assignment operator is left or right associative

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Operator precedence

Operator precedence determines how operators are parsed concerning each other. Operators with higher precedence become the operands of operators with lower precedence.

Precedence and associativity

Consider an expression describable by the representation below, where both OP1 and OP2 are fill-in-the-blanks for OPerators.

The combination above has two possible interpretations:

Which one the language decides to adopt depends on the identity of OP1 and OP2 .

If OP1 and OP2 have different precedence levels (see the table below), the operator with the higher precedence goes first and associativity does not matter. Observe how multiplication has higher precedence than addition and executed first, even though addition is written first in the code.

Within operators of the same precedence, the language groups them by associativity . Left-associativity (left-to-right) means that it is interpreted as (a OP1 b) OP2 c , while right-associativity (right-to-left) means it is interpreted as a OP1 (b OP2 c) . Assignment operators are right-associative, so you can write:

with the expected result that a and b get the value 5. This is because the assignment operator returns the value that is assigned. First, b is set to 5. Then the a is also set to 5 — the return value of b = 5 , a.k.a. right operand of the assignment.

As another example, the unique exponentiation operator has right-associativity, whereas other arithmetic operators have left-associativity.

Operators are first grouped by precedence, and then, for adjacent operators that have the same precedence, by associativity. So, when mixing division and exponentiation, the exponentiation always comes before the division. For example, 2 ** 3 / 3 ** 2 results in 0.8888888888888888 because it is the same as (2 ** 3) / (3 ** 2) .

For prefix unary operators, suppose we have the following pattern:

where OP1 is a prefix unary operator and OP2 is a binary operator. If OP1 has higher precedence than OP2 , then it would be grouped as (OP1 a) OP2 b ; otherwise, it would be OP1 (a OP2 b) .

If the unary operator is on the second operand:

Then the binary operator OP2 must have lower precedence than the unary operator OP1 for it to be grouped as a OP2 (OP1 b) . For example, the following is invalid:

Because + has higher precedence than yield , this would become (a + yield) 1 — but because yield is a reserved word in generator functions, this would be a syntax error. Luckily, most unary operators have higher precedence than binary operators and do not suffer from this pitfall.

If we have two prefix unary operators:

Then the unary operator closer to the operand, OP2 , must have higher precedence than OP1 for it to be grouped as OP1 (OP2 a) . It's possible to get it the other way and end up with (OP1 OP2) a :

Because await has higher precedence than yield , this would become (await yield) 1 , which is awaiting an identifier called yield , and a syntax error. Similarly, if you have new !A; , because ! has lower precedence than new , this would become (new !) A , which is obviously invalid. (This code looks nonsensical to write anyway, since !A always produces a boolean, not a constructor function.)

For postfix unary operators (namely, ++ and -- ), the same rules apply. Luckily, both operators have higher precedence than any binary operator, so the grouping is always what you would expect. Moreover, because ++ evaluates to a value , not a reference , you can't chain multiple increments together either, as you may do in C.

Operator precedence will be handled recursively . For example, consider this expression:

First, we group operators with different precedence by decreasing levels of precedence.

• The ** operator has the highest precedence, so it's grouped first.
• Looking around the ** expression, it has * on the right and + on the left. * has higher precedence, so it's grouped first. * and / have the same precedence, so we group them together for now.
• Looking around the * / / expression grouped in 2, because + has higher precedence than >> , the former is grouped.

Within the * / / group, because they are both left-associative, the left operand would be grouped.

Note that operator precedence and associativity only affect the order of evaluation of operators (the implicit grouping), but not the order of evaluation of operands . The operands are always evaluated from left-to-right. The higher-precedence expressions are always evaluated first, and their results are then composed according to the order of operator precedence.

If you are familiar with binary trees, think about it as a post-order traversal .

After all operators have been properly grouped, the binary operators would form a binary tree. Evaluation starts from the outermost group — which is the operator with the lowest precedence ( / in this case). The left operand of this operator is first evaluated, which may be composed of higher-precedence operators (such as a call expression echo("left", 4) ). After the left operand has been evaluated, the right operand is evaluated in the same fashion. Therefore, all leaf nodes — the echo() calls — would be visited left-to-right, regardless of the precedence of operators joining them.

Short-circuiting

In the previous section, we said "the higher-precedence expressions are always evaluated first" — this is generally true, but it has to be amended with the acknowledgement of short-circuiting , in which case an operand may not be evaluated at all.

