Race conditions occur in multi-threaded applications or multi-process systems. A race condition, at its most basic, is anything that makes the assumption that two things not in the same thread or process will happen in a particular order, without taking steps to ensure that they do. This happens commonly when two threads are passing messages by setting and checking member variables of a class both can access. There's almost always a race condition when one thread calls sleep to give another thread time to finish a task (unless that sleep is in a loop, with some checking mechanism).

防止竞争条件的工具依赖于语言和操作系统，但一些常见的工具是互斥锁、临界区和信号。互斥锁在你想确保你是唯一一个在做某事的时候很有用。当你想确保别人已经完成某件事时，信号是很好的。最小化共享资源还有助于防止意外行为

Detecting race conditions can be difficult, but there are a couple signs. Code which relies heavily on sleeps is prone to race conditions, so first check for calls to sleep in the affected code. Adding particularly long sleeps can also be used for debugging to try and force a particular order of events. This can be useful for reproducing the behavior, seeing if you can make it disappear by changing the timing of things, and for testing solutions put in place. The sleeps should be removed after debugging.

但是，如果某个问题只在某些机器上断断续续地发生，则是存在竞争条件的标志性标志。常见的错误是崩溃和死锁。使用日志记录，您应该能够找到受影响的区域并从那里返回。

竞态条件是一种bug，只会在特定的时间条件下发生。

例子: 假设您有两个线程，A和B。

在线程A中:

if( object.a != 0 )
    object.avg = total / object.a

线程B:

object.a = 0

如果线程A在检查完对象后被抢占。a不为空，B将执行a = 0，当线程a将获得处理器时，它将执行“除零”。

这个错误只发生在if语句之后的线程A被抢占时，这是非常罕见的，但它是有可能发生的。

竞态条件是并发编程中的一种情况，其中两个并发线程或进程争夺资源，最终状态取决于谁先获得资源。

当访问共享资源的多线程(或其他并行)代码可能以导致意外结果的方式访问共享资源时，就存在“竞争条件”。

举个例子:

for ( int i = 0; i < 10000000; i++ )
{
   x = x + 1; 
}

如果你有5个线程同时执行这段代码，x的值最终不会是50,000,000。事实上，它会随着每一次运行而变化。

这是因为，为了让每个线程增加x的值，它们必须做以下事情:(显然是简化的)

Retrieve the value of x
Add 1 to this value
Store this value to x

任何线程都可以在任何时间处于此进程的任何步骤，并且当涉及共享资源时，它们可以相互踩。在读取x和写回x之间的时间内，x的状态可以由另一个线程改变。

假设一个线程获取了x的值，但还没有存储它。另一个线程也可以检索相同的x值(因为还没有线程更改它)，然后它们都将在x中存储相同的值(x+1) !

例子:

Thread 1: reads x, value is 7
Thread 1: add 1 to x, value is now 8
Thread 2: reads x, value is 7
Thread 1: stores 8 in x
Thread 2: adds 1 to x, value is now 8
Thread 2: stores 8 in x

竞争条件可以通过在代码访问共享资源之前使用某种锁定机制来避免:

for ( int i = 0; i < 10000000; i++ )
{
   //lock x
   x = x + 1; 
   //unlock x
}

这里，答案每次都是50,000,000。

有关锁的更多信息，请搜索:互斥量，信号量，临界区，共享资源。

微软实际上已经发布了一篇关于竞态条件和死锁的非常详细的文章。最概括的摘要是标题段:

A race condition occurs when two threads access a shared variable at the same time. The first thread reads the variable, and the second thread reads the same value from the variable. Then the first thread and second thread perform their operations on the value, and they race to see which thread can write the value last to the shared variable. The value of the thread that writes its value last is preserved, because the thread is writing over the value that the previous thread wrote.

一个有点规范的定义是“当两个线程同时访问内存中的同一个位置，并且至少有一次访问是写操作。”在这种情况下，“reader”线程可能获得旧值或新值，这取决于哪个线程“赢得了比赛”。这并不总是一个bug——事实上，一些非常复杂的低级算法会故意这样做——但通常应该避免。@Steve Gury的例子很好地说明了这可能是个问题。