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A copy 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 without mutating the argument.

Syntax Explanation Implicitly-declared copy assignment operator Implicitly-defined copy assignment operator Deleted copy assignment operator Trivial copy assignment operator Eligible copy assignment operator Notes Example Defect reports See also

[ edit ] Syntax

For the formal copy assignment operator syntax, see function declaration . The syntax list below only demonstrates a subset of all valid copy assignment operator syntaxes.


return-type parameter-list	(1)

return-type parameter-list function-body	(2)

return-type parameter-list-no-default	(3)	(since C++11)

return-type parameter-list	(4)	(since C++11)

return-type class-name parameter-list function-body	(5)

return-type class-name parameter-list-no-default	(6)	(since C++11)

class-name	-	the class whose copy assignment operator is being declared, the class type is given as in the descriptions below
parameter-list	-	a of only one parameter, which is of type , , const T&, volatile T& or const volatile T&
parameter-list-no-default	-	a of only one parameter, which is of type , , const T&, volatile T& or const volatile T& and does not have a default argument
function-body	-	the of the copy assignment operator
return-type	-	any type, but is favored in order to allow chaining asssignments

[ edit ] Explanation

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, 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 B of T has a copy assignment operator whose parameters are B or const B & or const volatile B & ;
each non-static data member M of T of class type or array of class type has a copy assignment operator whose parameters are M or const M & or const volatile M & .

Otherwise the implicitly-declared copy assignment operator is declared as T & T :: operator = ( T & ) .

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 = ( 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 default . (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) noexcept 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 ] 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 or needed for constant evaluation (since C++14) . For union types, the implicitly-defined copy assignment copies the object representation (as by std::memmove ). For non-union class types, the operator performs member-wise copy assignment of the object's direct bases and non-static data members, in their initialization order, using built-in assignment for the scalars, memberwise copy-assignment for arrays, and copy assignment operator for class types (called non-virtually).

The implicitly-defined copy assignment operator for a class is if is a , and that is of class type (or array thereof), the assignment operator selected to copy that member is a constexpr function.	(since C++14) (until C++23)
The implicitly-defined copy assignment operator for a class is .	(since C++23)

The generation of the implicitly-defined copy assignment operator is deprecated if has a user-declared destructor or user-declared copy constructor.

(since C++11)

[ edit ] Deleted copy assignment operator

An implicitly-declared or explicitly-defaulted (since C++11) copy assignment operator for class T is undefined (until C++11) defined as deleted (since C++11) 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 copy 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.

The implicitly-declared copy assignment operator for class is defined as deleted if declares a or .

(since C++11)

[ edit ] Trivial copy assignment operator

The copy 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 copy assignment operator selected for every direct base of T is trivial;
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.

[ edit ] Eligible copy assignment operator

A copy assignment operator is eligible if it is either user-declared or both implicitly-declared and definable.	(until C++11)
A copy assignment operator is eligible if it is not deleted.	(since C++11) (until C++20)
A copy assignment operator is eligible if all following conditions are satisfied: (if any) are satisfied. than any other copy assignment operator.	(since C++20)

Triviality of eligible copy 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 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

[ edit ] defect reports.

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

Applied to	Behavior as published	Correct behavior
C++98	the conditions where implicitly-declared copy assignment operators are undefined did not consider multi-dimensional array types	consider these types
C++11	a volatile subobject made defaulted copy assignment operators non-trivial ( )	triviality not affected
C++11	operator=(X&) = default was non-trivial	made trivial
C++11	a defaulted copy assignment operator for class was not defined as deleted if is abstract and has non-copy-assignable direct virtual base classes	the operator is defined as deleted in this case
C++20	a copy assignment operator was not eligible if there is another copy assignment operator which is more constrained but does not satisfy its associated constraints	it can be eligible in this case

[ edit ] See also

converting constructor
copy constructor
copy elision
default constructor
aggregate initialization
constant initialization
copy initialization
default initialization
direct initialization
initializer list
list initialization
reference initialization
value initialization
zero initialization
move assignment
move constructor
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CopyAssignment Python Project: Master the Art of Replicating with Python!

CopyAssignment Python Project: Master the Art of Replicating with Python! 👩‍💻🐍

Ah, the thrill of venturing into the realm of CopyAssignment Python Project! 🚀 Let’s embark on an exciting journey to unravel the enchanting world of replicating with Python through this project guide.

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Exploring the Concept of Copying in Python

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Program Code – CopyAssignment Python Project: Master the Art of Replicating with Python!

Expected code output:, code explanation:.

The program demonstrates the use of shallow and deep copying in Python through a practical example of a project management context.