Short-circuiting is jargon for conditional evaluation. For example, in the expression a && (b + c) , if a is falsy , then the sub-expression (b + c) will not even get evaluated, even if it is grouped and therefore has higher precedence than && . We could say that the logical AND operator ( && ) is "short-circuited". Along with logical AND, other short-circuited operators include logical OR ( || ), nullish coalescing ( ?? ), and optional chaining ( ?. ).

When evaluating a short-circuited operator, the left operand is always evaluated. The right operand will only be evaluated if the left operand cannot determine the result of the operation.

Note: The behavior of short-circuiting is baked in these operators. Other operators would always evaluate both operands, regardless if that's actually useful — for example, NaN * foo() will always call foo , even when the result would never be something other than NaN .

The previous model of a post-order traversal still stands. However, after the left subtree of a short-circuiting operator has been visited, the language will decide if the right operand needs to be evaluated. If not (for example, because the left operand of || is already truthy), the result is directly returned without visiting the right subtree.

Consider this case:

Only C() is evaluated, despite && having higher precedence. This does not mean that || has higher precedence in this case — it's exactly because (B() && A()) has higher precedence that causes it to be neglected as a whole. If it's re-arranged as:

Then the short-circuiting effect of && would only prevent C() from being evaluated, but because A() && C() as a whole is false , B() would still be evaluated.

However, note that short-circuiting does not change the final evaluation outcome. It only affects the evaluation of operands , not how operators are grouped — if evaluation of operands doesn't have side effects (for example, logging to the console, assigning to variables, throwing an error), short-circuiting would not be observable at all.

The assignment counterparts of these operators ( &&= , ||= , ??= ) are short-circuited as well. They are short-circuited in a way that the assignment does not happen at all.

The following table lists operators in order from highest precedence (18) to lowest precedence (1).

Several general notes about the table:

• Not all syntax included here are "operators" in the strict sense. For example, spread ... and arrow => are typically not regarded as operators. However, we still included them to show how tightly they bind compared to other operators/expressions.
• Some operators have certain operands that require expressions narrower than those produced by higher-precedence operators. For example, the right-hand side of member access . (precedence 17) must be an identifier instead of a grouped expression. The left-hand side of arrow => (precedence 2) must be an arguments list or a single identifier instead of some random expression.
• Some operators have certain operands that accept expressions wider than those produced by higher-precedence operators. For example, the bracket-enclosed expression of bracket notation [ … ] (precedence 17) can be any expression, even comma (precedence 1) joined ones. These operators act as if that operand is "automatically grouped". In this case we will omit the associativity.
• The operand can be any expression.
• The "right-hand side" must be an identifier.
• The "right-hand side" can be any expression.
• The "right-hand side" is a comma-separated list of any expression with precedence > 1 (i.e. not comma expressions).
• The operand must be a valid assignment target (identifier or property access). Its precedence means new Foo++ is (new Foo)++ (a syntax error) and not new (Foo++) (a TypeError: (Foo++) is not a constructor).
• The operand must be a valid assignment target (identifier or property access).
• The operand cannot be an identifier or a private property access.
• The left-hand side cannot have precedence 14.
• The operands cannot be a logical OR || or logical AND && operator without grouping.
• The "left-hand side" must be a valid assignment target (identifier or property access).
• The associativity means the two expressions after ? are implicitly grouped.
• The "left-hand side" is a single identifier or a parenthesized parameter list.
• Only valid inside object literals, array literals, or argument lists.

Documentation

Operator precedence: we can do better.

Feb 26, 2019 • Jeff Walker

The longer I’ve thought about how to handle operator precedence and associatively in a programming language, the more convinced I’ve become that languages have fallen short. Because it was simple, easy and efficient, language designers have generally provided a total order for operator precedence and made all operators associative. This is typically expressed as a set of operator precedence levels and associativity for each operator. However, this often leads to unexpected or even confusing precedence between operators. Languages allowing programmers to define new operators from combinations of symbols are particularly hurt by forcing all operators to be placed in one of a few precedence levels. In reaction, some designers eschew operator precedence entirely. While simple, that violates deep-seated programmer intuitions opening the way for mistakes and surprise. I believe future languages should adopt intransitive operator precedence instead.

