the assignment operator is denoted by in dbms

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• 3.2 Relational algebra
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3.2. Relational algebra ¶

In the last chapter, we introduced the relational model of the database, and defined the fundamental mathematical object in the model, the relation . In this chapter, we discuss relational algebra , which is the set of algebraic operations that can be performed on relations. Relational algebra can be viewed as one mechanism for expressing queries on data stored in relations, and an understanding of relational algebra is important in understanding how relational databases represent and optimize queries. We will cover only basic relational algebra, excluding later extensions such as those for group and aggregate operations and those for outer joins.

A related topic, which we do not cover in this book, is relational calculus . Relational calculus provides another mathematical expression of queries on relations, and is equivalent in expressiveness to relational algebra.

3.2.1. Unary operations ¶

The unary operations in relational algebra act on one relation and result in another relation. The unary operations are selection , projection , and renaming , and their associated operators are typically written as the Greek letters which match the starting letters of the operation:

\(\sigma\) (sigma): selection

\(\pi\) (pi): projection

\(\rho\) (rho): renaming

We will explore each of these unary operators in application to the relation books shown below:

The books relation has primary key book_id , while author_id is a foreign key to another table we will use later in this chapter.

3.2.1.1. Selection ¶

Selection applies a Boolean condition to the tuples in a relation. The result of a selection operation is a relation containing exactly those tuples for which the selection condition is true. For example, if we are interested in books published after 1960, we can write the selection operation to retrieve just those books as:

The operator is written with the Boolean condition as a subscript, and then the operand (the input relation) is given in parentheses. Note that the Boolean condition refers to an attribute of the books relation, comparing it to a constant value. The result of this operation is a relation with the same schema as books , but with no name:

Simple Boolean expressions in the relational algebra usually involve comparisons of an attribute with a constant, using any comparison operator. More complex Boolean expressions can be constructed from simple expressions using AND , OR , and NOT . For instance, if we are interested in books published after 1960 as well as books by the author with author_id equal to 6, we could write:

Selection can result in a relation that has all of the tuples from the original (a relation equivalent to the original), some of the tuples from the original, or no tuples at all (an empty set). In the case of an empty relation, we still consider the relation to have the same schema as the original relation.

Since the result of a selection is a relation, we can apply another selection to the result. For example, we could find the books published after 1950, and then select from that result the books with author_id equal to 6:

This would give us a result with one tuple:

This composition of selection operations is equivalent to a single selection operation using a conjunction (AND) of the selection conditions:

3.2.1.2. Projection ¶

The projection operation creates a new relation which has a subset of the attributes of the input relation. We could use projection, for example, to get a set of tuples expressing just the title and publication year of our books. We write the projection operator with the list of attribute names in the subscript, followed by the operand in parentheses:

This result contains a tuple for each tuple in books , but the tuples in the result only have the attributes specified by the projection operation, thus the result relation has a different schema from the original:

At first glance, it might seem the result of a projection will always have the same number of tuples as the input relation, but this is not the case. Consider what happens if we project books onto the single attribute year . There are two tuples in books with the same year value of 1968. Since relations cannot contain duplicates, the result of our projection operation can contain only one tuple with year equal to 1968. Thus, the result has fewer tuples than the input relation:

Since the result of projection is a relation, we can apply selection to the result:

Note the order of operations here: first, we supply books as an input to the projection operation; second, the result of the projection is given as the input to the selection operation.

Similarly, since the result of a selection is a relation, we can apply projection after selection. The above expression is equivalent to:

The result in both cases is:

It is important to note, however, that you cannot always change the order of projection and selection for an equivalent result. Consider the following expressions:

In the first expression, we select the books which were published in 1968, and then project the resulting tuples onto the title attribute. This result is:

However, the second expression is not a correct expression. The projection occurs first, yielding a relation with just one attribute named title . The following selection is then incorrect, because it makes reference to an attribute, year , which does not exist in the input relation.

Projection can also be applied to the result of another projection; however, the result is equivalent to just performing the second projection. Compare:

Note that we cannot change the order of the two projection operations in the first expression above, as the expression would then be incorrect.

