仓库源文，站点原文

title: What is object slicing? toc: true cover: 'https://img.paulzzh.com/touhou/random?44' date: 2022-05-03 17:42:38 categories: C++ tags: [C++]

description: In this passage we will talk about Object Slicing in C++(A very common occurrence in OOP).

Source Code:

https://github.com/JasonkayZK/cpp-learn/blob/jupyter/oop/object_slicing.ipynb

In this passage we will talk about Object Slicing in C++(A very common occurrence in OOP).

What is object slicing?

Before we talk the Object Slicing, let’s see a example first:

#include <iostream>

class Base {
public:
    explicit Base(int mVal) : m_val_(mVal) {}

    [[nodiscard]] virtual const char *get_name() const { return "Base"; }

    [[nodiscard]] int get_m_val() const { return m_val_; }

protected:
    int m_val_;
};

class Derived : public Base {
public:
    explicit Derived(int mVal) : Base(mVal) {}

    [[nodiscard]] const char *get_name() const override { return "Derived"; }
};

int main() {
    Derived derived{5};
    std::cout << "derived is a " << derived.get_name() << " and has value " << derived.get_m_val() << '\n';

    Base &ref{derived};
    std::cout << "ref is a " << ref.get_name() << " and has value " << ref.get_m_val() << '\n';

    Base *ptr{&derived};
    std::cout << "ptr is a " << ptr->get_name() << " and has value " << ptr->get_m_val() << '\n';

    return 0;
}

In the example above, i created a Base class and Derived class (based on Base);

Then, i created a derived object from Derived class, and two other variable:

derived to Base type reference;
derived to Base type pointer;

Finally, we call get_name and get_m_val via these three variables. And it is obvious that:

All get_name returns "Derived";
All get_m_val returns 5;

This is because: dual to the inheritance from Base, the variable derived inherits the m_val_ from Base .

And even though we defined ref as &Base and ptr as *Base separately, but finally they all pointed to the variable: derived!

But this is an exception: Copy Constructor!

In the above example, we created reference and pointer from variable derived;

But what if we create a new object by Copy constructor:

Base base{derived};
std::cout << "base is a " << base.get_name() << " and has value " << base.get_m_val() << '\n';

The code above will finally print:

base is a Base and has value 5

Remember that:

**The derived object has a base part and a derived part.**

**When we assign the derived object to the base object, ONLY the base part will be copied, and the derived part will not (Via Copy Constructor).**

**In the above example, the base object receives a copy of the base part of a derived object, but ignores the derived part. The derived part is sliced off.**

**This is so called `Object Slicing`!**

Object Slicing in function call

In the above example you may think it stupid, because hardly nobody assigns derived classes to base classes.

But it is most likely when you are doing some function call:

void printName(const Base base) { // note: base pass by value, not reference
    std::cout << "I am a " << base.get_name() << '\n';
}

Derived d{ 5 };
printName(d);

This code will finally print:

I am a Base

Dual to we passed value (which will be copied), not reference to the function.

This can be simply fix that we just need to pass the reference:

void printName(const Base &base) { // note: base pass by reference
    std::cout << "I am a " << base.get_name() << '\n';
}

This will print:

I am a Derived

Object Slicing in Vector(Or Collection)

Another area where new programmers run into trouble with object slicing is trying to implement polymorphism with std::vector.

Consider the following program:

std::vector<Base> v{};
v.push_back(Base{ 5 }); // All a Base object into vector
v.push_back(Derived{ 6 }); // All a Derived object into vector
for (const auto& element : v)
  std::cout << "I am a " << element.get_name() << " with value " << element.get_m_val() << '\n';

This will print:

I am a Base with value 5
I am a Base with value 6

Obviously, because std::vector is declared as a base type, when Derived(6) is added to vector, it has been sliced!

Fix Object Slicing in Vector

Unfortunately, we can not just create a reference type std:vector like this:

std::vector<Base&> v{};

This will not compile:

**The object of `std::vector` must be assignable, but the reference cannot be assigned (the reference can be assigned only at initialization)!**

One solution to solve this problem is to create a pointer type std::vector:

std::vector<Base *> v2{};

Base b{5}; // b and d show be initiate explicitly!
Derived d{6};

v2.push_back(&b);
v2.push_back(&d);

for (const auto *element: v2)
  std::cout << "I am a " << element->get_name() << " with value " << element->get_m_val() << '\n';

This will print:

I am a Base with value 5
I am a Derived with value 6

It works!

But dual to the complexity of the pointer, there are more things we should care of:

Nullpter is now a legal option and may or may not be appropriate for your use scenario!
Now you have to operate with the pointer, which may be awkwardly!

The Frankenobject

In the above examples, we’ve seen cases where slicing lead to the wrong result because the derived class had been sliced off.

Now let’s take a look at another dangerous case where the derived object still exists!

Consider the following code:

Derived dd1{ 5 };
Derived dd2{ 6 };
Base &bb{ dd2 };

bb = dd1; // cause problem
std::cout << "I am a " << bb.get_name() << " with value " << bb.get_m_val() << '\n';

The first three lines in the function are pretty straightforward:

Create two Derived objects, and set a Base reference to the second one.

The fourth line is where things go astray: Since bb points at dd2, and we’re assigning dd1 to bb.

you might think that the result would be that dd1 would get copied into dd2 (it would, if b were a Derived!).

But b is a Base, and the operator= that C++ provides for classes isn’t virtual by default.

Consequently, only the Base portion of d1 is copied into d2.

As a result, you’ll discover that: d2 now has the Base portion of d1 and the Derived portion of d2!

In this particular example, that’s not a problem (because the Derived class has no data of its own).

But in most cases, you’ll have just created a Franken Object: composed of parts of multiple objects.

Worse, there’s no easy way to prevent this from happening (other than avoiding assignments like this as much as possible).

Google Style Guide

A wisable choice to simply avoid this problem is to forbid the Copy & Move Constructor explicitly, if your type do not need them:

class Base {
public:
    explicit Base(int mVal) : m_val_(mVal) {}

    [[nodiscard]] virtual const char *get_name() const { return "Base"; }

    [[nodiscard]] int get_m_val() const { return m_val_; }

    // Base is neither copyable nor movable.
    Base(const Base&) = delete;
    Base& operator=(const Base&) = delete;

protected:
    int m_val_;
};

By adding these code:

// Base is neither copyable nor movable.
Base(const Base&) = delete;
Base& operator=(const Base&) = delete;

we declared that: Base is neither copyable nor movable!

Then this program will be not compiled:

Derived dd1{ 5 };
Derived dd2{ 6 };
Base &bb{ dd2 };

bb = dd1; // cause problem

Because Base is neither copyable nor movable now!

Conclusion

In this passage, first i gave two examples to introduce what is Object Slicing.

Then, I gave three typical and common mistakes:

Object Slicing in function call
Object Slicing in vector
Object Slicing to cause Franken Object

And the corresponding solutions.

Finally, be referred Google Style Guide, a usual principle to avoid this mistake is to forbid the copy and move constructor.

Appendix

Source Code:

https://github.com/JasonkayZK/cpp-learn/blob/jupyter/oop/object_slicing.ipynb

Reference: