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c assignment operator default implementation

21.13 — Shallow vs. deep copying

While this code looks harmless enough, it contains an insidious problem that will cause the program to exhibit undefined behavior!

Deep copying

When the overloaded assignment operator is called, the item being assigned to may already contain a previous value, which we need to make sure we clean up before we assign memory for new values. For non-dynamically allocated variables (which are a fixed size), we don’t have to bother because the new value just overwrites the old one. However, for dynamically allocated variables, we need to explicitly deallocate any old memory before we allocate any new memory. If we don’t, the code will not crash, but we will have a memory leak that will eat away our free memory every time we do an assignment!

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  • Copy constructors, assignment operators,

Copy constructors, assignment operators, and exception safe assignment

*

MyClass& other ); MyClass( MyClass& other ); MyClass( MyClass& other ); MyClass( MyClass& other );
MyClass* other );
MyClass { x; c; std::string s; };
MyClass& other ) : x( other.x ), c( other.c ), s( other.s ) {}
);
print_me_bad( std::string& s ) { std::cout << s << std::endl; } print_me_good( std::string& s ) { std::cout << s << std::endl; } std::string hello( ); print_me_bad( hello ); print_me_bad( std::string( ) ); print_me_bad( ); print_me_good( hello ); print_me_good( std::string( ) ); print_me_good( );
, );
=( MyClass& other ) { x = other.x; c = other.c; s = other.s; * ; }
< T > MyArray { size_t numElements; T* pElements; : size_t count() { numElements; } MyArray& =( MyArray& rhs ); };
<> MyArray<T>:: =( MyArray& rhs ) { ( != &rhs ) { [] pElements; pElements = T[ rhs.numElements ]; ( size_t i = 0; i < rhs.numElements; ++i ) pElements[ i ] = rhs.pElements[ i ]; numElements = rhs.numElements; } * ; }
<> MyArray<T>:: =( MyArray& rhs ) { MyArray tmp( rhs ); std::swap( numElements, tmp.numElements ); std::swap( pElements, tmp.pElements ); * ; }
< T > swap( T& one, T& two ) { T tmp( one ); one = two; two = tmp; }
<> MyArray<T>:: =( MyArray tmp ) { std::swap( numElements, tmp.numElements ); std::swap( pElements, tmp.pElements ); * ; }

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3842 The compiler cannot automatically generate a deep-copying of the
objects because it has absolutely no way of knowing whether the member
pointers are pointing at memory allocated and owned by the object
itself, or whether they are just pointing to something else (for example
trying to deep-copy an element of a doubly-linked list would be quite a
bad idea).
The default assignment operator simply assigns each individual member
variable from the parameter to this, regardless of what it is. If it's
eg. a pointer, then a simple pointer assignment will be done.

In many cases the default assignment operator is ok, especially if the
class does not contain pointers.

It returns a const-reference to *this.

The compiler cannot automatically generate a deep-copying of the
objects because it has absolutely no way of knowing whether the member
pointers are pointing at memory allocated and owned by the object
itself, or whether they are just pointing to something else (for example
trying to deep-copy an element of a doubly-linked list would be quite a
bad idea).
The default assignment operator simply assigns each individual member
variable from the parameter to this, regardless of what it is. If it's
eg. a pointer, then a simple pointer assignment will be done.

In many cases the default assignment operator is ok, especially if the
class does not contain pointers.

It returns a const-reference to *this. Not a const reference.
It does member by member assignment. Whether this does a deep
copy or a shallow one depends on the assignment operators in the
members, for raw pointers, it is shallow, since pointer
assignment is shallow.

The signature of the operator, for a class T, is usually:
T& T::operator=( T const& )
If any of the base classes or members have a copy assignment
operator which requires a non-const reference, however, the
parameter will be a non-const reference.

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The compiler cannot automatically generate a deep-copying of the
objects because it has absolutely no way of knowing whether the member
pointers are pointing at memory allocated and owned by the object
itself, or whether they are just pointing to something else (for
example trying to deep-copy an element of a doubly-linked list would
be quite a bad idea).
The default assignment operator simply assigns each individual member
variable from the parameter to this, regardless of what it is. If it's
eg. a pointer, then a simple pointer assignment will be done.

In many cases the default assignment operator is ok, especially if
the class does not contain pointers.