3.2.1.3. Renaming ¶

The final unary operation allows for relations and their attributes to be renamed. As we will see, this operation is primarily useful in eliminating name conflicts in certain binary operations - that is, in expressions involving two relations in which the name of some attribute is the same in both relations. The general form of the renaming operator lets us provide new names for the relation and all of its attributes:

This results in a relation with the name mybooks with attributes b_id , a_id , title , and year . The tuples of the new relation have the same values as the tuples of the old relation, but the values are associated with the new attribute names.

As in this example, it is not necessary to alter the name of every attribute (we left unchanged the attribute names title and year ), but some name must be provided for every attribute. A non-standard alternative notation allows us to rename only the attributes we want to change:

We can optionally leave out either the relation name or the list of attributes. For example, the following expression is correct and results in a relation named books with attributes book_id , author_id , title , and publication_year :

3.2.2. Cross products and joins ¶

We now turn our attention to operations which extend tuples in one relation with tuples from another relation. For this section, we will be using books and a second relation, authors :

The authors relation has a primary key of author_id . The books relation is related to authors via a foreign key on author_id .

3.2.2.1. Cross product ¶

The cross product (or Cartesian product ) of two relations A and B is a new relation containing all tuples that can be created by concatenating some tuple from B onto some tuple from A [ 1 ] . Here we are using the definition of tuple as an ordered list of values. The attributes of the new relation are normally the attributes of A and B concatenated. However, if there is a name collision, e.g., if both A and B have some attribute x , we will disambiguate the attributes in the new relation by prepending the relation names, that is, the cross product will have attributes A.x and B.x ; we can avoid having to do this if we first apply renaming to one relation or the other.

The cross product operator is denoted \(\times\) , and is written between its two operands. To start, consider two rather abstract relations S and T :

We write the cross product of S and T as:

which gives us the relation containing every pairing of a tuple from S with every tuple from T :

From the definition, it is trivial to determine that the size of the cross product is the product of the sizes of the operands.

3.2.2.2. Join ¶

The cross product is a fundamental operation in relational algebra, but not a generally useful one when we consider actual data. Consider the cross product of books and authors :

The full set of tuples in this relation is large (the number of books multiplied by the number of authors), so we only show a subset below:

The author of The House of the Spirits is Isabel Allende. What meaning, then, can we make of a tuple that pairs The House of the Spirits with the author Ralph Ellison (the author of Invisible Man )?

We are typically interested in pairing only certain tuples of a relation with certain tuples of another. In the above example, we are interested in tuples where the author_id attribute from books agrees with the author_id attribute from authors . This relationship is indicated not only by the names we have used for attributes, but also by the foreign key constraint on books and authors . To retain only the tuples with matching author_id values, we can apply a selection operation to the result of our cross product:

This yields a useful result:

Since this pattern of applying a selection after a cross product is so common, we have an operator that combines the two into an operation known as a join [ 2 ] . Using the join operator, the above expression becomes:

or, you can instead format the expression as:

Note that one tuple from authors does not contribute to the join. This tuple’s author_id matches none of the tuples in books , and thus no combined tuple using it can appear in the join result. We call this tuple a dangling tuple . Dangling tuples may be an indication of a problem in the data; in this example, it may suggest that we are missing information about books by one author.

3.2.2.3. Theta-join and equijoin ¶

While an equality condition is typically used in joins, more generally any condition of the following form can be used:

where A.x is an attribute from one relation, B.y is an attribute from the other relation, and \(\Theta\) is a comparison operator (such as =, <, etc.). A condition of this form is known as a theta condition , and a join using such a condition or a conjunction (AND) of such conditions is known as a theta-join .

A theta-join using only equality comparisons (as in our example above) is further known as an equijoin .

This terminology is not especially important in understanding the algebra, but is something you may encounter if you intend a deeper study of relational algebra.

3.2.2.4. Natural join ¶

When we join books with authors we run into the issue that both relations contain an attribute named author_id . Since a relation cannot have more than one attribute with the same name, joining (or taking a cross product of) these two relations requires us to rename the attributes in some fashion. This can be done either by an explicit renaming operation prior to joining or by prepending the original relation name (as we did in our example). Because our join condition was equality on the author_id attributes, both the books.author_id and authors.author_id in the resulting relation always agree. This unnecessary redundancy can be removed using projection and renaming.

