Assignment Operator Not Inherited C++ Tutorials

Assignment Operators

What is “self assignment”?

Self assignment is when someone assigns an object to itself. For example,

Obviously no one ever explicitly does a self assignment like the above, but since more than one pointer or reference can point to the same object (aliasing), it is possible to have self assignment without knowing it:

This is only valid for copy assignment. Self-assignment is not valid for move assignment.

Why should I worry about “self assignment”?

If you don’t worry about self assignment, you’ll expose your users to some very subtle bugs that have very subtle and often disastrous symptoms. For example, the following class will cause a complete disaster in the case of self-assignment:

If someone assigns a object to itself, line #1 deletes both and since and are the same object. But line #2 uses , which is no longer a valid object. This will likely cause a major disaster.

The bottom line is that you the author of class are responsible to make sure self-assignment on a object is innocuous. Do not assume that users won’t ever do that to your objects. It is your fault if your object crashes when it gets a self-assignment.

Aside: the above has a second problem: If an exception is thrown while evaluating (e.g., an out-of-memory exception or an exception in ’s copy constructor), will be a dangling pointer — it will point to memory that is no longer valid. This can be solved by allocating the new objects before deleting the old objects.

This is only valid for copy assignment. Self-assignment is not valid for move assignment.

Okay, okay, already; I’ll handle self-assignment. How do I do it?

You should worry about self assignment every time you create a class. This does not mean that you need to add extra code to all your classes: as long as your objects gracefully handle self assignment, it doesn’t matter whether you had to add extra code or not.

We will illustrate the two cases using the assignment operator in the previous FAQ:

  1. If self-assignment can be handled without any extra code, don’t add any extra code. But do add a comment so others will know that your assignment operator gracefully handles self-assignment:

    Example 1a:

    Example 1b:

  2. If you need to add extra code to your assignment operator, here’s a simple and effective technique:

    Or equivalently:

By the way: the goal is not to make self-assignment fast. If you don’t need to explicitly test for self-assignment, for example, if your code works correctly (even if slowly) in the case of self-assignment, then do not put an test in your assignment operator just to make the self-assignment case fast. The reason is simple: self-assignment is almost always rare, so it merely needs to be correct - it does not need to be efficient. Adding the unnecessary statement would make a rare case faster by adding an extra conditional-branch to the normal case, punishing the many to benefit the few.

In this case, however, you should add a comment at the top of your assignment operator indicating that the rest of the code makes self-assignment is benign, and that is why you didn’t explicitly test for it. That way future maintainers will know to make sure self-assignment stays benign, or if not, they will need to add the test.

This is only valid for copy assignment. Self-assignment is not valid for move assignment.

I’m creating a derived class; should my assignment operators call my base class’s assignment operators?

Yes (if you need to define assignment operators in the first place).

If you define your own assignment operators, the compiler will not automatically call your base class’s assignment operators for you. Unless your base class’s assignment operators themselves are broken, you should call them explicitly from your derived class’s assignment operators (again, assuming you create them in the first place).

However if you do not create your own assignment operators, the ones that the compiler create for you will automatically call your base class’s assignment operators.

Example:

A copy assignment operator of class is a non-template non-static member function with the name operator= that takes exactly one parameter of type T, T&, const T&, volatile T&, or constvolatile T&. For a type to be , it must have a public copy assignment operator.

[edit]Syntax

class_nameclass_name ( class_name ) (1)
class_nameclass_name ( const class_name ) (2)
class_nameclass_name ( const class_name ) = default; (3) (since C++11)
class_nameclass_name ( const class_name ) = delete; (4) (since C++11)

[edit]Explanation

  1. Typical declaration of a copy assignment operator when copy-and-swap idiom can be used.
  2. Typical declaration of a copy assignment operator when copy-and-swap idiom cannot be used (non-swappable type or degraded performance).
  3. Forcing a copy assignment operator to be generated by the compiler.
  4. Avoiding implicit copy assignment.

The copy assignment operator is called whenever selected by overload resolution, e.g. when an object appears on the left side of an assignment expression.

[edit]Implicitly-declared copy assignment operator

If no user-defined copy assignment operators are provided for a class type (struct, class, or union), the compiler will always declare one as an inline public member of the class. This implicitly-declared copy assignment operator has the form T& T::operator=(const T&) if all of the following is true:

  • each direct base of has a copy assignment operator whose parameters are B or const B& or constvolatile B&;
  • each non-static data member of of class type or array of class type has a copy assignment operator whose parameters are M or const M& or constvolatile M&.

Otherwise the implicitly-declared copy assignment operator is declared as T& T::operator=(T&). (Note that due to these rules, the implicitly-declared copy assignment operator cannot bind to a volatile lvalue argument.)

