使用函数指针和静态变量时是否需要readlock？

Question

我们是否需要一个读锁来读取除变量小于memalign之外的整数以及某些缓存情况？

Answer 1

The analyser assumes that because the address of f1() has been taken, it could be called through that function pointer from any context. If, for example, you have one thread updating x :

lock xlock;
update x;
unlock xlock;

and another thread calls f1() simultaneously through the function pointer, the access to f1() in that second thread won't be protected by the lock. It could therefore see a partial update of x , or see the update incorrectly ordered with respect to other updates.

The reason why the function pointer matters is that if the address of f1() is never taken, then the analyser can determine exactly where the function is called from. When the address has been taken, it has to assume the worst case.

Answer 2

From the text you quoted (emphasis mine):

If a function has had its address taken, then we presume that we do not know the context of every call made upon that function .

Because lint cannot determine every context that a call to f1 could be made in, it assumes that it could be called in any context, including those in which accessing x results in some data race.

Reasonably, you probably won't hit anything dire by returning x . But it's a probability, and the whole point of using lint is to reduce uncertainty with respect to how your code functions with unexpected inputs.