当前草案的以下摘录说明了我的意思:
namespace std {
typedef struct atomic_bool {
bool is_lock_free() const volatile;
bool is_lock_free() const;
void store(bool, memory_order = memory_order_seq_cst) volatile;
void store(bool, memory_order = memory_order_seq_cst);
bool load(memory_order = memory_order_seq_cst) const volatile;
bool load(memory_order = memory_order_seq_cst) const;
operator bool() const volatile;
operator bool() const;
bool exchange(bool, memory_order = memory_order_seq_cst) volatile;
bool exchange(bool, memory_order = memory_order_seq_cst);
bool compare_exchange_weak(bool&, bool, memory_order, memory_order) volatile;
bool compare_exchange_weak(bool&, bool, memory_order, memory_order);
bool compare_exchange_strong(bool&, bool, memory_order, memory_order) volatile;
bool compare_exchange_strong(bool&, bool, memory_order, memory_order);
bool compare_exchange_weak(bool&, bool, memory_order = memory_order_seq_cst) volatile;
bool compare_exchange_weak(bool&, bool, memory_order = memory_order_seq_cst);
bool compare_exchange_strong(bool&, bool, memory_order = memory_order_seq_cst) volatile;
bool compare_exchange_strong(bool&, bool, memory_order = memory_order_seq_cst);
atomic_bool() = default;
constexpr atomic_bool(bool);
atomic_bool(const atomic_bool&) = delete;
atomic_bool& operator=(const atomic_bool&) = delete;
atomic_bool& operator=(const atomic_bool&) volatile = delete;
bool operator=(bool) volatile;
} atomic_bool;
}
易失性是可传递的。因此,您不能从易失性对象调用非易失性成员函数。另一方面,允许从非易失性对象调用易失性成员函数。
那么,原子类中的 volatile 和 non-volatile 成员函数在实现上有什么区别吗?换句话说,是否需要非易失性重载?