Note: I am focused here only on language with infix operators. Languages using prefix notation , such as Lisp variants, and languages using postfix notation , such as Forth, can be unambiguous without operator precedence.

Existing Practise

Most programming languages with infix operators fall into one of four categories:

• Total Order Precedence and Total Associativity: Every operator has a precedence relative to every other operator. Every operator is either left- or right-associative. Example Languages: C, C++, C♯, Java, Go, Lua, Kotlin
• Total Order Precedence with Partial Associativity: Every operator has a precedence relative to every other operator. Some operators are neither left- nor right-associative. In some languages, there are non-associative operators. For example, in Rust x <= y == z is illegal and would need to have parentheses added. In other languages, chained operators are interpreted differently. For example, in Python x < y < z is equivalent to x < y and y < z . Example Languages: Python, Rust, Prolog, Haskell, Perl
• Single Precedence and Associativity: Every infix operator has the same precedence and associativity. Unary operators may or may not have higher precedence than binary operators. Example Languages: Smalltalk, APL, Mary
• Single Precedence and Non-associative: Every infix operator has the same precedence and is non-associative. Thus all expressions must be fully disambiguated with parentheses. Unary operators may or may not have higher precedence than binary operators. Example Languages: occam , Pony , RELAX NG

Unfortunately, each these options has shortcomings. A set of test expressions best illustrates this.

• x + y * z is almost universally read as x + (y * z) because this is the convention everyone is taught from elementary school onward. Breaking this convention will only lead to confusion and frustration. Requiring explicit parentheses, in this case, isn’t as bad, but is still annoying.
• x < y < z is probably either a bug or meant to mean x < y and y < z . Treating relational operators as left-associative has led to hard to spot bugs in C code.
• By mathematical convention, logical-and has higher precedence than logical-or, so a or b and c should be parsed as a or (b and c) . However, there is no convention for the relative precedence of logical-xor. Any precedence assigned to it will be arbitrary. Yet, all logical connective should have lower precedence than equality. Thus we need an operator that has no precedence relative to some operators, but precedence relative to others so that a xor x == y parses as a xor (x == y) , but a xor b or c is an error.

Let’s consider how each of the approaches fairs on our test cases. Of course, we don’t want to evaluate a single language, but an idealized version of each approach. Single precedence and associativity requires that all operators be either left- or right-associative; which should we pick? Regardless of which is chosen, it will be easy to construct examples where it is incorrect for the operators involved. To simplify the test, I’ve always assumed the worst case for the given test.

Partial Order

Of the existing options, total order precedence with partial associativity scores the best. However, it fails to treat a xor b or c as an error. How can we fix this? Well, we could make operator precedence a partial order instead of a total order. We could then include in our precedence or ≺ and , xor ≺ == , or ≺ == , and and ≺ == . That would correctly handle both a xor x == y and a xor b or c .

However, using a partial order for operator precedence can still lead to problems. Consider the expression x and y + z . Since this mixes logical and arithmetic operators, there isn’t an obvious precedence. We want to force the developer to add parentheses. One might think this is not a problem for a partial order. Yet, logical operators are lower precedence than equality ( and ≺ == ) and equality is lower precedence than arithmetic ( == ≺ + ) . Since partial order relations are transitive, those imply that and ≺ + . That isn’t what we want, so we need a precedence relation that is intransitive .

Intransitive Precedence

Let’s define the kind of precedence we want. I’ll call this an intransitive operator precedence . We’ll define both an equivalence relation “≐” for operators at the same precedence and a compatible order relation “⋖” for operators with different precedence. However, our precedence relation will be intransitive. Additionally, we’ll require that the precedence form a DAG . We can then use them to define the precedence relationships between our operators. Associativity will be specified separately.

For the mathematically inclined, the relations have the following properties:

• a ≐ a ( reflexivity )
• if a ≐ b then b ≐ a ( symmetry )
• if a ≐ b and b ≐ c then a ≐ c ( transitivity )
• It is never the case that a ⋖ a ( irreflexivity )
• If a ⋖ b , then it is not the case that b ⋖ a ( asymmetry )
• If a ⋖ b and b ⋖ c , it does not follow that a ⋖ c (but it could be the case) ( intransitivity )
• There does not exist a 0 , … , a n such that a 0 ⋖ a 1 , … , a n -1 ⋖ a n and a n ⋖ a 0 ( acyclic )
• If a ≐ b and a ⋖ c , then b ⋖ c . Likewise if a ≐ b and d ⋖ a , then d ⋖ b .

