问题描述
挥发性似乎是每个人无休止的问题。我想我知道它的一切，但后来我遇到过这样的：
volatile seems to be a never ending question of every one. I thought I knew everything about it, but then I encountered this:
所以，我有一块线程之间共享内存和我定义它是这样的：
So, I have a piece of memory shared between threads and I defined it like this:
volatile type *name;
如果它让你感觉更好，你可以想像键入只是一个 INT 。
If it makes you feel better, you can imagine type is just an int.
这意味着我有一个指针（即不挥发），一些数据是易失。因此，例如，当涉及到优化，编译器可以缓存名称的价值，但不名称[0] 。我说得对不对？
This means I have a pointer (that is not volatile) to some data that are volatile. So, for example when it comes to optimizing, the compiler can cache the value of name but not name[0]. Am I right?
所以，现在我 vfree ING这个指针（这是一个Linux内核模块），它告诉我， vfree 预计常量无效* 当我向它传递挥发型* 。
So, now I am vfreeing this pointer (it's in a Linux kernel module) and it tells me that vfree expects const void * while I am passing it volatile type *.
我明白怎么可以是危险传递一个挥发型* 为键入* ，因为该功能，名称的值[I] 可缓存（作为优化的结果），这是不可取的。
I understand how it can be dangerous to pass a volatile type * as a type * because in that function, the values of name[i] could be cached (as a result of optimization) which is not desirable.
我不明白为什么，虽然， vfree 希望我的指针不一定发送到非易失性数据。是否有我丢失的东西吗？或者是谁写的没有考虑这种情况，它只是球员 vfree ？
I don't understand why though, vfree expects me to send it a pointer necessarily to non-volatile data. Is there something I am missing there? Or is it just the guys who wrote vfree not thinking about this situation?
我想我只是铸造我指针无效* 将不会造成任何伤害，是这样吗？
I assume me simply casting my pointer to void * would not cause any harm, is that right?
推荐答案
我的结论是，只是投放指针无效* 不会造成问题，事实上， 免费和 vfree 不直接接受指针挥发性数据只是一些被忽视了。
My conclusion was that just casting the pointer to void * would not cause a problem and the fact that free and vfree don't directly accept pointers to volatile data is just something that was overlooked.
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