In this special situation in which we wish to join specifically by equating the attributes with the same names in both relations - subsequently removing the “duplicate” attributes - we can instead do a natural join . We can indicate a natural join using the join operator with no conditions [ 3 ] :

which yields the simplified relation:

3.2.3. Set operations ¶

Unsurprisingly, given that relations are sets, relational algebra includes the usual set operations - union , intersection , and set difference - with some restrictions. These binary operations are denoted by:

\(\cup\) : union

\(\cap\) : intersection

\(-\) : set difference

Given two relations A and B , the union \(\text{A} \cup \text{B}\) is the set of all tuples that exist in A , or exist in B , or both. The intersection \(\text{A} \cap \text{B}\) is the set of all tuples that exist in both A and B . Finally, the set difference \(\text{A} - \text{B}\) is the set of all tuples that exist in A but do not exist in B .

For example, let A and B be the relations below:

Then we have:

Note that union and intersection are commutative, but set difference is not.

The important restriction on set operations in relational algebra is that the relations must be compatible in terms of their schemas. The meaning of “compatible” varies, but for our purposes, assume we view the tuples in a relation as ordered lists, where each position in the list is associated with a particular attribute and type domain. Then, if we have two relations, we require that, for a given position in the tuples in either relation, the attribute and type domain are the same. For A and B shown above, we might assert that the first position corresponds to attribute x and contains character strings, while the second position ( y ) contains integers.

A looser requirement allows attribute names (but not type domains) to differ between relations. This requirement aligns less closely with the second definition of tuple given in the previous chapter, but it eliminates the occasional need for renaming operations prior to applying set operations. If the attribute names do not match in the two relations, we adopt the attribute names from the left-hand operand for the result relation.

While intersection is a useful operation, it is not strictly needed for the algebra, as the same result can be obtained using set difference:

3.2.4. Division ¶

The operations described above are sufficient for most query needs. However, one other binary operation, division , is typically included in the basic relational algebra. To divide a relation P by another relation R , we write:

Division is the most difficult operation to describe; in a very loose sense it acts as a kind of inverse to a cross product. That is, if P , Q , and R are relations and

then it is true that

However, the reverse is not necessarily true. Rather, let P be some relation, with attributes x and y [ 4 ] . We require that R has attribute y . Then \(\text{P} \div \text{R}\) will contain the values of x which are paired (in P ) with every value of y listed in R .

We will start with an abstract example. Let P be the relation pictured below:

Then \(\text{Q} = \text{P} \div \text{R}\) is

because only the values 1 and 3 are paired with blue, green, and yellow in P . The value 2 is not paired with green, so it does not appear in the quotient. The value 3 is also paired with red, but red is not in R and thus does not affect the result.

For a more tangible example, consider the following relation, named authors_awards :

and the relation science_fiction_awards :

We might ask the question, “Which authors have received all of the science fiction book awards?” The answer is given by

Like the join and set intersection operations, division can be accomplished using other relational algebra operations; however, the construction is fairly complex. If we have relation P with attributes x and y , and relation R with attribute y , then

By carefully applying the right-hand side expression above to one of our examples, you can verify that the desired result is obtained, but the basic intuition is that we must first find the values of x in P which are not paired (in P ) with one or more y values listed in R , and then subtract that list of x values from the list of all x values in P :

Create a relation containing every x value in P paired with every y value in R :

Subtract (using set difference) P from the cross product result above. These are the possible pairings of x (in P ) and y (in R ) that do not exist in P :

Project the last result onto attribute x . These are the x values that are not paired with some value from R :

Subtract the last result from the set of all x values in P for the final solution:

3.2.5. Queries ¶

As we have seen, the operations of the relational algebra act on relations and result in relations, and thus we can apply relational operations sequentially to obtain a final desired result. With the operations we have discussed, we can express a very wide array of queries (questions to be answered by the data). We have seen examples of simple queries throughout this chapter, mostly involving one or two basic operations.

Even simple questions, however, can require the application of multiple operations. Consider the question, “What books by J.R.R. Tolkien were published after 1950?”. This is similar to a question we asked earlier, using the author ID value rather than the author’s name. With only the author’s name, we have to do a bit more work.

