c++ - G++ CAS 的奇怪行为

Question

考虑这段代码:

#include <iostream>
using namespace std;

int main()
{
   bool lock = false;
   lock = __sync_val_compare_and_swap( &lock, false, true );
   cout << lock << endl;
}

我希望结果显示为 1 但 o/p 为 0。只需调用 __sync_val_compare_and_swap( &lock, false, true );(因此不会捕获返回值)然后显示锁定结果显示 1。

我在这里错过了什么？

Answer 1

来自 GCC doco:

bool __sync_bool_compare_and_swap (type *ptr, type oldval type newval, ...)type __sync_val_compare_and_swap (type *ptr, type oldval type newval, ...)

These builtins perform an atomic compare and swap. That is, if the current value of *ptr is oldval, then write newval into *ptr.

The “bool” version returns true if the comparison is successful and newval was written. The “val” version returns the contents of *ptr before the operation.

在我看来 0 是正确的值。我认为您错误地将“...操作前 *ptr 的内容”分配给 lock。

这应该输出合理的结果:

#include <iostream>
using namespace std;

int main()
{
   bool lock = false;
   bool oldvalue = __sync_val_compare_and_swap( &lock, false, true );
   cout << lock << ", " << oldvalue << endl;
}