A class can have multiple copy assignment operators, e.g. both T& T::operator=(const T&) and T& T::operator=(T). If some user-defined copy assignment operators are present, the user may still force the generation of the implicitly declared copy assignment operator with the keyword .(since C++11)

The implicitly-declared (or defaulted on its first declaration) copy assignment operator has an exception specification as described in dynamic exception specification(until C++17)exception specification(since C++17)

Because the copy assignment operator is always declared for any class, the base class assignment operator is always hidden. If a using-declaration is used to bring in the assignment operator from the base class, and its argument type could be the same as the argument type of the implicit assignment operator of the derived class, the using-declaration is also hidden by the implicit declaration.

[edit]Deleted implicitly-declared copy assignment operator

A implicitly-declared copy assignment operator for class is defined as deleted if any of the following is true:

  • has a user-declared move constructor;
  • has a user-declared move assignment operator.

Otherwise, it is defined as defaulted.

A defaulted copy assignment operator for class is defined as deleted if any of the following is true:

  • has a non-static data member of non-class type (or array thereof) that is const;
  • has a non-static data member of a reference type;
  • has a non-static data member or a direct or virtual base class that cannot be copy-assigned (overload resolution for the copy assignment fails, or selects a deleted or inaccessible function);
  • is a union-like class, and has a variant member whose corresponding assignment operator is non-trivial.

[edit]Trivial copy assignment operator

The copy assignment operator for class is trivial if all of the following is true:

  • it is not user-provided (meaning, it is implicitly-defined or defaulted) , , and if it is defaulted, its signature is the same as implicitly-defined(until C++14);
  • has no virtual member functions;
  • has no virtual base classes;
  • the copy assignment operator selected for every direct base of is trivial;
  • the copy assignment operator selected for every non-static class type (or array of class type) member of is trivial;
  • has no non-static data members of volatile-qualified type.
(since C++14)

A trivial copy assignment operator makes a copy of the object representation as if by std::memmove. All data types compatible with the C language (POD types) are trivially copy-assignable.

[edit]Implicitly-defined copy assignment operator

If the implicitly-declared copy assignment operator is neither deleted nor trivial, it is defined (that is, a function body is generated and compiled) by the compiler if odr-used. For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove). For non-union class types (class and struct), the operator performs member-wise copy assignment of the object's bases and non-static members, in their initialization order, using built-in assignment for the scalars and copy assignment operator for class types.

The generation of the implicitly-defined copy assignment operator is deprecated(since C++11) if has a user-declared destructor or user-declared copy constructor.

[edit]Notes

If both copy and move assignment operators are provided, overload resolution selects the move assignment if the argument is an rvalue (either a prvalue such as a nameless temporary or an xvalue such as the result of std::move), and selects the copy assignment if the argument is an lvalue (named object or a function/operator returning lvalue reference). If only the copy assignment is provided, all argument categories select it (as long as it takes its argument by value or as reference to const, since rvalues can bind to const references), which makes copy assignment the fallback for move assignment, when move is unavailable.

It is unspecified whether virtual base class subobjects that are accessible through more than one path in the inheritance lattice, are assigned more than once by the implicitly-defined copy assignment operator (same applies to move assignment).

See assignment operator overloading for additional detail on the expected behavior of a user-defined copy-assignment operator.

[edit]Example

Run this code

Output:

#include <iostream>#include <memory>#include <string>#include <algorithm>   struct A {int n;std::string s1;// user-defined copy assignment, copy-and-swap form A& operator=(A other){std::cout<<"copy assignment of A\n";std::swap(n, other.n);std::swap(s1, other.s1);return*this;}};   struct B : A {std::string s2;// implicitly-defined copy assignment};   struct C {std::unique_ptr<int[]> data;std::size_t size;// non-copy-and-swap assignment C& operator=(const C& other){// check for self-assignmentif(&other == this)return*this;// reuse storage when possibleif(size != other.size){ data.reset(new int[other.size]); size = other.size;}std::copy(&other.data[0], &other.data[0]+ size, &data[0]);return*this;}// note: copy-and-swap would always cause a reallocation};   int main(){ A a1, a2;std::cout<<"a1 = a2 calls "; a1 = a2;// user-defined copy assignment   B b1, b2; b2.s1="foo"; b2.s2="bar";std::cout<<"b1 = b2 calls "; b1 = b2;// implicitly-defined copy assignmentstd::cout<<"b1.s1 = "<< b1.s1<<" b1.s2 = "<< b1.s2<<'\n';}
a1 = a2 calls copy assignment of A b1 = b2 calls copy assignment of A b1.s1 = foo b1.s2 = bar

[edit]Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.

DR Applied to Behavior as published Correct behavior
CWG 2171 C++14 operator=(X&)=default was non-trivial made trivial
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