Postfix Operator Overloading Order [duplicate]

Question

Possible Duplicate:
Undefined Behavior and Sequence Points

I'm having trouble understanding the order of actions when overloading the postfix operator. Let's examine the two small examples below:

int i = 0;
std::cout << std::endl << "i: " << i;
i = ++i;
std::cout << std::endl << "i: " << i;
i = i++;
std::cout << std::endl << "i: " << i;

MyClass myObject;
std::cout << std::endl << "myObject: " << myObject.getMyValue();
myObject = ++myObject;
std::cout << std::endl << "myObject: " << myObject.getMyValue();
myObject = myObject++;
std::cout << std::endl << "myObject: " << myObject.getMyValue();

Two very different behaviors emerge. The output is as follows:

i: 0
i: 1
i: 2
myObject: 0
myObject: 1
myObject: 1

Different behavior, you see. Here's the outline of my overloaded-operator methods.

MyClass & MyClass::operator++ ()
{
    ++myValue;
    return *this;
}

MyClass MyClass::operator++ (int postfixFlag)
{
    MyClass myTemp(*this);
    ++myValue;
    return myTemp;
}

Alright. Prefix makes sense. You increment whatever you need to, then return the same object, now modified, in case of assignment. But postfix is what's tripping me up. It's supposed to assign, then increment. Here we're self assigning. So with the built-in integer type, it makes sense. I assign i's value to itself, then i gets incremented. Fair enough. But let's say MyClass is a recreation of the int. It starts out at 0, gets prefix-incremented, and becomes 1. Then, the key line. myObject = myObject++. That's the same thing as myObject = myObject.operator++(int postfixFlag). It gets called. myTemp gets initialized with the value 1. It's incremented to 2. Then we return the temp. That works, if we're assigning to another object. But here I'm self-assigning, so after the increment to 2, myObject is set equal to the returned temp object initialized with the initial value, and we're back to 1! That makes sense. But it's a fundamentally different behavior.

How do I work around it? How does int do it? How is this method generally written? Do you have any comments about C++ behavior and design relating to this? Etc. I'm a little perplexed right now, since books and online examples always seem to use a variant on the method above.

Thanks for reading, and any input will be appreciated!

Answer 1

As others have said, with int the behaviour is undefined. But I thought I'd try to explain why for your MyClass it is not ever getting to 2.

The trick is that you are taking the following three steps in the postfix version:

1. Making a copy of this called myTemp (with myValue == 1).
2. Incrementing this->myValue (so myTemp.myValue == 1; this->myValue == 2).
3. Returning myTemp (with myValue == 1).

So you are modifying this, but the code that calls myObject++ is never going to see this again. It's only going to look at the value returned, which is a copy of the old myObject.

The code for operator++ is fine. The problem is how you are using it -- you shouldn't be writing the result of a pre-increment or post-increment back to the same variable (behaviour is undefined). Here is some code that might be more instructive:

int i = 0;
std::cout << "i: " << i << std::endl;
int j = ++i;
std::cout << "i: " << i << ", j: " << j << std::endl;
int k = i++;
std::cout << "i: " << i << ", k: " << k << std::endl;

MyClass myObject;
std::cout << "myObject: " << myObject.getMyValue() << std::endl;
MyClass myObject1 = ++myObject;
std::cout << "myObject: " << myObject.getMyValue()
    << ", myObject1: " << myObject1.getMyValue() << std::endl;
MyClass myObject2 = myObject++;
std::cout << "myObject: " << myObject.getMyValue()
    << ", myObject2: " << myObject2.getMyValue() << std::endl;

This prints:

i: 0
i: 1, j: 1
i: 2, k: 1
myObject: 0
myObject: 1, myObject1: 1
myObject: 2, myObject2: 1

I changed your code so that rather than assigning back to itself, it assigns to a fresh variable each time. Note that in both the int and the MyClass cases, the main variable (i/myObject) is incremented both times. However, in the pre-increment case, the fresh variable (j/myObject1) takes on the new value, while in the post-increment case, the fresh variable (k/myObject2) takes on the old value.

Edit: Just answering another part of the question, "How does int do it?" I assume this question means "what does the pre-increment and post-increment code look like in the int class, and how can I make mine the same?" The answer is, there is no "int class". int is a special built-in type in C++ and the compiler treats it specially. These types aren't defined with ordinary C++ code, they are hard-coded into the compiler.

Note: For anyone who wants to try this themselves, here is the code for MyClass that the question didn't include:

class MyClass
{
private:
    int myValue;
public:
    MyClass() : myValue(0) {}
    int getMyValue() { return myValue; }
    MyClass& operator++();
    MyClass operator++(int postfixFlag);
};