[英]Static initialization in c++ and thread safety
类实例的静态初始化不是线程安全的。 下面的代码是不应该做的事情的一个例子:
extern int computesomething();
class cachedcomputation
{
public:
cachedcomputation()
{
result = computesomething();
}
int result;
};
void usecached()
{
static cachedcomputation c;
// use of c.result - may break
}
但是,下面的代码是否是线程安全的? (忽略解决方案的丑陋)何时或为何会破坏?
extern int computesomething();
class cachedcomputation
{
public:
cachedcomputation()
{
if(0==InterlockedCompareExchange(&mutex, 1, 0))
{
// first thread
result = computesomething();
InterlockedExchange(&mutex, 2);
}
else
{
// other thread - spinlock until mutex==2
while(2!=InterlockedCompareExchange(&mutex, 2, 2)){}
}
}
int result;
private:
long mutex;
};
void usecached()
{
static cachedcomputation c;
// use of c.result - ???
}
你需要:
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