There are many ways to get to our desired result. One possible approach might begin with the conditions presented: the author is J.R.R. Tolkien, and the publication year is greater than 1950. Author names are in the authors relation, while publication years are in the books relation. So we might guess we need two selection operations, one on each relation:

This gives us two relations which are related by the author_id attribute present in both. So a natural join might be our next step:

Finally, we are only interested in the book titles (or possibly titles and publication years), so we finish with a projection operation:

This is only one of many possible expressions that yield identical results. Here are some equivalent expressions:

3.2.5.1. Operation sequences ¶

As queries become more complex, expressions like the ones shown above can become quite long and difficult to understand. An alternative approach is to use intermediate variables to decompose and label the parts of our expression. The result is a more sequential view of the operations.

We will demonstrate this approach with one of the queries from the last section:

Using variables, we can write this as a sequence of operations:

with R holding our final result.

3.2.5.2. Expression trees ¶

Another representation of relational algebra expressions is in the form of a tree. Expression trees are a useful visual representation of a query.

We will make use of them again in Chapter XXX, which is concerned with how database software considers different action plans for executing a query.

We will again demonstrate using the query:

The tree representation of this query looks like:

Operations start at the bottom of the tree, with the relations authors and books , and proceed upwards. We can apply either selection operation first, then the other; both must be applied before we can perform the join, and we finish with the projection.

Here is another example, corresponding to the expression:

The tree is:

Ramblings of a Crafty DBA

Assignment Operator

I am going to continue the series of very short articles or tidbits on Transaction SQL Operators. An operator is a symbol specifying an action that is performed on one or more expressions.

I will exploring the Assignment Operator today. This equality symbol = in mathematics.

There are two ways in which the assignment operator can be used: alias – associating a column heading with a expression or storage – placing the results of a expression into a variable.

The TSQL example below compares the assignment operator against old and new ANSI standards for aliasing.

The output of each calculation is listed below.

The TSQL example below is a good example of using the assignment operator to store the results of a expression into a variable. It creates a temporary table, declares a local variable and sums up the numbers in the temporary table.

The output of the summation is listed below.

Next time, I will be exploring the Bitwise Operators .

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Relational Algebra in DBMS: Operations with Examples

Relational Algebra

RELATIONAL ALGEBRA is a widely used procedural query language. It collects instances of relations as input and gives occurrences of relations as output. It uses various operations to perform this action. SQL Relational algebra query operations are performed recursively on a relation. The output of these operations is a new relation, which might be formed from one or more input relations.

Basic SQL Relational Algebra Operations

Unary relational operations.

• SELECT (symbol: σ)
• PROJECT (symbol: π)
• RENAME (symbol: ρ)

Relational Algebra Operations From Set Theory

• INTERSECTION ( ),
• DIFFERENCE (-)
• CARTESIAN PRODUCT ( x )

Binary Relational Operations

Let’s study them in detail with solutions:

The SELECT operation is used for selecting a subset of the tuples according to a given selection condition. Sigma(σ)Symbol denotes it. It is used as an expression to choose tuples which meet the selection condition. Select operator selects tuples that satisfy a given predicate.

σ is the predicate

r stands for relation which is the name of the table

p is prepositional logic

Output – Selects tuples from Tutorials where topic = ‘Database’.

Output – Selects tuples from Tutorials where the topic is ‘Database’ and ‘author’ is guru99.

Output – Selects tuples from Customers where sales is greater than 50000

Projection(π)

The projection eliminates all attributes of the input relation but those mentioned in the projection list. The projection method defines a relation that contains a vertical subset of Relation.

This helps to extract the values of specified attributes to eliminates duplicate values. (pi) symbol is used to choose attributes from a relation. This operator helps you to keep specific columns from a relation and discards the other columns.

Example of Projection:

Consider the following table

Here, the projection of CustomerName and status will give

Rename is a unary operation used for renaming attributes of a relation.

ρ (a/b)R will rename the attribute ‘b’ of relation by ‘a’.

Union operation (υ)

UNION is symbolized by ∪ symbol. It includes all tuples that are in tables A or in B. It also eliminates duplicate tuples. So, set A UNION set B would be expressed as:

The result <- A ∪ B

For a union operation to be valid, the following conditions must hold –

• R and S must be the same number of attributes.
• Attribute domains need to be compatible.
• Duplicate tuples should be automatically removed.