It returns a const-reference to *this. This behavior is sometimes refered to as a "bitwise copy" although that is
not strictly true. If it was a bitwise copy, copying the bits of the class
ignoring constructors, then there would be different values then there are.

Basically it will take the base copy of whatever types are being copied.
For most things this is fine. This will work for stl::containers ,
std::string, std::vector, etc... but it usually does not give you want you
want for pointers. Your copied class will have it's pointers copied by
value, pointing to the same things they were pointing to before. In some
cases this may be what you want. In most cases it is not. One of the
biggest dangers of this behavior is when the destructor of the class
destroyes it's pointed to value (delete). Then your copied class may point
to data that is no longer valid.

Consider:

class Foo
{
public:
Foo() { MyData = new char[100]; }
~Foo() { delete[] MyData; }
private:
char* MyData;
}

If this class is copied, MyData will be copied by value, the pointer will be
copied, and you will have 2 instances pointing to the same portion of
memory, which is not what you want in most cases. Also, if the destructor
ever gets invoked, such as a temporary being created and destroyed, then
MyData will become invalid in the copied class. The normal way of dealing
with this is with a copy constructor that assigns it's own memory.

#include <algorithm>

class Foo
{
public:
Foo() { MyData = new char[100]; }
~Foo() { delete[] MyData; }
Foo( const Foo& rhs )
{
MyData = new char[100];
std::copy( &rhs.MyData[0], &rhs.MyData[99], MyData );
}
private:
char* MyData;
};

--
Jim Langston


The compiler cannot automatically generate a deep-copying of the
objects because it has absolutely no way of knowing whether the
member pointers are pointing at memory allocated and owned by the
object itself, or whether they are just pointing to something else
(for example trying to deep-copy an element of a doubly-linked list
would be quite a bad idea).
The default assignment operator simply assigns each individual
member variable from the parameter to this, regardless of what it
is. If it's eg. a pointer, then a simple pointer assignment will be
done. In many cases the default assignment operator is ok, especially if
the class does not contain pointers.

It returns a const-reference to *this.
This behavior is sometimes refered to as a "bitwise copy" although
that is not strictly true. If it was a bitwise copy, copying the
bits of the class ignoring constructors, then there would be
different values then there are.
Basically it will take the base copy of whatever types are being
copied. For most things this is fine. This will work for
stl::containers , std::string, std::vector, etc... but it usually does
not give you want you want for pointers. Your copied class will have
it's pointers copied by value, pointing to the same things they were
pointing to before. In some cases this may be what you want. In most
cases it is not. One of the biggest dangers of this behavior is when
the destructor of the class destroyes it's pointed to value (delete).
Then your copied class may point to data that is no longer valid.

Consider:

class Foo
{
public:
Foo() { MyData = new char[100]; }
~Foo() { delete[] MyData; }
private:
char* MyData;
}

If this class is copied, MyData will be copied by value, the pointer
will be copied, and you will have 2 instances pointing to the same
portion of memory, which is not what you want in most cases. Also,
if the destructor ever gets invoked, such as a temporary being
created and destroyed, then MyData will become invalid in the copied
class. The normal way of dealing with this is with a copy
constructor that assigns it's own memory.
#include <algorithm>

class Foo
{
public:
Foo() { MyData = new char[100]; }
~Foo() { delete[] MyData; }
Foo( const Foo& rhs )
{
MyData = new char[100];
std::copy( &rhs.MyData[0], &rhs.MyData[99], MyData );
}
private:
char* MyData;
}; A few things, this is the copy constructor, not the assignment operator.
Both would need to be specified (see the rule of three). The assignment
operator would look just about the same.
Second,
std::copy( &rhs.MyData[0], &rhs.MyData[99], MyData );
probably needs to be
std::copy( &rhs.MyData[0], &rhs.MyData[100], MyData );
because it is my understanding that the end iterator needs to point 1 past
the data.
--
Jim Langston


The compiler cannot automatically generate a deep-copying of the
objects because it has absolutely no way of knowing whether the member
pointers are pointing at memory allocated and owned by the object
itself, or whether they are just pointing to something else (for
example trying to deep-copy an element of a doubly-linked list would
be quite a bad idea).
The default assignment operator simply assigns each individual member
variable from the parameter to this, regardless of what it is. If it's
eg. a pointer, then a simple pointer assignment will be done.

In many cases the default assignment operator is ok, especially if
the class does not contain pointers.

