将数组作为 `params` 参数传递

问题描述

我有以下方法:

void MyMethod(params object[] args)
{
}

我试图用 object[] 类型的参数调用它:

which I am trying to call with a parameter of type object[]:

object[] myArgs = GetArgs();
MyMethod(myArgs);

它编译得很好，但在 MyMethod 内部我 args == { myArgs}，即一个数组，其中一个元素是我的原始参数.显然我想要 args = myArgs，我做错了什么?

It compiles fine, but inside MyMethod I args == { myArgs}, i.e. an array with one element that is my original arguments. Obviously I wanted to have args = myArgs, what am I doing wrong?

Jon Skeet 实际上是对的，GetArgs() 确实将这个东西包装在一个单元素数组中，抱歉问了一个愚蠢的问题.

Jon Skeet was actually right, the GetArgs() did wrap the thing in an one element array, sorry for stupid question.

推荐答案

你所描述的根本没有发生.除非需要，编译器不会不创建包装器数组.这是一个简短但完整的程序来演示这一点:

What you've described simply doesn't happen. The compiler does not create a wrapper array unless it needs to. Here's a short but complete program demonstrating this:

using System;

class Test
{
    static void MyMethod(params object[] args)
    {
        Console.WriteLine(args.Length);
    }

    static void Main()
    {
        object[] args = { "foo", "bar", "baz" };
        MyMethod(args);
    }
}

根据您的问题，这会打印 1 - 但它不会打印 3.args 的值直接传递给 MyMethod，没有进一步扩展.

According to your question, this would print 1 - but it doesn't, it prints 3. The value of args is passed directly to MyMethod, with no further expansion.

要么您的代码与您发布的不同，要么包装"发生在 GetArgs 中.

Either your code isn't as you've posted it, or the "wrapping" occurs within GetArgs.

您可以通过将 args 强制转换为 object 来强制包裹它.例如，如果我将 Main 的最后一行更改为:

You can force it to wrap by casting args to object. For example, if I change the last line of Main to:

MyMethod((object) args);

... then 它打印 1，因为它有效地调用了 MyMethod(new object[] { args }).

... then it prints 1, because it's effectively calling MyMethod(new object[] { args }).

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