将未设置的变量传递给函数 [英] Passing unset variables to functions

问题描述

我的代码:

function Check($Variable, $DefaultValue) {
    if(isset($Variable) && $Variable != "" && $Variable != NULL) {
        return $Variable;
    }
    else {
        return $DefaultValue;
    }
}

$a = Check(@$foo, false);
$b = Check(@$bar, "Hello");

//$a now equals false because $foo was not set.
//$b now equals "Hello" because $bar was not set.

1. 当变量不存在并传递给函数(抑制错误)时，实际传递的是什么?
2. 此功能可能表现出任何未定义的行为吗?
3. 是否有更好的方法包装测试变量是否存在并提供函数的默认值?
4. isset()在测试变量时会在后台检查什么?

1. When a variable doesn't exist and is passed to the function (suppressing the error) what is actually passed?
2. Is there any undefined behaviour that this function could exhibit?
3. Is there a better way of wrapping the testing for variable existence and supplying a default value from a function?
4. What does isset() check under the hood when testing a variable?

---

默认值由用户定义.有时是数字，有时是字符串.

The default value is there to be user defined. Sometimes it will be a number, sometimes a string.

推荐答案

1. 传递了NULL，引发通知错误.
2. 不，函数只能看到它的参数(它不在乎如何调用)
3. 您可以轻松指定默认值-函数func($ mandatory，$ optional ='default value');
4. 在函数的参数上进行设置是没有意义的，因为参数已在函数签名中设置.

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