It returns a const-reference to *this.
This behavior is sometimes refered to as a "bitwise copy" although that is
not strictly true. If it was a bitwise copy, copying the bits of the class
ignoring constructors, then there would be different values then there are.
That's a bit confusing, because of dangling references. So here's the
deal: the compiler-generated assignment operator does member-by-member
assignment, not bitwise assignment. Bitwise assignment copies the bits
(think memcpy). Member-by-member assignment uses each type's assignment
operator, and bitwise copying for builtin types.

--
Pete
Roundhouse Consulting, Ltd. ( ) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
( )


Not a const reference. You mean it's possible, when default assignment operators are used, to
do this: (a = b) = c; ?

I think I'm pretty sure it's not, even without trying.

Not a const reference.
You mean it's possible, when default assignment operators are used, to
do this: (a = b) = c; ?

I think I'm pretty sure it's not, even without trying. Are you still sure after trying it?

--
Pete
Roundhouse Consulting, Ltd. ( ) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
( )


The compiler cannot automatically generate a deep-copying of the
objects because it has absolutely no way of knowing whether the member
pointers are pointing at memory allocated and owned by the object
itself, or whether they are just pointing to something else (for
example trying to deep-copy an element of a doubly-linked list would be
quite a bad idea).
The default assignment operator simply assigns each individual member
variable from the parameter to this, regardless of what it is. If it's
eg. a pointer, then a simple pointer assignment will be done.

In many cases the default assignment operator is ok, especially if
the class does not contain pointers.

It returns a const-reference to *this.
This behavior is sometimes refered to as a "bitwise copy" although that
is not strictly true. If it was a bitwise copy, copying the bits of the
class ignoring constructors, then there would be different values then
there are.

Basically it will take the base copy of whatever types are being copied.
For most things this is fine. This will work for stl::containers ,
std::string, std::vector, etc... but it usually does not give you want
you want for pointers. Your copied class will have it's pointers copied
by value, pointing to the same things they were pointing to before. In
some cases this may be what you want. In most cases it is not. One of
the biggest dangers of this behavior is when the destructor of the class
destroyes it's pointed to value (delete). Then your copied class may
point to data that is no longer valid. Well, generally if an object owns some pointers, they should be wrapped
in some std::auto_ptr, boost::scoped_p tr or something like this.
Otherwise you loose exception safety and gain many headaches. OTOH
compiler generated assignment operator for class with std::auto_ptr is
even more surprising, so I recommend const std::auto_ptr.

--
Tadeusz B. Kopec (tk****@NOSPAMP LEASElife.pl)
<casMercury: gpm isn't a very good web browser. fix it.

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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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default implementation of assignment operator

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Hello everyone, I am wondering the default implementation of assignment operator (e.g. when we do not implement assignment operator in user defined class, what will be returned? temporary object? reference or const reference? deep copy or shallow copy is used in default assignment operator?)? I have the C++ Programming Book at hand, but can not find it from Index page. thanks in advance, George
It's mainly necessary when your object contains pointers, since you probably want to create a copy of the object being pointed to, not just the pointer.
I think (not positive) that it works like: Code: MyClass MyClass::operator = (const MyClass& right) {     value1 = right.value1;     value2 = right.value2;     value3 = right.value3;     // etc...     return this; } That may not be totally correct but I think I got the main idea.
The compiler generated assignment operator for a class invokes the assignment operators of any base classes, and then does a member-wise copy of any non-static data members specific to the class. The return value is a reference to the object that is being assigned (effectively a reference to *this). This allows chaining of assignments (eg a = b = c = d;). The behaviour described is modified in some circumstances (eg by attributes of or existence of assignment operators for base classes or data members). For example, if a base class has a private assignment operator, it cannot be called (implicitly or explicitly) in the content of a derived class.
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C++ Assignment Operator Overloading

Prerequisite: Operator Overloading

The assignment operator,”=”, is the operator used for Assignment. It copies the right value into the left value. Assignment Operators are predefined to operate only on built-in Data types.

  • Assignment operator overloading is binary operator overloading.
  • Overloading assignment operator in C++ copies all values of one object to another object.
  • Only a non-static member function should be used to overload the assignment operator.

We can’t directly use the Assignment Operator on objects. The simple explanation for this is that the Assignment Operator is predefined to operate only on built-in Data types. As the class and objects are user-defined data types, so the compiler generates an error.

here, a and b are of type integer, which is a built-in data type. Assignment Operator can be used directly on built-in data types.

c1 and c2 are variables of type “class C”.

