“对易失性字段的引用不会被视为易失性” 影响
以下代码
using System.Threading;
class Test
{
volatile int counter = 0;
public void Increment()
{
Interlocked.Increment(ref counter);
}
}
引发以下编译器警告：
"A reference to a volatile field will not be treated as volatile"
我在这里做错了什么来引发此警告吗？ 为什么编译器会对此发出警告？
The following code
using System.Threading;
class Test
{
volatile int counter = 0;
public void Increment()
{
Interlocked.Increment(ref counter);
}
}
Raises the following compiler warning:
"A reference to a volatile field will not be treated as volatile"
Am I doing something wrong here to raise this warning? Why does the compiler me warn about this?
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评论（4）
你没有做错任何事。 根据文档：
You are not doing anything wrong. According to the documentation:
基本上警告是，当您通过引用传递易失性字段时，调用代码不知道以易失性方式处理它。 对于 Interlocked.Increment 来说，由于该方法的性质，这可能并不重要 - 但如果您使用 Interlocked，则无论如何您都不需要变量是易失性的。
一般来说，我认为我会避免混合两者 - 如果您使用 Interlocked，请在任何地方（使用
Interlocked.CompareExchange(ref counter, 0, 0)来阅读它）。 我个人不能说我经常使用 volatile。 对于简单的计数器，我可能使用互锁，但对于大多数任务我更有可能使用锁。Basically the warning is that when you pass a volatile field by reference, the calling code doesn't know to treat it in a volatile manner. For Interlocked.Increment that probably doesn't matter, due to the nature of the method - but then you don't need the variable to be volatile anyway if you're using Interlocked.
In general, I think I'd avoid mixing the two - if you're using Interlocked, do it everywhere (using
Interlocked.CompareExchange(ref counter, 0, 0)to read it). I can't say I use volatile very often, personally. For simple counters I might use Interlocked, but I'm more likely to use a lock for most tasks.用这个：
Use this:
您收到错误是因为您通过引用传递字段。 我认为这意味着目标方法不知道该字段被标记为易失性，因此不会将其视为易失性。
You're getting the error because you're passing the field by reference. I think what this means is that the target method has no idea the field is marked as
volatile, and therefore will not treat it as such.