C ++对volatile对象的引用 - 原因和影响

Question

我正在研究具有对象的volatile引用的现有C ++代码

volatile vClass & vobj;

我来自C所以我熟悉volatile用于访问内存映射的IO，如下所示：

*((volatile unsigned int *) 0x12345678) = 0x5

题

将volatile应用于（引用）对象有什么影响？

我猜它的所有数据成员都继承了volatile但是如果一个成员函数有非易失性内存访问，那该怎么办呢

void vClass::AccessMem() {*((unsigned int *) 0x12345678) = 0x5;}

内存访问是否会变得不稳定？

Answer 1

成员函数必须是volatile才能从volatile对象调用：

int not_a_member;

struct vClass {
  int i;
  void set(int j) {
     i=j; //i is not accessed as a volatile
     static_assert(std::is_same_v<decltype((i)),int&>);
     }
  void set_v(int j) volatile{
     i=j; //here i is accessed as a volatile 
     static_assert(std::is_same_v<decltype((i)),volatile int&>);
     not_a_member=j;//here not a volatile access because "not_a_member" is
                    // not a member.
     //To understand what happen, since you are a C programmer it is going to be simple.
     //The volatile qualifier is actualy apply to the "this" pointer, 
     // which is a pointer to the object from which this member function is
     // called. So inside this function "this" as the type "volatile vClass"
     //Inside a member function, all access to member data, as "i" are
     //actualy short ends for "this->i". So in this access, "i" 
     //adopt the volatile qualifier of "this".
     //So the volatile qualifier just applies to the data member of the object.
     }
  }

void use_vClass(){
   volatile vClass x;
   x.set(10); //Do not compile: "Error:try to access volatile object as non volatile"
   x.set_v(10); //Compile
  }

因此，从volatile对象或引用开始，您只需调用volatile限定成员函数，所有数据成员访问都将是“volatile”。

Answer 2

它是对易失性而非易变性引用的引用（后者无论如何都没有意义）。

它类似于：

volatile char * ptr = ...;

ptr的内存可以在没有通知的情况下改变，但是ptr本身是稳定的（不像它是

char * volatile ptr = ...;