ABA问题

ABA问题分析

  /* Naive lock-free stack which suffers from ABA problem.*/  class Stack {    std::atomic<Obj*> top_ptr;    //    // Pops the top object and returns a pointer to it.    //    Obj* Pop() {      while(1) {        Obj* ret_ptr = top_ptr;        if (!ret_ptr) return nullptr;        // For simplicity, suppose that we can ensure that this dereference is safe        // (i.e., that no other thread has popped the stack in the meantime).        Obj* next_ptr = ret_ptr->next;        // If the top node is still ret, then assume no one has changed the stack.        // (That statement is not always true because of the ABA problem)        // Atomically replace top with next.        if (top_ptr.compare_exchange_weak(ret_ptr, next_ptr)) {          return ret_ptr;        }        // The stack has changed, start over.      }    }    //    // Pushes the object specified by obj_ptr to stack.    //    void Push(Obj* obj_ptr) {      while(1) {        Obj* next_ptr = top_ptr;        obj_ptr->next = next_ptr;        // If the top node is still next, then assume no one has changed the stack.        // (That statement is not always true because of the ABA problem)        // Atomically replace top with obj.        if (top_ptr.compare_exchange_weak(next_ptr, obj_ptr)) {          return;        }        // The stack has changed, start over.      }    }  };

举个例子：当前全局变量s ，s的数据是 top->A->B->C

  { // Thread 2 runs pop:    ret = A;    next = B;    compare_exchange_weak(A, B)  // Success, top = B    return A;  } // Now the stack is top → B → C  { // Thread 2 runs pop again:    ret = B;    next = C;    compare_exchange_weak(B, C)  // Success, top = C    return B;  } // Now the stack is top → C  delete B;  { // Thread 2 now pushes A back onto the stack:    A->next = C;    compare_exchange_weak(C, A)  // Success, top = A  }

由例子可以看出 假色全局变量 stack s；加入当前存在两个线程对这个全局变量s进行操作，线程1 调用了pop 当执行到

 Obj* next_ptr = ret_ptr->next;

的时候线程1 被挂起，此时线程2进行pop操作 结果变为 top->B->C 然而2继续pop  则s变为 top->C 此时线程2进行push A操作则 堆栈变为top->A->C. 线程2执行完， 线程1被唤醒，那么线程1进行

top_ptr.compare_exchange_weak(ret_ptr, next_ptr)

操作 发现 ret_ptr的确就是 A，替换成 nest_ptr（即B） 但是当前B已经被销毁了，所以程序是崩毁的。

有人使用过的一种解决办法就是修改地址法：目的就是为了使

top_ptr.compare_exchange_weak(next_ptr, obj_ptr)

这个操作失败，那么怎么样才可以使这个操作失败呢？因为 他成功的前提是 top_ptr 与 next_ptr相等，所以只要使next_ptr发生变化就可以了，因为当非本线程对 next_ptr做过手脚的话 就应该留下一点痕迹 这样就可以解决该问题了。

通过考察各种基于CAS原子操作的无锁数据结构实现，目前公认可实现无锁安全的数据结构是数组和单向队列。其他实现都一定程度上受到复杂度和ABA问题的威胁。