As of PHP 5, function declarations may define a default value for argument passed by reference.

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