This allows us to declare our desired precedence reasonably easily. First, we declare which operators have equal precedence, for example * ≐ / . Then we declare the relative precedence of operators, for example or ⋖ and . Operators of equal precedence share in the precedence we define. However, because precedence is intransitive, there can still be a lot of relations to specify. To simplify, we adopt two notational conveniences. First, that a precedence chain relates every operator to every other operator before and after it so that or ⋖ and ⋖ not states that or ⋖ not as well and second, that groups of operators can be related by using sets. For example, { and , or , not } ⋖ == relates all the boolean operators to the equality operator.

It’s easy to get lost in the math and notation. Let’s look at a concrete example to see how this might play out in a real language. Below I’ve defined a simple expression language over integers and booleans. To be clear, I’m not arguing for this particular set of operator precedences. Other language designers may prefer slightly different ones. I am arguing that languages should use this kind of flexible precedence to avoid undesirable precedence relationships.

I’ve used a form of EBNF augmented with additional notation to represent associativity and intransitive operator precedence. Without these additional annotations, the grammar would be an ambiguous expression grammar . The intent is that a hypothetical parser generator could directly use this grammar. The grammar notation section below gives a detailed explanation of the additional notation used.

This grammar tries to follow mathematical conventions without relating operators that have no conventional relationship. Powers are right associative and higher precedence than negation. The division slash is non-associative to avoid confusion. It does follow the C convention and make equality lower precedence than relations. The test cases below demonstrate the grammar has the desired properties.

Added Grammar Notation

In the grammar above (E) = indicates that E is a “parenthesized” nonterminal. Normally, the declaration of E would be ambiguous, but a parenthesized nonterminal defaults to disallowing alternatives containing recursive uses of the nonterminal from being immediate children of the production. Thus (P) = P "~" P | P "$" P | ID; is effectively transformed to P = P' "~" P' | P' "$" P'; P' = ID; . This has the effect of making the operators non-associative. The intuition here is that parenthesized nonterminals will have to be fully parenthesized unless additional associativity and precedence rules are declared.

Associativity is indicated by enclosing a recursive use of the nonterminal in parentheses. A recursive use enclosed in parentheses allows the same alternative to occur as a direct child of that nonterminal. Thus (E) = (E) "+" E is left-associative and (E) = E "^" (E) is right-associative. The rule (P) = (P) "~" (P) is ambiguous. Again, non-associative is the default for parenthesized nonterminals, i.e. (E) = E "<" E . Intuitively, the parentheses indicate which side expressions should be grouped on. One wrinkle this creates is that to allow nesting of parentheses in parentheses, the nonterminal must be enclosed in parentheses as (E) = "(" (E) ")" or else ((x)) is illegal.

Labels are applied to each alternative by placing them after the alternative. Labels are prefixed with a pound sign. The ANTLR parser generator uses the same notation. Labels provide a way to refer to alternatives later. They can also be used by a parser generator to name the AST node for that alternative.

Operator precedence is declared after the production rules using the precedence relation applied to the alternative labels. Using the labels makes it easy to give binary and unary versions of the same operator different precedences. Two operators are given the same precedence using the =.= operator. Relative precedence is established using the <. operator. As described in the previous section, chains and sets can be used to simplify the declaration of precedence. A precedence declaration affects recursive uses of the nonterminal in the alternatives it relates. Alternatives with higher precedence may be direct children at any use of the nonterminal. Alternatives with equal precedence may only be direct children where the nonterminal is enclosed in parentheses.

In some instances, more complex operators have different precedence for different subexpressions. An array indexing operator (P) = P "[" P "]" #Index; would be such a situation. Here, the bracketed P could be of any precedence while the left P must be higher precedence. In such situations, we can refer to the precedence of the subexpressions using a bracket notation listing the indexes of nonterminals in the alternative. For example, #Index[1] refers to the first subexpression, #Index[2] refers to the second, and #Index[1,2] refers to both. For convenience, four shorthands are provided. The names left and right refer to the leftmost and rightmost nonterminal not bracketed by a terminal. In the example, #Index[left] is the same as #Index[1] while #Index[right] is an error because the rightmost P has the terminal "]" to its right. The name outer refers to both the left and right so #X[outer] would be equal to #X[left, right] . The name inner refers to every subexpression that is not an outer subexpression. Thus #Index[inner] would be equal to #Index[2] . In the example grammar above, this is used to allow a negative sign in the exponent while giving exponentiation higher precedence and to allow logical but not arithmetic expressions in the condition of a conditional expression.