Class Definition : The Project class is defined with three attributes: name , duration , and team_members . Methods are defined to add team members and retrieve project details.
Shallow Copy ( copy.copy() ): This creates a new Project object ( shallow_copied_project ) that shares the same team_members list with the original object ( original_project ). Modification in team_members of the shallow copied version affects the original as well, and vice versa.
Deep Copy ( copy.deepcopy() ): This creates a new Project object ( deep_copied_project ) with a completely independent copy of the team_members list. Modifications in team_members of the deep copied project does not affect the original project list.
Adding Team Members : We add a new member to each of the shallow and deep copied projects and display the details.
Print Statements : Results are printed showing the differing effects of shallow and deep copying on the Project class members.

The final output illustrates how only the deep copied project remains independent of changes in the team_members attribute, maintaining data integrity for use cases where such separation is necessary.

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What is a copyassignment python project.

A CopyAssignment Python Project is a programming project that focuses on mastering the art of replicating data structures and objects in Python. This project involves creating functions and classes to copy or clone elements in Python efficiently.

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Copy constructors and copy assignment operators (C++)

8 contributors

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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Why must the copy assignment operator return a reference/const reference? [duplicate]

In C++, the concept of returning reference from the copy assignment operator is unclear to me. Why can't the copy assignment operator return a copy of the new object? In addition, if I have class A , and the following:

The operator= is defined as follows:

operator-overloading
copy-constructor
assignment-operator

4 There's no such requirement. But if you want to stick to the principle of least surprize you'll return A& just like a=b is an lvalue expression referring to a in case a and b are ints. – sellibitze Commented Jun 24, 2010 at 6:23
@MattMcNabb Thank you for letting me know! Will do that – bks Commented Feb 20, 2015 at 6:35
Why cant we return A* from the copy assignment operator I guess the chaining assignment would still work properly. Can anyone help understand the perils of returning A* if there are any. – krishna Commented Mar 19, 2015 at 7:35
1 Note: Since C++11 there is also the move-assignment operator , all the same logic in this Q & A also applies to the move-assignment operator. In fact they could both be the same function if declared as A & operator=(A a); , i.e. taking the argument by value. – M.M Commented Mar 27, 2015 at 2:28
1 There is now also a C++ Core Guideline regarding this, explaining the historical context as well. – MakisH Commented Feb 22, 2022 at 20:38

8 Answers 8

A bit of clarification as to why it's preferable to return by reference for operator= versus return by value --- as the chain a = b = c will work fine if a value is returned.

If you return a reference, minimal work is done. The values from one object are copied to another object.

However, if you return by value for operator= , you will call a constructor AND destructor EACH time that the assignment operator is called!!

In sum, there is nothing gained by returning by value, but a lot to lose.

( Note : This isn't meant to address the advantages of having the assignment operator return an lvalue. Read the other posts for why that might be preferable)

What I find annoying is that you're allowed to return a stupid type. I think it should enforce the non-const ref of the type... Although to delete the function, whatever it returns doesn't matter much. – Alexis Wilke Commented Feb 4, 2021 at 23:50

Strictly speaking, the result of a copy assignment operator doesn't need to return a reference, though to mimic the default behavior the C++ compiler uses, it should return a non-const reference to the object that is assigned to (an implicitly generated copy assignment operator will return a non-const reference - C++03: 12.8/10). I've seen a fair bit of code that returns void from copy assignment overloads, and I can't recall when that caused a serious problem. Returning void will prevent users from 'assignment chaining' ( a = b = c; ), and will prevent using the result of an assignment in a test expression, for example. While that kind of code is by no means unheard of, I also don't think it's particularly common - especially for non-primitive types (unless the interface for a class intends for these kinds of tests, such as for iostreams).

I'm not recommending that you do this, just pointing out that it's permitted and that it doesn't seem to cause a whole lot of problems.

These other SO questions are related (probably not quite dupes) that have information/opinions that might be of interest to you.

Has anyone found the need to declare the return parameter of a copy assignment operator const?
Overloading assignment operator in C++

I found it useful to return void on the assignment operator when I needed to prevent the automatic destruction of objects as they came of the stack. For ref-counted objects, you don't want destructors being called when you don't know about them. – cjcurrie Commented Jan 12, 2013 at 13:54

When you overload operator= , you can write it to return whatever type you want. If you want to badly enough, you can overload X::operator= to return (for example) an instance of some completely different class Y or Z . This is generally highly inadvisable though.

In particular, you usually want to support chaining of operator= just like C does. For example:

That being the case, you usually want to return an lvalue or rvalue of the type being assigned to. That only leaves the question of whether to return a reference to X, a const reference to X, or an X (by value).