Consider the following tables.

A ∪ B gives

Set Difference (-)

– Symbol denotes it. The result of A – B, is a relation which includes all tuples that are in A but not in B.

• The attribute name of A has to match with the attribute name in B.
• The two-operand relations A and B should be either compatible or Union compatible.
• It should be defined relation consisting of the tuples that are in relation A, but not in B.

Intersection

An intersection is defined by the symbol ∩

Defines a relation consisting of a set of all tuple that are in both A and B. However, A and B must be union-compatible.

Cartesian Product(X) in DBMS

Cartesian Product in DBMS is an operation used to merge columns from two relations. Generally, a cartesian product is never a meaningful operation when it performs alone. However, it becomes meaningful when it is followed by other operations. It is also called Cross Product or Cross Join.

Example – Cartesian product

σ column 2 = ‘1’ (A X B)

Output – The above example shows all rows from relation A and B whose column 2 has value 1

Join Operations

Join operation is essentially a cartesian product followed by a selection criterion.

Join operation denoted by ⋈.

JOIN operation also allows joining variously related tuples from different relations.

Types of JOIN:

Various forms of join operation are:

Inner Joins:

• Natural join

Outer join:

• Left Outer Join
• Right Outer Join
• Full Outer Join

In an inner join, only those tuples that satisfy the matching criteria are included, while the rest are excluded. Let’s study various types of Inner Joins:

The general case of JOIN operation is called a Theta join. It is denoted by symbol θ

Theta join can use any conditions in the selection criteria.

For example:

When a theta join uses only equivalence condition, it becomes a equi join.

EQUI join is the most difficult operations to implement efficiently using SQL in an RDBMS and one reason why RDBMS have essential performance problems.

NATURAL JOIN (⋈)

Natural join can only be performed if there is a common attribute (column) between the relations. The name and type of the attribute must be same.

Consider the following two tables

In an outer join, along with tuples that satisfy the matching criteria, we also include some or all tuples that do not match the criteria.

Left Outer Join(A ⟕ B)

In the left outer join, operation allows keeping all tuple in the left relation. However, if there is no matching tuple is found in right relation, then the attributes of right relation in the join result are filled with null values.

Consider the following 2 Tables

Right Outer Join ( A ⟖ B )

In the right outer join, operation allows keeping all tuple in the right relation. However, if there is no matching tuple is found in the left relation, then the attributes of the left relation in the join result are filled with null values.

Full Outer Join ( A ⟗ B)

In a full outer join, all tuples from both relations are included in the result, irrespective of the matching condition.

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Related MCQs

• State true or false. i) Select operator is not a unary operator. ii) Project operator chooses subset of attributes or columns of a relation.
• The                        operation, denoted by −, allows us to find tuples that are in one relation but are not in another.
• A unit of storage that can store one or more records in a hash file organization is denoted as
• The ___________ operation, denoted by −, allows us to find tuples that are in one relation but are not in another.
• The                operator takes the results of two queries and returns only rows that appear in both result sets.
• The intersection operator is used to get the            tuples.
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• Which of the following is the comparison operator in tuple relational calculus
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• In query processing, the                        is the lowest-level operator to access data.

• ▼Operators

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What are SQL operators?

An operator performs on separate data items and returns a result. The data items are called operands or arguments. Operators are mentioned by special characters or by keywords. For example , the plus operator is represented by a plus (+) sign and the operator that checks for nulls are represented by the keywords IS NULL or IS NOT NULL.

Types of SQL operators

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How to use Assignment Operator in SQL Server - SQL Server / TSQL Tutorial Part 128

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Why should the assignment operator return a reference to the object?

I'm doing some revision of my C++, and I'm dealing with operator overloading at the minute, specifically the "="(assignment) operator. I was looking online and came across multiple topics discussing it. In my own notes, I have all my examples taken down as something like

In all the references I found online, I noticed that the operator returns a reference to the source object. Why is the correct way to return a reference to the object as opposed to the nothing at all?