The above example can be done by implementing methods or functions inside the class, but we choose operator overloading instead. The reason for this is, operator overloading gives the functionality to use the operator directly which makes code easy to understand, and even code size decreases because of it. Also, operator overloading does not affect the normal working of the operator but provides extra functionality to it.

Now, if the user wants to use the assignment operator “=” to assign the value of the class variable to another class variable then the user has to redefine the meaning of the assignment operator “=”.  Redefining the meaning of operators really does not change their original meaning, instead, they have been given additional meaning along with their existing ones.

Also, always check if the object is not being assigned to itself (e.g., if (this != &other)), as assigning an object to itself does not make sense and may cause runtime issues.

While managing dynamic resources, the above approach of assignment overloading have few flaws and there is more efficient approach that is recommended. See this article for more info – Copy-and-Swap Idiom in C++

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Implement C++ assignment operator in terms of constructor

Suppose you want to implement a resource-managing class in C++. You cannot use the Rule of Zero or Rule of Five Defaults , so you actually need to implement copy and move constructor, copy and move assignment operator, and destructor. In this question, Iʼll use Box as an example, but this can apply to lots of different types.

(Note that a better Box implementation would have more features like noexcept and constexpr functions, explicit forwarding constructors based on value_type ʼs constructors, allocator support, etc; here Iʼm implementing just whatʼs necessary for the question and the tests. It might also save some code with std::unique_ptr , but that would make it a less-clear example)

Note that the assignment operators share lots of code with each other, with their respective constructors, and with the destructor. This would be somewhat lessened if I didnʼt want to allow assignment to moved-from Box en, but would be more apparent in a more complicated class.

Digression: Copy-and-Swap

One standard way of handling this is with the Copy-And-Swap Idiom (also in this context called the Rule of Four and a Half) , which also gets you a strong exception guarantee if swap is nothrow :

This allows you to write only one assignment operator (taking other by value lets the compiler move the other to the parameter if possible, and does the copying for you if necessary), and that assignment operator is simple (assuming you already have swap ). However, as the linked article says, that has issues such as doing an extra allocation and keeping extra copies of the contents around during the operation.

What I havenʼt seen before is something Iʼll call a Destroy-and-Initialize assignment operator. Since weʼre already doing all the work in the constructor, and an assigned-to version of the object should be identical to a copy-constructed one, why not use the constructor, as follows:

This still does an extra allocation like Copy-and-Swap does, but only in the copy assignment case instead of the move assignment case, and it does it after destroying one of the copies of T , so it doesnʼt fail under resource constraints.

  • Has this been proposed before, and if so where can I read more about it?
  • Is this UB in some cases, such as if the Box is a subobject of something else, or is it allowed to destroy and re-construct subobjects?
  • Does this have any downsides I havenʼt mentioned, like not being constexpr compatible?
  • Are there other options to avoid assignment operator code reuse, like this and the Rule of Four and a Half, when you canʼt just = default them?
  • copy-and-swap
  • rule-of-zero
  • rule-of-five

Daniel H's user avatar

  • 2 This approach will have UB if any of the non-static member of the class are const qualified or of a reference type: stackoverflow.com/a/58415092/4342498 –  NathanOliver Commented Nov 30, 2021 at 21:55
  • @NathanOliver Yeah, but nothing else would work either , unless you want to special-case the handling of that member and it makes sense to have that not change on assignment. –  Daniel H Commented Nov 30, 2021 at 22:34
  • That approach will cause invalid object if copy constructor throws. –  Phil1970 Commented Dec 1, 2021 at 0:18
  • @Phil1970 Ah, youʼre right, and at least in my example Box thereʼs no way to avoid it without just specifying that it requires T be nothrow copy constructible, which isnʼt otherwise required. And even then since the copy constructor is allocating memory, that might fail. It could work for move assignment in this case, which avoids some of the code reuse but not as much. –  Daniel H Commented Dec 1, 2021 at 0:47
  • I think that Copy and swap is adequate 99% of the time. The extra copy (when required) make the assignment transactional which is generally a good thing. If you really need more control, you can always write the operator yourself. –  Phil1970 Commented Dec 2, 2021 at 1:54

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c assignment operator default implementation

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COMMENTS

  1. Default assignment operator= in c++ is a shallow copy?

    23. Yes, default operator= is a shallow copy. By the way, the actual difference between shallow copy and deep copy becomes visible when the class has pointers as member fields. In the absence of pointers, there is no difference (to the best of my knowledge)! To know the difference between them, see these topics (on stackoverflow itself): What ...