Returning a const reference to X is generally a poor idea. In particular, a const reference is allowed to bind to a temporary object. The lifetime of the temporary is extended to the lifetime of the reference to which it's bound--but not recursively to the lifetime of whatever that might be assigned to. This makes it easy to return a dangling reference--the const reference binds to a temporary object. That object's lifetime is extended to the lifetime of the reference (which ends at the end of the function). By the time the function returns, the lifetime of the reference and temporary have ended, so what's assigned is a dangling reference.

Of course, returning a non-const reference doesn't provide complete protection against this, but at least makes you work a little harder at it. You can still (for example) define some local, and return a reference to it (but most compilers can and will warn about this too).

Returning a value instead of a reference has both theoretical and practical problems. On the theoretical side, you have a basic disconnect between = normally means and what it means in this case. In particular, where assignment normally means "take this existing source and assign its value to this existing destination", it starts to mean something more like "take this existing source, create a copy of it, and assign that value to this existing destination."

From a practical viewpoint, especially before rvalue references were invented, that could have a significant impact on performance--creating an entire new object in the course of copying A to B was unexpected and often quite slow. If, for example, I had a small vector, and assigned it to a larger vector, I'd expect that to take, at most, time to copy elements of the small vector plus a (little) fixed overhead to adjust the size of the destination vector. If that instead involved two copies, one from source to temp, another from temp to destination, and (worse) a dynamic allocation for the temporary vector, my expectation about the complexity of the operation would be entirely destroyed. For a small vector, the time for the dynamic allocation could easily be many times higher than the time to copy the elements.

The only other option (added in C++11) would be to return an rvalue reference. This could easily lead to unexpected results--a chained assignment like a=b=c; could destroy the contents of b and/or c , which would be quite unexpected.

That leaves returning a normal reference (not a reference to const, nor an rvalue reference) as the only option that (reasonably) dependably produces what most people normally want.

I don't see which danger situation you are referring to in the "Returning a const reference" part; if someone writes const T &ref = T{} = t; then it is a dangling reference regardless of whether operator= returned T& or T const & . Ironically it is fine if the operator= returned by value! – M.M Commented Feb 19, 2015 at 3:44
@MattMcNabb: Oops--that was supposed to say lvalue reference . Thanks for pointing it out (because yes, an rvalue reference is clearly a bad idea here). – Jerry Coffin Commented Mar 27, 2015 at 2:45

It's partly because returning a reference to self is faster than returning by value, but in addition, it's to allow the original semantics that exist in primitive types.

Not going to vote down, but I'd like to point out that returning by value would make no sense. Imagine (a = b = c) if (a = b) returned 'a' by value. Your latter point is very legit. – stinky472 Commented Jun 25, 2010 at 12:45
5 You'd get (a = (b = c)), I believe, which would still produce the intended result. Only if you did (a = b) = c would it be broken. – Puppy Commented Jun 25, 2010 at 12:51

operator= can be defined to return whatever you want. You need to be more specific as to what the problem actually is; I suspect that you have the copy constructor use operator= internally and that causes a stack overflow, as the copy constructor calls operator= which must use the copy constructor to return A by value ad infinitum.

That would be a lame (and unusual) copy-ctor implementation. The reason to return A& from A::operator= is different in most cases. – jpalecek Commented Jun 25, 2010 at 11:53
@jpalecek, I agree, but given the original post and lack of clarity when stating the actual problem, it is most likely that the assignment operator executing results in a stackoverflow due to infinite recursion. If there is another explanation for this question I would love to know it. – MSN Commented Jun 28, 2010 at 21:18
@MSN I dont know it was his problem or not . But surely your post here has addressed my problem +1 for that – Invictus Commented Apr 1, 2012 at 18:39

There is no core language requirement on the result type of a user-defined operator= , but the standard library does have such a requirement:

C++98 §23.1/3:

” The type of objects stored in these components must meet the requirements of CopyConstructible types (20.1.3), and the additional requirements of Assignable types.

C++98 §23.1/4:

” In Table 64, T is the type used to instantiate the container, t is a value of T , and u is a value of (possibly const ) T .

Returning a copy by value would still support assignment chaining like a = b = c = 42; , because the assignment operator is right-associative, i.e. this is parsed as a = (b = (c = 42)); . But returning a copy would prohibit meaningless constructions like (a = b) = 666; . For a small class returning a copy could conceivably be most efficient, while for a larger class returning by reference will generally be most efficient (and a copy, prohibitively inefficient).

Until I learned about the standard library requirement I used to let operator= return void , for efficiency and to avoid the absurdity of supporting side-effect based bad code.

