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Quick Q: Most concise way to disable copy and move semantics
By Adrien Hamelin | Feb 20, 2018 08:40 AM | Tags: c++11 advanced
Quick A: Delete the move assignment.
Recently on SO:
Most concise way to disable copy and move semantics According to this chart (by Howard Hinnant): The most concise way is to =delete move assignment operator (or move constructor, but it can cause problems mentioned in comments). Though, in my opinion the most readable way is to =delete both copy constructor and copy assignment operator.
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Copy Constructor vs Assignment Operator in C++
- How to Create Custom Assignment Operator in C++?
- Assignment Operators In C++
- Why copy constructor argument should be const in C++?
- Advanced C++ | Virtual Copy Constructor
- Move Assignment Operator in C++ 11
- Self assignment check in assignment operator
- Is assignment operator inherited?
- Copy Constructor in C++
- How to Implement Move Assignment Operator in C++?
- Default Assignment Operator and References in C++
- Can a constructor be private in C++ ?
- When is a Copy Constructor Called in C++?
- C++ Assignment Operator Overloading
- std::move in Utility in C++ | Move Semantics, Move Constructors and Move Assignment Operators
- C++ Interview questions based on constructors/ Destructors.
- Assignment Operators in C
- Copy Constructor in Python
- Copy Constructor in Java
- Constructors in Objective-C
Copy constructor and Assignment operator are similar as they are both used to initialize one object using another object. But, there are some basic differences between them:
Consider the following C++ program.
Explanation: Here, t2 = t1; calls the assignment operator , same as t2.operator=(t1); and Test t3 = t1; calls the copy constructor , same as Test t3(t1);
Must Read: When is a Copy Constructor Called in C++?
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Copy constructors and copy assignment operators (C++)
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Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment . In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see Move Constructors and Move Assignment Operators (C++) .
Both the assignment operation and the initialization operation cause objects to be copied.
Assignment : When one object's value is assigned to another object, the first object is copied to the second object. So, this code copies the value of b into a :
Initialization : Initialization occurs when you declare a new object, when you pass function arguments by value, or when you return by value from a function.
You can define the semantics of "copy" for objects of class type. For example, consider this code:
The preceding code could mean "copy the contents of FILE1.DAT to FILE2.DAT" or it could mean "ignore FILE2.DAT and make b a second handle to FILE1.DAT." You must attach appropriate copying semantics to each class, as follows:
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x); .
Use the copy constructor.
If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you. Similarly, if you don't declare a copy assignment operator, the compiler generates a member-wise copy assignment operator for you. Declaring a copy constructor doesn't suppress the compiler-generated copy assignment operator, and vice-versa. If you implement either one, we recommend that you implement the other one, too. When you implement both, the meaning of the code is clear.
The copy constructor takes an argument of type ClassName& , where ClassName is the name of the class. For example:
Make the type of the copy constructor's argument const ClassName& whenever possible. This prevents the copy constructor from accidentally changing the copied object. It also lets you copy from const objects.
Compiler generated copy constructors
Compiler-generated copy constructors, like user-defined copy constructors, have a single argument of type "reference to class-name ." An exception is when all base classes and member classes have copy constructors declared as taking a single argument of type const class-name & . In such a case, the compiler-generated copy constructor's argument is also const .
When the argument type to the copy constructor isn't const , initialization by copying a const object generates an error. The reverse isn't true: If the argument is const , you can initialize by copying an object that's not const .
Compiler-generated assignment operators follow the same pattern for const . They take a single argument of type ClassName& unless the assignment operators in all base and member classes take arguments of type const ClassName& . In this case, the generated assignment operator for the class takes a const argument.
When virtual base classes are initialized by copy constructors, whether compiler-generated or user-defined, they're initialized only once: at the point when they are constructed.
The implications are similar to the copy constructor. When the argument type isn't const , assignment from a const object generates an error. The reverse isn't true: If a const value is assigned to a value that's not const , the assignment succeeds.
For more information about overloaded assignment operators, see Assignment .
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cppreference.com
Move assignment operator.
A move assignment operator is a non-template non-static member function with the name operator = that can be called with an argument of the same class type and copies the content of the argument, possibly mutating the argument.
[ edit ] Syntax
For the formal move assignment operator syntax, see function declaration . The syntax list below only demonstrates a subset of all valid move assignment operator syntaxes.