Don’t Mix Associativity

To consider the issues involved in mixing operators with different associativity at the same precedence level, imagine adding the following to the above grammar.

By the rules stated before, what would the effect of this be? Let’s look at each case.

Given that this is almost certainly not what one wants, it is best to simply make it illegal to have operators with the same precedence but different associativity.

Assignment Example

In C style languages the assignment operator is right-associative and evaluates to the value of the left-hand variable after it is assigned. Assignment has lower precedence than addition, so the expression a+b=c+d parses to (a+b)=(c+d) which is illegal. One might prefer that it parse as a+(b=(c+d)) . Setting aside whether that is a good idea, it can be achieved with this scheme. The example expression grammar could be extended with assignment by adding the rule and precedences below. By splitting the precedence of the left and right, we can make assignment bind very tightly on the left, but very loosely on the right.

I’m not the first one to propose something like intransitive operator precedence. The Fortress language has an elaborate operator precedence scheme that is similar. Check out The Fortress Language Specification v1.0 , chapter 16 for more information. However, it was difficult to find much else. The precedence level approach seems to have completely dominated. Hopefully, I’ve convinced you of the value of intransitive operator precedence or at least given you something to think about. I’d love to see future programming languages adopt this approach. Unfortunately, algorithms for parsing such precedence schemes are lacking. If you want to implement such a scheme or are interested in learning more, check out these sources:

• Parsing Fortress Syntax by Sukyoung Ryu.
• Parsing Mixfix Operators by Nils Anders Danielsson and Ulf Norell which proposes a similar scheme for the Agda language.
• SDF and SDF3 which provide both position based priority and intransitive priority for parsing.

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Assignment operators (C# reference)

• 11 contributors

The assignment operator = assigns the value of its right-hand operand to a variable, a property , or an indexer element given by its left-hand operand. The result of an assignment expression is the value assigned to the left-hand operand. The type of the right-hand operand must be the same as the type of the left-hand operand or implicitly convertible to it.

The assignment operator = is right-associative, that is, an expression of the form

is evaluated as

The following example demonstrates the usage of the assignment operator with a local variable, a property, and an indexer element as its left-hand operand:

The left-hand operand of an assignment receives the value of the right-hand operand. When the operands are of value types , assignment copies the contents of the right-hand operand. When the operands are of reference types , assignment copies the reference to the object.

This is called value assignment : the value is assigned.

ref assignment

Ref assignment = ref makes its left-hand operand an alias to the right-hand operand, as the following example demonstrates:

In the preceding example, the local reference variable arrayElement is initialized as an alias to the first array element. Then, it's ref reassigned to refer to the last array element. As it's an alias, when you update its value with an ordinary assignment operator = , the corresponding array element is also updated.

The left-hand operand of ref assignment can be a local reference variable , a ref field , and a ref , out , or in method parameter. Both operands must be of the same type.

Compound assignment

For a binary operator op , a compound assignment expression of the form

is equivalent to

except that x is only evaluated once.

Compound assignment is supported by arithmetic , Boolean logical , and bitwise logical and shift operators.

Null-coalescing assignment

You can use the null-coalescing assignment operator ??= to assign the value of its right-hand operand to its left-hand operand only if the left-hand operand evaluates to null . For more information, see the ?? and ??= operators article.

Operator overloadability

A user-defined type can't overload the assignment operator. However, a user-defined type can define an implicit conversion to another type. That way, the value of a user-defined type can be assigned to a variable, a property, or an indexer element of another type. For more information, see User-defined conversion operators .

A user-defined type can't explicitly overload a compound assignment operator. However, if a user-defined type overloads a binary operator op , the op= operator, if it exists, is also implicitly overloaded.

C# language specification

For more information, see the Assignment operators section of the C# language specification .

• C# operators and expressions
• ref keyword
• Use compound assignment (style rules IDE0054 and IDE0074)

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C# Operator Precedence

Operator precedence is a set of rules which defines how an expression is evaluated. In C#, each C# operator has an assigned priority and based on these priorities, the expression is evaluated.

