比预期更多的参数:PHP 中没有错误? [英] More parameter than expected: no error in PHP?

问题描述

请看这段代码，我向自定义函数传递了比预期更多的参数:

Please see this code, where I pass more parameters than expected to a custom function:

error_reporting(E_ALL);
ini_set("display_errors", 1);

daidai("asassa", "asgfasfas", "asassa");

function daidai($aa)
{
    echo $aa;
}

这根本不会发出错误，而我期待 警告:函数 daidai() 最多需要 1 个参数，给定 3 个

This doesn't emit an error at all, while I was expecting Warning: function daidai() expects at most 1 parameter, 3 given

令我困惑的是，这会按预期发出上述错误:

What puzzles me is that this emits said error as expected:

$Odate=new DateTime();
$sfasfasf=$Odate->setTime("23", "59", "30", "unexpected");

为什么?

推荐答案

那是因为 setTime() 方法本身会检查参数的数量，如果你想在你的函数中使用它，你必须自己实现它.

That is because the setTime() method itself performs a check on the number of parameters, if you want it in your functions, you have to implement it yourself.

作为参考，您可以查看 文档 的 func_num_args(), func_get_arg() 和 func_get_args()

For reference you might look into the docs for func_num_args(), func_get_arg() and func_get_args()

来自文档:

不需要特殊语法来说明函数是可变参数的；但是访问函数的参数必须使用func_num_args()，func_get_arg() 和 func_get_args().

No special syntax is required to note that a function is variadic; however access to the function's arguments must use func_num_args(), func_get_arg() and func_get_args().

因此，php 确实将每个函数都视为可变参数.

Hence, php really treats every function as a variadic one.

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