[ edit ] Explanation
The move assignment operator is called whenever it is selected by overload resolution , e.g. when an object appears on the left-hand side of an assignment expression, where the right-hand side is an rvalue of the same or implicitly convertible type.
Move assignment operators typically "steal" the resources held by the argument (e.g. pointers to dynamically-allocated objects, file descriptors, TCP sockets, I/O streams, running threads, etc.), rather than make copies of them, and leave the argument in some valid but otherwise indeterminate state. For example, move-assigning from a std::string or from a std::vector may result in the argument being left empty. This is not, however, a guarantee. A move assignment is less, not more restrictively defined than ordinary assignment; where ordinary assignment must leave two copies of data at completion, move assignment is required to leave only one.
[ edit ] Implicitly-declared move assignment operator
If no user-defined move assignment operators are provided for a class type, and all of the following is true:
- there are no user-declared copy constructors ;
- there are no user-declared move constructors ;
- there are no user-declared copy assignment operators ;
- there is no user-declared destructor ,
then the compiler will declare a move assignment operator as an inline public member of its class with the signature T & T :: operator = ( T && ) .
A class can have multiple move assignment operators, e.g. both T & T :: operator = ( const T && ) and T & T :: operator = ( T && ) . If some user-defined move assignment operators are present, the user may still force the generation of the implicitly declared move assignment operator with the keyword default .
The implicitly-declared (or defaulted on its first declaration) move assignment operator has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17) .
Because some assignment operator (move or copy) 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 ] Implicitly-defined move assignment operator
If the implicitly-declared move 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 or needed for constant evaluation (since C++14) .
For union types, the implicitly-defined move assignment operator copies the object representation (as by std::memmove ).
For non-union class types, the move assignment operator performs full member-wise move assignment of the object's direct bases and immediate non-static members, in their declaration order, using built-in assignment for the scalars, memberwise move-assignment for arrays, and move assignment operator for class types (called non-virtually).
As with copy assignment, 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 move assignment operator:
[ edit ] Deleted move assignment operator
The implicitly-declared or defaulted move assignment operator for class T is defined as deleted if any of the following conditions is satisfied:
- T has a non-static data member of a const-qualified non-class type (or possibly multi-dimensional array thereof).
- T has a non-static data member of a reference type.
- T has a potentially constructed subobject of class type M (or possibly multi-dimensional array thereof) such that the overload resolution as applied to find M 's move assignment operator
- does not result in a usable candidate, or
- in the case of the subobject being a variant member , selects a non-trivial function.
A deleted implicitly-declared move assignment operator is ignored by overload resolution .
[ edit ] Trivial move assignment operator
The move assignment operator for class T is trivial if all of the following is true:
- It is not user-provided (meaning, it is implicitly-defined or defaulted);
- T has no virtual member functions;
- T has no virtual base classes;
- the move assignment operator selected for every direct base of T is trivial;
- the move assignment operator selected for every non-static class type (or array of class type) member of T is trivial.
A trivial move assignment operator performs the same action as the trivial copy assignment operator, that is, makes a copy of the object representation as if by std::memmove . All data types compatible with the C language (POD types) are trivially move-assignable.
[ edit ] Eligible move assignment operator
Triviality of eligible move assignment operators determines whether the class is a trivially copyable type .
[ 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 move assignment operator (same applies to copy assignment ).
See assignment operator overloading for additional detail on the expected behavior of a user-defined move-assignment operator.
[ edit ] Example
[ edit ] defect reports.
The following behavior-changing defect reports were applied retroactively to previously published C++ standards.
[ edit ] See also
- constructor
- converting constructor
- copy assignment
- copy constructor
- default constructor
- aggregate initialization
- constant initialization
- copy initialization
- default initialization
- direct initialization
- list initialization
- reference initialization
- value initialization
- zero initialization
- move constructor
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13. Deleting the copy-constructor and copy-assignment operator is the simplest and clearest way to disable copying: class X. {. X(X const &) = delete; void operator=(X const &x) = delete; }; I don't follow what you are talking about with virtual destructors in the question body .
The alternative is to declare a copy-constructor private and leave it undefined, or define it with a BOOST_STATIC_ASSERT(false). If you are working with C++11 you can also delete your copy constructor: class nocopy. {. nocopy( nocopy const& ) = delete; }; answered Oct 19, 2011 at 15:37. K-ballo.