For example , the precedence of multiplication (*) operator is higher than the precedence of addition (+) operator. Therefore, operation involving multiplication is carried out before addition.

Take a look at the statement below.

What will be the value of x after executing this statement?

The operand 3 is associated with + and * . As stated earlier, multiplication has a higher precedence than addition. So, the operation 3 * 5 is carried out instead of 4 + 3 . The value of variable x will be 19 .

If addition would have a higher precedence, 4 + 3 would be evaluated first and the value of x would be 35 .

• Operator Precedence Table

The higher the precedence of operator is, the higher it appears in the table

The assignment operators have the lowest precedence while the postfix increment and decrement operators have the highest precedence.

• Example 1: Operator Precedence

When we run the program, the output will be:

Let's understand how the expression is evaluated in the program.

The precedence of -- and ++ is higher than * , and precedence of * is higher than - . Hence the statement,

is equivalent to

The expression inside parentheses is always evaluated first no matter what the precedence of operators outside it is.

• Hence the final value of result1 will be 19 .

In the next expression, the precedence of + is higher than >= . So, c and a is added first and the sum is compared with b to produce false .

Associativity of Operators in C#

In the previous section, we discussed about operator precedence. If two operators with different precedence are used, the operator with higher precedence is evaluated first.

But what if both the operators have same precedence?

In such case, the expression is evaluated based on the associativity of operator (left to right or right to left).

For example:

Here, both * and / have the same precedence. But since the associativity of these operators is from left to right , a * b is evaluated first and then division is carried out. The final result of this expression will be 10 .

In this particular example, the associativity does not really matter. Because even if division was carried out before multiplication, the result would be unaffected.

Let's take a look at another example.

The associativity of = operator is from right to left . So the value of c (i.e. 3 ) is assigned to b , and then the value of b is assigned to a . So after executing this statement, the values of a , b and c will be 3 .

The table below shows the associativity of C# operators:

Almost all the operators have associativity from left to right. The operators having associativity from right to left are:

• Unary operators
• Prefix Increment and Decrement Operators
• Ternary Operator
• Assignment Operators

Example 2: Associativity of Operators

Table of contents.

• Operator Associativity Table
• Example 1: Operator Associativity

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Java Operator Associativity

Operators with the same precedence follow the operator group's operator associativity. Operators in Java can be left-associative, right-associative, or have no associativity at all. Left-associative operators are assessed from left to right, right-associative operators are reviewed from right to left, and operators with no associativity are evaluated in any order.

Operator Precedence Vs. Operator Associativity

The operator's precedence refers to the order in which operators are evaluated within an expression whereas associativity refers to the order in which the consecutive operators within the same group are carried out.

Precedence rules specify the priority (which operators will be evaluated first) of operators.

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• Java Variables
• Scope of Variables In Java

Wrapper Classes in Java

Input/output in java.

• How to Take Input From User in Java?
• Scanner Class in Java
• Java.io.BufferedReader Class in Java
• Difference Between Scanner and BufferedReader Class in Java
• Ways to read input from console in Java
• System.out.println in Java
• Difference between print() and println() in Java
• Formatted Output in Java using printf()
• Fast I/O in Java in Competitive Programming

Flow Control in Java

• Decision Making in Java (if, if-else, switch, break, continue, jump)
• Java if statement with Examples
• Java if-else
• Java if-else-if ladder with Examples
• Loops in Java
• For Loop in Java
• Java while loop with Examples
• Java do-while loop with Examples
• For-each loop in Java
• Continue Statement in Java
• Break statement in Java
• Usage of Break keyword in Java
• return keyword in Java
• Java Arithmetic Operators with Examples
• Java Unary Operator with Examples

Java Assignment Operators with Examples

• Java Relational Operators with Examples
• Java Logical Operators with Examples
• Java Ternary Operator with Examples
• Bitwise Operators in Java
• Strings in Java
• String class in Java
• Java.lang.String class in Java | Set 2
• Why Java Strings are Immutable?
• StringBuffer class in Java
• StringBuilder Class in Java with Examples
• String vs StringBuilder vs StringBuffer in Java
• StringTokenizer Class in Java
• StringTokenizer Methods in Java with Examples | Set 2
• StringJoiner Class in Java
• Arrays in Java
• Arrays class in Java
• Multidimensional Arrays in Java
• Different Ways To Declare And Initialize 2-D Array in Java
• Jagged Array in Java
• Final Arrays in Java
• Reflection Array Class in Java
• util.Arrays vs reflect.Array in Java with Examples