  2. Copy assignment 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.

  3. Assignment operators

    for assignments to class type objects, the right operand could be an initializer list only when the assignment is defined by a user-defined assignment operator. removed user-defined assignment constraint. CWG 1538. C++11. E1 ={E2} was equivalent to E1 = T(E2) ( T is the type of E1 ), this introduced a C-style cast. it is equivalent to E1 = T{E2}

  4. 21.12

    21.12 — Overloading the assignment operator. Alex July 22, 2024. The copy assignment operator (operator=) is used to copy values from one object to another already existing object. As of C++11, C++ also supports "Move assignment". We discuss move assignment in lesson 22.3 -- Move constructors and move assignment .

  5. 21.13

    Shallow copying. Because C++ does not know much about your class, the default copy constructor and default assignment operators it provides use a copying method known as a memberwise copy (also known as a shallow copy).This means that C++ copies each member of the class individually (using the assignment operator for overloaded operator=, and direct initialization for the copy constructor).

  6. Default Assignment Operator and References in C++

    Output: can't use default assignment operator. The compiler doesn't create default assignment operator in the following cases: 1. Class has a non-static data member of a const type or a reference type. 2. Class has a non-static data member of a type that has an inaccessible copy assignment operator. 3.

  7. default implementation of assignment operator

    On Dec 15, 7:06 am, George2 <george4acade.. [email protected] e: I am wondering the default implementation of assignment operator (e.g. when we do not implement assignment operator

  8. PDF Copy Constructors and Assignment Operators

    In situations like these, you'll need to override C++'s default behavior by providing your own copy constructors and assignment operators. The Rule of Three There's a well-established C++ principle called the "rule of three" that almost always identifies the spots where you'll need to write your own copy constructor and assignment operator.

  9. Copy constructors, assignment operators,

    Note that none of the following constructors, despite the fact that. they could do the same thing as a copy constructor, are copy. constructors: 1. 2. MyClass( MyClass* other ); MyClass( const MyClass* other ); or my personal favorite way to create an infinite loop in C++: MyClass( MyClass other );

  10. default implementation of assignment operator

    The default assignment operator simply assigns each individual member variable from the parameter to this, regardless of what it is. If it's eg. a pointer, then a simple pointer assignment will be done. In many cases the default assignment operator is ok, especially if the class does not contain pointers.

  11. Copy constructors and copy assignment operators (C++)

    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.

  12. Move assignment operator

    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 ...

  13. default implementation of assignment operator

    The compiler generated assignment operator for a class invokes the assignment operators of any base classes, and then does a member-wise copy of any non-static data members specific to the class. The return value is a reference to the object that is being assigned (effectively a reference to *this).

  14. Copy Constructor vs Assignment Operator in C++

    C++ compiler implicitly provides a copy constructor, if no copy constructor is defined in the class. A bitwise copy gets created, if the Assignment operator is not overloaded. 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 ...

  15. operator overloading

    In those situations where copy assignment cannot benefit from resource reuse (it does not manage a heap-allocated array and does not have a (possibly transitive) member that does, such as a member std::vector or std::string), there is a popular convenient shorthand: the copy-and-swap assignment operator, which takes its parameter by value (thus working as both copy- and move-assignment ...

  16. c++

    The copy assignment operator, often just called the "assignment operator", is a special case of assignment operator where the source (right-hand side) and destination (left-hand side) are of the same class type. It is one of the special member functions, which means that a default version of it is generated automatically by the compiler if the ...

  17. C++ Assignment Operator Overloading

    The assignment operator,"=", is the operator used for Assignment. It copies the right value into the left value. Assignment Operators are predefined to operate only on built-in Data types. Assignment operator overloading is binary operator overloading. Overloading assignment operator in C++ copies all values of one object to another object.

  18. Copy constructors

    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 ...

  19. Implement C++ assignment operator in terms of constructor

    You cannot use the Rule of Zero or Rule of Five Defaults, so you actually need to implement copy and move constructor, copy and move assignment operator, and destructor. In this question, Iʼll use Box as an example, but this can apply to lots of different types. using value_type = T; // Default constructor. Box() : elem{new value_type} {}