请确认这是一个MSFT错误 [英] Please confirm this is a MSFT bug

问题描述

你好，


在我通知MSFT之前，有人可以确认这实际上是一个编译器错误（它好像b $ b似乎很明显，但我''对C＃来说相当新。以下代码

（错误地）生成编译器错误CS0177（在控制离开当前方法之前，必须将out参数''Whatever''

分配给。）


public void SomeFunc（out string Whatever）

{

bool Continue = ShouldWeContinue（）;

if （继续）

{

Whatever =" Continue";

}


if（b） ！继续）

{

无论什么=没有继续;

}

}

解决方案

Jake，


编译错误是正确的 - 您必须为在离开方法之前''无论''

。您的代码可以很容易地更新来执行此操作

如此：


bool Continue = ShouldWeContinue（）;

if（继续）

Whatever =继续;

其他

无论什么=没有继续;


希望这会有所帮助。

Dan Manges


Jake Forson写道：

你好，
在我通知MSFT之前，有人可以确认这实际上是一个编译器错误（它看起来非常明显，但我对C＃来说相当新）。以下代码
（错误地）生成编译器错误CS0177（在控制离开当前方法之前必须指定out参数''Whatever''
。

public void SomeFunc （out string Whatever）
{
bool Continue = ShouldWeContinue（）;
if（Continue）
{
Whatever =" Continue";
}

如果（！继续）
{
随便=没有继续;
}
}

> Jake，

编译错误是正确的 - 在离开方法之前，必须为''Whatever''
分配一个值。您的代码可以很容易地更新来执行此操作
如此：

bool Continue = ShouldWeContinue（）;
if（继续）
Whatever =" Continue" ;;

Whatever =没有继续;

希望这会有所帮助。

谢谢你反馈。它确实在所有情况下分配它，但这就是为什么

我认为它是一个错误。

正如所写，是的 - 我们可以观察到它它总是会有一个值。

它退出的时间。


但是：如果变量是一个字段，或者变量是用于一个（2.0） ）

匿名委托（作为/ capture /变量），那么

（在代码中没有显而易见）另一个线程可以改变值

两次通话之间继续，顺便说一下可能会或可能不会发现

，具体取决于波动性/记忆障碍。鉴于此，对于我来说，编译器遵循保持简单的意义对我来说很有意义。原则，并且

平等地考虑所有三种情况 - 否则你必须记住愚蠢的

规则，了解每种情况何时适用，并做出改变（比如在
anonymous delegate）会突然导致无关的代码无法编译。


这有意义吗？


Marc

Hi there,

Before I notify MSFT, can someone confirm this is in fact a compiler bug (it
seems pretty obvious but I''m fairly new to C#). The following code
(erroneously) generates compiler error CS0177 (The out parameter ''Whatever''
must be assigned to before control leaves the current method).

public void SomeFunc(out string Whatever)
{
bool Continue = ShouldWeContinue();
if (Continue)
{
Whatever = "Continue";
}

if (!Continue)
{
Whatever = "Didn''t continue";
}
}

解决方案

Jake,

The compiler error is correct - you must assign a value to ''Whatever''
before leaving the method. Your code could easily be updated to do this
as so:

bool Continue = ShouldWeContinue();
if (Continue)
Whatever = "Continue";
else
Whatever = "Didn''t continue";

Hope this helps.
Dan Manges

Jake Forson wrote:

Hi there,

Before I notify MSFT, can someone confirm this is in fact a compiler bug (it
seems pretty obvious but I''m fairly new to C#). The following code
(erroneously) generates compiler error CS0177 (The out parameter ''Whatever''
must be assigned to before control leaves the current method).

public void SomeFunc(out string Whatever)
{
bool Continue = ShouldWeContinue();
if (Continue)
{
Whatever = "Continue";
}

if (!Continue)
{
Whatever = "Didn''t continue";
}
}

> Jake,

The compiler error is correct - you must assign a value to ''Whatever''
before leaving the method. Your code could easily be updated to do this
as so:

bool Continue = ShouldWeContinue();
if (Continue)
Whatever = "Continue";
else
Whatever = "Didn''t continue";

Hope this helps.

Thanks for the feedback. It does assign it in all cases however which is why
I consider it a bug.

As written, yes - we can observe that it will always have a value by the
time it exits.

However: if the variable was a field, or if the variable was used in a (2.0)
anonymous delegate (as a /captured/ variable), then it would be possible for
(without it being obvious in the code) another thread to alter the value of
Continue between the two calls, which incidentally may or may-not be spotted
depending on volatility / memory-barriers. Given this, it makes a lot of
sense for me for the compiler to follow the "keep it simple" principle, and
consider all three cases equally - otherwise you would have to remember daft
rules about when each applies, and making a change (like using it in an
anonymous delegate) would suddenly cause unrelated code to not compile.

Does that make sense?

Marc

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