为什么 yield return 不能出现在带有 catch 的 try 块中?

问题描述

以下是可以的:

try
{
    Console.WriteLine("Before");

    yield return 1;

    Console.WriteLine("After");
}
finally
{
    Console.WriteLine("Done");
}

finally 块在整个事情执行完毕后运行(IEnumerator 支持 IDisposable 以提供一种方法来确保这当枚举在完成之前被放弃时).

The finally block runs when the whole thing has finished executing (IEnumerator<T> supports IDisposable to provide a way to ensure this even when the enumeration is abandoned before it finishes).

但这不行:

try
{
    Console.WriteLine("Before");

    yield return 1;  // error CS1626: Cannot yield a value in the body of a try block with a catch clause

    Console.WriteLine("After");
}
catch (Exception e)
{
    Console.WriteLine(e.Message);
}

假设(为了论证)try 块内的一个或另一个 WriteLine 调用抛出异常.在 catch 块中继续执行有什么问题?

Suppose (for the sake of argument) that an exception is thrown by one or other of the WriteLine calls inside the try block. What's the problem with continuing the execution in catch block?

当然，yield return 部分(目前)无法抛出任何东西，但是为什么这会阻止我们使用封闭的 try/catch 来处理异常在 yield return 之前或之后抛出?

Of course, the yield return part is (currently) unable to throw anything, but why should that stop us from having an enclosing try/catch to deal with exceptions thrown before or after a yield return?

更新:有一个来自 Eric Lippert 的有趣评论 - 似乎他们已经在正确实施 try/finally 行为时遇到了足够多的问题！

Update: There's an interesting comment from Eric Lippert here - seems that they already have enough problems implementing the try/finally behaviour correctly!

有关此错误的 MSDN 页面是:http://msdn.microsoft.com/en-us/library/cs1x15az.aspx.不过，它并没有解释原因.

The MSDN page on this error is: http://msdn.microsoft.com/en-us/library/cs1x15az.aspx. It doesn't explain why, though.

推荐答案

我怀疑这是一个实用性问题而不是可行性问题.我怀疑在极少数情况下，此限制实际上是一个无法解决的问题 - 但是编译器中增加的复杂性将非常重要.

I suspect this is a matter of practicality rather than feasibility. I suspect there are very, very few times where this restriction is actually an issue that can't be worked around - but the added complexity in the compiler would be very significant.

我已经遇到过一些这样的事情:

There are a few things like this that I've already encountered:

• 属性不能通用
• X 无法从 X.Y(X 中的嵌套类)派生
• 使用生成类中的公共字段的迭代器块

在每种情况下，都有可能获得更多的自由，但代价是编译器的额外复杂性.团队做出了务实的选择，为此我为他们鼓掌 - 我宁愿拥有一种具有 99.9% 准确编译器的稍微限制性的语言(是的，存在错误；我前几天在 SO 上遇到了一个)而不是更多无法正确编译的灵活语言.

In each of these cases it would be possible to gain a little bit more freedom, at the cost of extra complexity in the compiler. The team made the pragmatic choice, for which I applaud them - I'd rather have a slightly more restrictive language with a 99.9% accurate compiler (yes, there are bugs; I ran into one on SO just the other day) than a more flexible language which couldn't compile correctly.

这是它为什么可行的伪证明.

Here's a pseudo-proof of how it why it's feasible.

考虑:

• 您可以确保 yield return 部分本身不会抛出异常(预先计算值，然后您只需设置一个字段并返回true")
• 您可以在迭代器块中使用不使用 yield return 的 try/catch.
• 迭代器块中的所有局部变量都是生成类型中的实例变量，因此您可以自由地将代码移动到新方法中

现在变换:

try
{
    Console.WriteLine("a");
    yield return 10;
    Console.WriteLine("b");
}
catch (Something e)
{
    Console.WriteLine("Catch block");
}
Console.WriteLine("Post");

进入(某种伪代码):

case just_before_try_state:
    try
    {
        Console.WriteLine("a");
    }
    catch (Something e)
    {
        CatchBlock();
        goto case post;
    }
    __current = 10;
    return true;

case just_after_yield_return:
    try
    {
        Console.WriteLine("b");
    }
    catch (Something e)
    {
        CatchBlock();
    }
    goto case post;

case post;
    Console.WriteLine("Post");


void CatchBlock()
{
    Console.WriteLine("Catch block");
}

唯一的重复是设置 try/catch 块 - 但这是编译器当然可以做的事情.

The only duplication is in setting up try/catch blocks - but that's something the compiler can certainly do.

我可能在这里遗漏了一些东西 - 如果是，请告诉我！

I may well have missed something here - if so, please let me know!

这篇关于为什么 yield return 不能出现在带有 catch 的 try 块中?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，WP2