我认为没有比Eric Lippert更好的人来回答这个问题了(在原文中强调):

In C#, "volatile" means not only "make sure that the compiler and the jitter do not perform any code reordering or register caching optimizations on this variable". It also means "tell the processors to do whatever it is they need to do to ensure that I am reading the latest value, even if that means halting other processors and making them synchronize main memory with their caches". Actually, that last bit is a lie. The true semantics of volatile reads and writes are considerably more complex than I've outlined here; in fact they do not actually guarantee that every processor stops what it is doing and updates caches to/from main memory. Rather, they provide weaker guarantees about how memory accesses before and after reads and writes may be observed to be ordered with respect to each other. Certain operations such as creating a new thread, entering a lock, or using one of the Interlocked family of methods introduce stronger guarantees about observation of ordering. If you want more details, read sections 3.10 and 10.5.3 of the C# 4.0 specification. Frankly, I discourage you from ever making a volatile field. Volatile fields are a sign that you are doing something downright crazy: you're attempting to read and write the same value on two different threads without putting a lock in place. Locks guarantee that memory read or modified inside the lock is observed to be consistent, locks guarantee that only one thread accesses a given chunk of memory at a time, and so on. The number of situations in which a lock is too slow is very small, and the probability that you are going to get the code wrong because you don't understand the exact memory model is very large. I don't attempt to write any low-lock code except for the most trivial usages of Interlocked operations. I leave the usage of "volatile" to real experts.

欲进一步阅读，请参阅:

理解低锁技术在多线程应用中的影响 再见不稳定

编译器有时会改变代码中语句的顺序来优化它。通常这在单线程环境中不是问题，但在多线程环境中可能是问题。请看下面的例子:

 private static int _flag = 0;
 private static int _value = 0;

 var t1 = Task.Run(() =>
 {
     _value = 10; /* compiler could switch these lines */
     _flag = 5;
 });

 var t2 = Task.Run(() =>
 {
     if (_flag == 5)
     {
         Console.WriteLine("Value: {0}", _value);
     }
 });

如果运行t1和t2，您将期望没有输出或结果为“Value: 10”。可能是编译器在t1函数内部切换了行。如果t2执行，可能是_flag值为5，而_value值为0。因此，预期的逻辑可能会被打破。

为了解决这个问题，你可以使用volatile关键字，你可以应用到字段。此语句禁用编译器优化，以便您可以强制代码中的正确顺序。

private static volatile int _flag = 0;

只有在真正需要时才应该使用volatile，因为它会禁用某些编译器优化，这会影响性能。它也不是所有。net语言都支持(Visual Basic不支持)，因此它阻碍了语言互操作性。

只需查看volatile关键字的官方页面，您就可以看到典型用法的示例。

public class Worker
{
    public void DoWork()
    {
        bool work = false;
        while (!_shouldStop)
        {
            work = !work; // simulate some work
        }
        Console.WriteLine("Worker thread: terminating gracefully.");
    }
    public void RequestStop()
    {
        _shouldStop = true;
    }
    
    private volatile bool _shouldStop;
}

将volatile修饰符添加到_shouldStop的声明中，您将总是得到相同的结果。但是，如果_shouldStop成员上没有这个修饰符，行为是不可预测的。

所以这绝对不是完全疯狂的事情。

存在缓存一致性，负责CPU缓存的一致性。

如果CPU采用强内存模型(如x86)

因此，volatile字段的读写在x86上不需要特殊的指令:普通的读写(例如，使用MOV指令)就足够了。

示例来自c# 5.0规范(第10.5.3章)

using System;
using System.Threading;
class Test
{
    public static int result;   
    public static volatile bool finished;
    static void Thread2() {
        result = 143;    
        finished = true; 
    }
    static void Main() {

        finished = false;
        new Thread(new ThreadStart(Thread2)).Start();

        for (;;) {
            if (finished) {
                Console.WriteLine("result = {0}", result);
                return;
            }
        }
    }
}

产生输出:result = 143

如果finished字段没有被声明为volatile，那么在store to finished之后，允许主线程可以看到store to result，因此主线程可以从字段结果中读取值0。

Volatile行为依赖于平台，所以你应该在需要的时候考虑使用Volatile，以确保它能满足你的需求。

即使是volatile也不能阻止(所有类型的)重排序(c# - c#内存模型的理论与实践，第2部分)