OOPS in Java

• Object Oriented Programming (OOPs) Concept in Java
• Why Java is not a purely Object-Oriented Language?
• Classes and Objects in Java
• Naming Conventions in Java
• Java Methods

Access Modifiers in Java

• Java Constructors
• Four Main Object Oriented Programming Concepts of Java

Inheritance in Java

Abstraction in java, encapsulation in java, polymorphism in java, interfaces in java.

• 'this' reference in Java
• Inheritance and Constructors in Java
• Java and Multiple Inheritance
• Interfaces and Inheritance in Java
• Association, Composition and Aggregation in Java
• Comparison of Inheritance in C++ and Java
• abstract keyword in java
• Abstract Class in Java
• Difference between Abstract Class and Interface in Java
• Control Abstraction in Java with Examples
• Difference Between Data Hiding and Abstraction in Java
• Difference between Abstraction and Encapsulation in Java with Examples
• Difference between Inheritance and Polymorphism
• Dynamic Method Dispatch or Runtime Polymorphism in Java
• Difference between Compile-time and Run-time Polymorphism in Java

Constructors in Java

• Copy Constructor in Java
• Constructor Overloading in Java
• Constructor Chaining In Java with Examples
• Private Constructors and Singleton Classes in Java

Methods in Java

• Static methods vs Instance methods in Java
• Abstract Method in Java with Examples
• Overriding in Java
• Method Overloading in Java
• Difference Between Method Overloading and Method Overriding in Java
• Differences between Interface and Class in Java
• Functional Interfaces in Java
• Nested Interface in Java
• Marker interface in Java
• Comparator Interface in Java with Examples
• Need of Wrapper Classes in Java
• Different Ways to Create the Instances of Wrapper Classes in Java
• Character Class in Java
• Java.Lang.Byte class in Java
• Java.Lang.Short class in Java
• Java.lang.Integer class in Java
• Java.Lang.Long class in Java
• Java.Lang.Float class in Java
• Java.Lang.Double Class in Java
• Java.lang.Boolean Class in Java
• Autoboxing and Unboxing in Java
• Type conversion in Java with Examples

Keywords in Java

• Java Keywords
• Important Keywords in Java
• Super Keyword in Java
• final Keyword in Java
• static Keyword in Java
• enum in Java
• transient keyword in Java
• volatile Keyword in Java
• final, finally and finalize in Java
• Public vs Protected vs Package vs Private Access Modifier in Java
• Access and Non Access Modifiers in Java

Memory Allocation in Java

• Java Memory Management
• How are Java objects stored in memory?
• Stack vs Heap Memory Allocation
• How many types of memory areas are allocated by JVM?
• Garbage Collection in Java
• Types of JVM Garbage Collectors in Java with implementation details
• Memory leaks in Java
• Java Virtual Machine (JVM) Stack Area

Classes of Java

• Understanding Classes and Objects in Java
• Singleton Method Design Pattern in Java
• Object Class in Java
• Inner Class in Java
• Throwable Class in Java with Examples

Packages in Java

• Packages In Java
• How to Create a Package in Java?
• Java.util Package in Java
• Java.lang package in Java
• Java.io Package in Java
• Java Collection Tutorial

Exception Handling in Java

• Exceptions in Java
• Types of Exception in Java with Examples
• Checked vs Unchecked Exceptions in Java
• Java Try Catch Block
• Flow control in try catch finally in Java
• throw and throws in Java
• User-defined Custom Exception in Java
• Chained Exceptions in Java
• Null Pointer Exception In Java
• Exception Handling with Method Overriding in Java
• Multithreading in Java
• Lifecycle and States of a Thread in Java
• Java Thread Priority in Multithreading
• Main thread in Java
• Java.lang.Thread Class in Java
• Runnable interface in Java
• Naming a thread and fetching name of current thread in Java
• What does start() function do in multithreading in Java?
• Difference between Thread.start() and Thread.run() in Java
• Thread.sleep() Method in Java With Examples
• Synchronization in Java
• Importance of Thread Synchronization in Java
• Method and Block Synchronization in Java
• Lock framework vs Thread synchronization in Java
• Difference Between Atomic, Volatile and Synchronized in Java
• Deadlock in Java Multithreading
• Deadlock Prevention And Avoidance
• Difference Between Lock and Monitor in Java Concurrency
• Reentrant Lock in Java