With C++11 there is additionally the requirement of T& result type for default -ing the assignment operator, because

C++11 §8.4.2/1:

” A function that is explicitly defaulted shall […] have the same declared function type (except for possibly differing ref-qualifiers and except that in the case of a copy constructor or copy assignment operator, the parameter type may be “reference to non-const T ”, where T is the name of the member function’s class) as if it had been implicitly declared

Copy assignment should not be void , otherwise assignment chain will not work
Copy assignment should not return a value, otherwise unnecessary copy constructor and destructor will be called
Copy assignment should not return rvalue reference cos it may have the assigned object moved . Again take the assignment chain for example

I guess, because user defined object should behave like builtin types. For example:

Not the answer you're looking for? Browse other questions tagged c++ operator-overloading copy-constructor assignment-operator or ask your own question .

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Machine Learning: A Gentle Introduction

Introduction to Machine Learning

Machine Learning is probably one of the most interesting and hyped branches of computer science. The thing that separates humans from machines is the fact that humans learn from their experiences. But is it possible to make a machine learn? And The answer is Yes ! It is possible through Machine Learning. But, What is ML? What does it do? And what makes Machine Learning so popular? Don’t worry we’ve got you covered. But before you proceed with the topic, we suggest you have an overview of Python from our Python Tutorial Series.

What is Machine Learning?

Machine Learning is a subdomain of Artificial Intelligence where humans train the model to learn, analyze and make decisions. The Machine Learning model improves its efficiency over time by learning from its experience. Data is the lifeline of ML, and the quality and quantity of data can significantly affect the performance of your model. That’s why it a good to know how to manipulate data and build a suitable model with the data.

How does Machine Learning work?

Let’s understand the working of the Machine Learning model with an example. Consider there are 5 houses with the following specification.


1000	120000
500	60000
800	80000
900	100000
1500	180000

Let’s assume, that the price of a house is solely determined by the area of the House. We have the area of the five houses and their corresponding price. So, what would the price of a house with an area of 1200 square feet be? Based on the data we have taken above, we can say that the price falls somewhere between ₹140000 to ₹150000. After thinking for a while we might be able to pinpoint the price too. But how will you make a machine make a guess?

To understand that let’s plot the data on a graph

Now once the machine has this data plotted on the graph. The ML model will improve itself and find the best fit for the line in the curve such that each point on the line, i.e. (x,y) where x is the area of the house and y is the predicted price of that house, this line will be fitted in the plot in such a way that the predicted values are nearer to the actual values. The Line will look something like this.

Now for the above line the value corresponding to x(Area) = 1200 is y(Price) = 140451.

The above method is something called Linear Regression . We’ll learn more in this series. But how did our model learn that this particular line is the best fit? Now before learning how did it find the line let’s understand what a line is? A Line is exactly what it says, A LINE . But Mathematically, Line is an equation of the form

Now, what the model will do is it’ll start with random values of m and c and since these are random values it could be anything, let’s say m = 300 and c = 9000.

Now the line will take our data as input and update the value of m and c such that the line is the best fit for the scatter plot. The in-depth math on how to get the best values for m and c will be explained in the linear regression section but to feed your curiosity we find the best m and c (or sometimes referred to as a and b) by a method called Ordinary Least Squares.

What makes Machine Learning so special?

Now you might be wondering what’s so special about ML, all it is is just mathematics at play. And you are mostly right, Machine Learning is Math at its core. But even though the above algorithm is the most basic one, the fact that the same algorithm can improve itself to fit the given data is what makes ML special. You just feed the data to the algorithm and the algorithm will improve. And that is why ML is special, you don’t need to redesign your algorithm according to data, you just need the data and leave the improvement to ML.

Many of today’s market-leading brands such as Facebook , Google , Netflix , and Amazon , have Machine learning a central part of their operations. Machine learning has several real-time applications that drive many businesses and results in time and money savings. In particular, there is a tremendous impact occurring within customer care and product recommendations , whereby machine learning is allowing people to get things done more quickly and efficiently.

Applications of Machine Learning

The applications of Machine learning systems are also widespread in the fields of health care, banking, e-commerce, etc., Some of the widely used applications are Email spam filtering, Web search results, Pattern and Image recognition, Video recommendation, Fraud detection, Malware threat detection, Business process automation (BPA), and so on. Most social media platforms and content-delivering networks use Machine Learning algorithms to provide a more personalized and enjoyable experience.

Prerequisites to becoming a Machine Learning Engineer

Basic knowledge of programming and scripting languages.
Knowledge of programming languages like Python or R(Python is recommended)
Intermediate knowledge of statistics and probability
Knowledge of data structures and algorithms.

It is no doubt that Machine Learning becoming a more widely accepted and adapted technology in many fields. In this article, we have got a brief introduction to Machine Learning and its types. As a next step, you can check out the important steps to building a machine learning model.

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