C++ volatile 成员函数

Question

class MyClass
{
    int x, y;
    void foo() volatile {
        // do stuff with x
        // do stuff with y
    }   
};

Do I need to declare x and y as volatile or will be all member variables treated as volatile automatically?

I want to make sure that "stuff with x " is not reordered with "stuff with y " by the compiler.

EDIT: What happens if I'm casting a normal type to a volatile type? Would this instruct the compiler to not reorder access to that location? I want to pass a normal variable in a special situation to a function which parameter is volatile. I must be sure compiler doesn't reorder that call with prior or followed reads and writes.

Answer 1

Marking a member function volatile is like marking it const ; it means that the receiver object is treated as though it were declared as a volatile T* . Consequentially, any reference to x or y will be treated as a volatile read in the member function. Moreover, a volatile object can only call volatile member functions.

That said, you may want to mark x and y volatile anyway if you really do want all accesses to them to be treated as volatile .

Answer 2

You don't have to declare the member variables explicitly..

From Standard docs 9.3.2.3 ,

Similarly, volatile semantics (7.1.6.1) apply in volatile member functions when accessing the object and its nonstatic data members.

Answer 3

The following code:

#include <iostream>

class Bar
{
    public:

        void test();
};

class Foo
{
    public:

        void test() volatile { x.test(); }

    private:

        Bar x;
};

int main()
{
    Foo foo;

    foo.test();

    return 0;
}

Raises an error upon compilation with gcc:

main.cpp: In member function 'void Foo::test() volatile':
main.cpp:14:33: error: no matching function for call to 'Bar::test() volatile'
main.cpp:7:8: note: candidate is: void Bar::test() <near match>

And since a volatile instance can't call a non-volatile method, we can assume that, yes, x and y will be volatile in the method, even if the instance of MyClass is not declared volatile .

Note: you can remove the volatile qualifier using a const_cast<> if you ever need to; however be careful because just like const doing so can lead to undefined behavior under some cases.

Answer 4

So using the original example:

class MyClass
{
    int x, y;
    void foo() volatile {
        // do stuff with x
        // do stuff with y
        // with no "non-volatile" optimization of the stuff done with x, y (or anything else)
    }   
    void foo() {
        // do stuff with x
        // do stuff with y
        // the stuff done with x, y (and anything else) may be optimized
    } 
};

Answer 5

IBM暗示它的工作方式与const函数完全相同。