Operators of Point class using non-member friend Functions 4. Overloading Binary Operators 5. Overloading Unary Operators. So we have to intervene and overload the assignment operator. Here's how. First we modify the class definition: class mystring { private: char* c; int lnth, capacity; public: mystring(); mystring& operator=(const mystring& rhs); bool append(char x ); };. We are going to write a new function, operator= ; this special syntax tells C++ . I would rather use different letter for template parameter: template Number& operator=(const Number& number) { m_value = number. m_value; //I would also directly access the member variable! return *this; }. I think, it is better to use explicit cast, if you want to use class type as template.">
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Overload Assignment Operator C++ Template Method

Coercion by Member Template[edit]


To increase the flexibility of a class template's interface by allowing the class template to participate in the same implicit type conversions (coercion) as its parameterizing types enjoy.

Also Known As[edit]


It is often useful to extend a relationship between two types to class templates specialized with those types. For example, suppose that class D derives from class B. A pointer to an object of type D can be assigned to a pointer to B; C++ supports that implicitly. However, types composed of these types do not share the relationship of the composed types. That applies to class templates as well, so a object normally cannot be assigned to a object.

classB{};classD:publicB{};template<classT>classHelper{};B*bptr;D*dptr;bptr=dptr;// OK; permitted by C++Helper<B>hb;Helper<D>hd;hb=hd;// Not allowed but could be very useful

There are cases where such conversions are useful, such as allowing conversion from to . That is quite intuitive, but isn't supported without using the Coercion by Member Template Idiom.

Solution and Sample Code[edit]

Define member template functions, in a class template, which rely on the implicit type conversions supported by the parameter types. In the following example, the templated constructor and assignment operator work for any type U, for which initialization or assignment of a from a is allowed.

template<classT>classPtr{public:Ptr(){}Ptr(Ptrconst&p):ptr(p.ptr){std::cout<<"Copy constructor\n";}// Supporting coercion using member template constructor.// This is not a copy constructor, but behaves similarly.template<classU>Ptr(Ptr<U>const&p):ptr(p.ptr)// Implicit conversion from U to T required{std::cout<<"Coercing member template constructor\n";}// Copy assignment operator.Ptr&operator=(Ptrconst&p){ptr=p.ptr;std::cout<<"Copy assignment operator\n";return*this;}// Supporting coercion using member template assignment operator.// This is not the copy assignment operator, but works similarly.template<classU>Ptr&operator=(Ptr<U>const&p){ptr=p.ptr;// Implicit conversion from U to T requiredstd::cout<<"Coercing member template assignment operator\n";return*this;}T*ptr;};intmain(void){Ptr<D>d_ptr;Ptr<B>b_ptr(d_ptr);// Now supportedb_ptr=d_ptr;// Now supported}

Another use for this idiom is to permit assigning an array of pointers to a class to an array of pointers to that class' base. Given that D derives from B, a D object is-a B object. However, an array of D objects is-not-an array of B objects. This is prohibited in C++ because of slicing. Relaxing this rule for an array of pointers can be helpful. For example, an array of pointers to D should be assignable to an array of pointers to B (assuming B's destructor is virtual). Applying this idiom can achieve that, but extra care is needed to prevent copying arrays of pointers to one type to arrays of pointers to a derived type. Specializations of the member function templates or SFINAE can be used to achieve that.

The following example uses a templated constructor and assignment operator expecting to only allow copying Arrays of pointers when the element types differ.


Many smart pointers such as std::unique_ptr, std::shared_ptr employ this idiom.


A typical mistake in implementing the Coercion by Member Template Idiom is failing to provide the non-template copy constructor or copy assignment operator when introducing the templated copy constructor and assignment operator. A compiler will automatically declare a copy constructor and a copy assignment operator if a class does not declare them, which can cause hidden and non-obvious faults when using this idiom.

Known Uses[edit]

  • std::unique_ptr
  • std::shared_ptr

Related Idioms[edit]


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.


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)


  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.


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.


Run this code


#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