尽管对A的写操作是不稳定的，从A_Won的读操作也是不稳定的，但栅栏都是单向的，实际上允许这种重新排序。

所以我相信，如果你想知道什么时候使用volatile (vs lock vs Interlocked)，你应该熟悉内存围栏(full, half)和同步的需求。那为了你自己，你自己去找答案吧。

有时候，编译器会优化一个字段并使用寄存器来存储它。如果线程1写了字段，而另一个线程访问了它，因为更新存储在寄存器(而不是内存)中，第二个线程将得到陈旧的数据。

你可以把volatile关键字看作是对编译器说“我想让你把这个值存储在内存中”。这保证了第二个线程检索到最新的值。

如果你想稍微了解一下volatile关键字的功能，可以考虑以下程序(我使用的是DevStudio 2005):

#include <iostream>
void main()
{
  int j = 0;
  for (int i = 0 ; i < 100 ; ++i)
  {
    j += i;
  }
  for (volatile int i = 0 ; i < 100 ; ++i)
  {
    j += i;
  }
  std::cout << j;
}

使用标准的优化(发布)编译器设置，编译器创建以下汇编器(IA32):

void main()
{
00401000  push        ecx  
  int j = 0;
00401001  xor         ecx,ecx 
  for (int i = 0 ; i < 100 ; ++i)
00401003  xor         eax,eax 
00401005  mov         edx,1 
0040100A  lea         ebx,[ebx] 
  {
    j += i;
00401010  add         ecx,eax 
00401012  add         eax,edx 
00401014  cmp         eax,64h 
00401017  jl          main+10h (401010h) 
  }
  for (volatile int i = 0 ; i < 100 ; ++i)
00401019  mov         dword ptr [esp],0 
00401020  mov         eax,dword ptr [esp] 
00401023  cmp         eax,64h 
00401026  jge         main+3Eh (40103Eh) 
00401028  jmp         main+30h (401030h) 
0040102A  lea         ebx,[ebx] 
  {
    j += i;
00401030  add         ecx,dword ptr [esp] 
00401033  add         dword ptr [esp],edx 
00401036  mov         eax,dword ptr [esp] 
00401039  cmp         eax,64h 
0040103C  jl          main+30h (401030h) 
  }
  std::cout << j;
0040103E  push        ecx  
0040103F  mov         ecx,dword ptr [__imp_std::cout (40203Ch)] 
00401045  call        dword ptr [__imp_std::basic_ostream<char,std::char_traits<char> >::operator<< (402038h)] 
}
0040104B  xor         eax,eax 
0040104D  pop         ecx  
0040104E  ret              

Looking at the output, the compiler has decided to use the ecx register to store the value of the j variable. For the non-volatile loop (the first) the compiler has assigned i to the eax register. Fairly straightforward. There are a couple of interesting bits though - the lea ebx,[ebx] instruction is effectively a multibyte nop instruction so that the loop jumps to a 16 byte aligned memory address. The other is the use of edx to increment the loop counter instead of using an inc eax instruction. The add reg,reg instruction has lower latency on a few IA32 cores compared to the inc reg instruction, but never has higher latency.

Now for the loop with the volatile loop counter. The counter is stored at [esp] and the volatile keyword tells the compiler the value should always be read from/written to memory and never assigned to a register. The compiler even goes so far as to not do a load/increment/store as three distinct steps (load eax, inc eax, save eax) when updating the counter value, instead the memory is directly modified in a single instruction (an add mem,reg). The way the code has been created ensures the value of the loop counter is always up-to-date within the context of a single CPU core. No operation on the data can result in corruption or data loss (hence not using the load/inc/store since the value can change during the inc thus being lost on the store). Since interrupts can only be serviced once the current instruction has completed, the data can never be corrupted, even with unaligned memory.

Once you introduce a second CPU to the system, the volatile keyword won't guard against the data being updated by another CPU at the same time. In the above example, you would need the data to be unaligned to get a potential corruption. The volatile keyword won't prevent potential corruption if the data cannot be handled atomically, for example, if the loop counter was of type long long (64 bits) then it would require two 32 bit operations to update the value, in the middle of which an interrupt can occur and change the data.

因此，volatile关键字只适用于小于或等于本机寄存器大小的对齐数据，这样操作总是原子的。

volatile关键字被设想用于IO操作，其中IO将不断变化，但有一个恒定的地址，例如内存映射的UART设备，编译器不应该一直重用从地址中读取的第一个值。

如果要处理大数据或有多个cpu，则需要更高级别(OS)的锁定系统来正确处理数据访问。

多个线程可以访问一个变量。 最新的更新将在变量上