Keeping the Copy Constructor and Copy assignment operator as private in the class. Below is the C++ implementation to illustrate how this can be done. #include <iostream> ... Above two ways are quite complex, C++11 has come up with a simpler solution i.e. just delete the copy constructor and assignment operator. Below is the C++ implementation ...
Use an assignment operator operator= that returns a reference to the class type and takes one parameter that's passed by const reference—for example ClassName& operator=(const ClassName& x);. Use the copy constructor. If you don't declare a copy constructor, the compiler generates a member-wise copy constructor for you.
The implicitly-declared (or defaulted on its first declaration) copy constructor has an exception specification as described in dynamic exception specification (until C++17) noexcept specification (since C++17). [] Implicitly-defined copy constructoIf the implicitly-declared copy constructor is not deleted, it is defined (that is, a function body is generated and compiled) by the compiler if ...
Copy constructors, assignment operators, and exception safe assignment. Score: 4.3/5 (3169 votes) What is a copy constructor? A copy constructor is a special constructor for a class/struct that is used to make a copy of an existing instance. According to the C++ ... 11: template<> MyArray<T>::operator= ...
the copy assignment operator selected for every non-static class type (or array of class type) member of T is trivial. 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.
Most concise way to disable copy and move semantics. According to this chart (by Howard Hinnant): The most concise way is to =delete move assignment operator (or move constructor, but it can cause problems mentioned in comments). Though, in my opinion the most readable way is to =delete both copy constructor and copy assignment operator.
Benefits of explicitly defaulted and deleted functions. In C++, the compiler automatically generates the default constructor, copy constructor, copy-assignment operator, and destructor for a type if it doesn't declare its own. These functions are known as the special member functions, and they're what make simple user-defined types in C++ ...
But, there are some basic differences between them: Copy constructor. Assignment operator. It is called when a new object is created from an existing object, as a copy of the existing object. This operator is called when an already initialized object is assigned a new value from another existing object. It creates a separate memory block for ...
Rule of three. If a class requires a user-defined destructor, a user-defined copy constructor, or a user-defined copy assignment operator, it almost certainly requires all three.. Because C++ copies and copy-assigns objects of user-defined types in various situations (passing/returning by value, manipulating a container, etc), these special member functions will be called, if accessible, and ...
Starting in C++11, two kinds of assignment are supported in the language: copy assignment and move assignment. In this article "assignment" means copy assignment unless explicitly stated otherwise. For information about move assignment, see
The copy constructor is for creating a new object. It copies an existing object to a newly constructed object.The copy constructor is used to initialize a new instance from an old instance. It is not necessarily called when passing variables by value into functions or as return values out of functions. The assignment operator is to deal with an ...
Copy Assignment The copy and swap idiom works as follows: We have an existing value we want to modify, and an existing value to read data from Use the copy constructor to create a temporary value from the value we're reading data from Swap the contents of the value to modify and the temporary
Triviality of eligible move assignment operators determines whether the class is a trivially copyable type. [] NoteIf 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 ...
8. The problem with the copy constructor and copy-assignment operator is that the compiler generates implementations automatically if they're not explicitly declared. This can easily cause unintended problems. If a class has a non-trivial destructor, it almost always needs to provide its own implementations for the copy constructor and copy ...
Preventing compilers from defining copy constructors and operator = overload for C++ Classes 4 Preventing copy construction and assignment of a return value reference
A copy constructor is used to initialize a previously uninitialized object from some other object's data. A(const A& rhs) : data_(rhs.data_) {} For example: A aa; A a = aa; //copy constructor An assignment operator is used to replace the data of a previously initialized object with some other object's data.
In C++ class with copy & move ctor and copy assignment: no need for move assignment? 2 Why object could be "moved" even lacks move constructor and move assignment operator?
in this code On_heap is a wrapper that manages an object on heap. is deleting move constructor and assignment here are mandatory?. template<typename T> struct On_heap { On_heap() :p(new T) { } // allocate ˜On_heap() { delete p; } // deallocate T& operator∗() { return ∗p; } T∗ operator−>() { return p; } On_heap(const On_heap&) = delete; // prevent copying On_heap operator=(const On ...