• operator-overloading
• return-type
• assignment-operator

• The correct way is whatever way implements the semantics you want; the idiomatic way is certainly to return T& ( Foo& in your example). –  ildjarn Commented Jan 30, 2012 at 23:08
• @MooingDuck, I guess I phrased the question wrong. I was going on the assumption that my notes were wrong, but wanted to know why more than which was correct. –  maccard Commented Jan 30, 2012 at 23:12
• 1 possible duplicate of assignment operator return a reference to *this in C++ ; also Returning *this with an assignment operator –  Rob Kennedy Commented Jan 30, 2012 at 23:26

4 Answers 4

The usual form returns a reference to the target object to allow assignment chaining. Otherwise, it wouldn't be possible to do:

Still, keep in mind that getting right the assigment operator is tougher than it might seem .

• 1 Never knew about the Copy and Swap part. I've always just checked for self assignment, assigned values, and returned void, I guess there's more to this than I was expecting. Accepting your answer for pointing out the Copy&Swap Thanks for the response. –  maccard Commented Jan 30, 2012 at 23:20

The return type doesn't matter when you're just performing a single assignment in a statement like this:

It starts to matter when you do this:

... and really matters when you do this:

That's why you return the current object: to allow chaining assignments with the correct associativity. It's a good general practice.

• 1 I don't understand why you say "it starts to matter". Either it matters or it doesn't. Can you please elaborate? –  balajeerc Commented Jun 20, 2015 at 6:15
• 3 @balajeerc: "It starts to matter" in English is read to mean "it matters in the latter situation but not the former". In other words, "when changing from situation A to B, importance ('mattering') goes from zero to nonzero". In straight assignment the return does not matter. Inside a conditional it matters if the thing you return is true or false, but not exactly which object it is. In the chained assignments case, you really want the return to be the current object because the results would be counterintuitive otherwise. –  Borealid Commented Jun 23, 2015 at 19:42

Your assignment operator should always do these three things:

Take a const-reference input (const MyClass &rhs) as the right hand side of the assignment. The reason for this should be obvious, since we don't want to accidentally change that value; we only want to change what's on the left hand side.

Always return a reference to the newly altered left hand side, return *this . This is to allow operator chaining, e.g. a = b = c; .

Always check for self assignment (this == &rhs) . This is especially important when your class does its own memory allocation.

• 2 Checking for self-assignment is a naive solution, the correct one is copy-and-swap. –  Matteo Italia Commented Jan 30, 2012 at 23:18
• Thanks for the response, but I was only trying to make a simple example by leaving out the self assignment check. I understood everything bar the returning a reference. –  maccard Commented Jan 30, 2012 at 23:23
• 1 @MatteoItalia Copy-and-swap can be expensive. For example, assignment of one large vector to another cannot reuse the target's memory if copy-and-swap is used. –  Sebastian Redl Commented Apr 1, 2016 at 10:04
• Check out the reference given in the accepted answer. It tells you why passing rhs by value may be a perfectly valid option. In Copy assignment operator - cppreference.com , both options of non-trivial implementations are distinguished (your answer reflects only the second one). This unfortunately makes your seemingly simple recipe a bit misleading. –  Wolf Commented Aug 29, 2022 at 9:13

When you overload an operator and use it, what really happens at compilation is this:

So you can see that the object b of type FOO is passed by value as argument to the object a's assignment function of the same type. Now consider this, what if you wanted to cascade assignment and do something like this:

It would be efficient to pass the objects by reference as argument to the function call because we know that NOT doing that we pass by value which makes a copy inside a function of the object which takes time and space.

The statement 'b.operator=(c)' will execute first in this statement and it will return a reference to the object had we overloaded the operator to return a reference to the current object:

Now our statement:

Where 'this' is the reference to the object that was returned after the execution of 'b.operator=(c)'. Object's reference is being passed here as the argument and the compiler doesn't have to create a copy of the object that was returned.

Had we not made the function to return Foo object or its reference and had made it return void instead:

The statement would've become something like:

And this would've thrown compilation error.

TL;DR You return the object or the reference to the object to cascade(chain) assignment which is:

• If the assignment operator returns by value, calling assginment function will invoke copy constructor which will consume more time and space. What I want to ask is, will there be some other side effects, or time and space is the only side effect of returning by value other than returning by reference? –  Lake_Lagunita Commented Jan 29, 2023 at 1:44

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