File Handling in Java

• Java.io.File Class in Java
• Java Program to Create a New File
• Different ways of Reading a text file in Java
• Java Program to Write into a File
• Delete a File Using Java
• File Permissions in Java
• FileWriter Class in Java
• Java.io.FileDescriptor in Java
• Java.io.RandomAccessFile Class Method | Set 1
• Regular Expressions in Java
• Regex Tutorial - How to write Regular Expressions?
• Matcher pattern() method in Java with Examples
• Pattern pattern() method in Java with Examples
• Quantifiers in Java
• java.lang.Character class methods | Set 1
• Java IO : Input-output in Java with Examples
• Java.io.Reader class in Java
• Java.io.Writer Class in Java
• Java.io.FileInputStream Class in Java
• FileOutputStream in Java
• Java.io.BufferedOutputStream class in Java
• Java Networking
• TCP/IP Model
• User Datagram Protocol (UDP)
• Differences between IPv4 and IPv6
• Difference between Connection-oriented and Connection-less Services
• Socket Programming in Java
• java.net.ServerSocket Class in Java
• URL Class in Java with Examples

JDBC - Java Database Connectivity

• Introduction to JDBC (Java Database Connectivity)
• JDBC Drivers
• Establishing JDBC Connection in Java
• Types of Statements in JDBC
• JDBC Tutorial
• Java 8 Features - Complete Tutorial

Operators constitute the basic building block of any programming language. Java too provides many types of operators which can be used according to the need to perform various calculations and functions, be it logical, arithmetic, relational, etc. They are classified based on the functionality they provide.

Types of Operators: 

• Arithmetic Operators
• Unary Operators
• Assignment Operator
• Relational Operators
• Logical Operators
• Ternary Operator
• Bitwise Operators
• Shift Operators

This article explains all that one needs to know regarding Assignment Operators. 

Assignment Operators

These operators are used to assign values to a variable. The left side operand of the assignment operator is a variable, and the right side operand of the assignment operator is a value. The value on the right side must be of the same data type of the operand on the left side. Otherwise, the compiler will raise an error. This means that the assignment operators have right to left associativity, i.e., the value given on the right-hand side of the operator is assigned to the variable on the left. Therefore, the right-hand side value must be declared before using it or should be a constant. The general format of the assignment operator is, 

Types of Assignment Operators in Java

The Assignment Operator is generally of two types. They are:

1. Simple Assignment Operator: The Simple Assignment Operator is used with the “=” sign where the left side consists of the operand and the right side consists of a value. The value of the right side must be of the same data type that has been defined on the left side.

2. Compound Assignment Operator: The Compound Operator is used where +,-,*, and / is used along with the = operator.

Let’s look at each of the assignment operators and how they operate: 

1. (=) operator: 

This is the most straightforward assignment operator, which is used to assign the value on the right to the variable on the left. This is the basic definition of an assignment operator and how it functions. 

Syntax:  

Example:  

2. (+=) operator: 

This operator is a compound of ‘+’ and ‘=’ operators. It operates by adding the current value of the variable on the left to the value on the right and then assigning the result to the operand on the left. 

Note: The compound assignment operator in Java performs implicit type casting. Let’s consider a scenario where x is an int variable with a value of 5. int x = 5; If you want to add the double value 4.5 to the integer variable x and print its value, there are two methods to achieve this: Method 1: x = x + 4.5 Method 2: x += 4.5 As per the previous example, you might think both of them are equal. But in reality, Method 1 will throw a runtime error stating the “i ncompatible types: possible lossy conversion from double to int “, Method 2 will run without any error and prints 9 as output.

Reason for the Above Calculation

Method 1 will result in a runtime error stating “incompatible types: possible lossy conversion from double to int.” The reason is that the addition of an int and a double results in a double value. Assigning this double value back to the int variable x requires an explicit type casting because it may result in a loss of precision. Without the explicit cast, the compiler throws an error. Method 2 will run without any error and print the value 9 as output. The compound assignment operator += performs an implicit type conversion, also known as an automatic narrowing primitive conversion from double to int . It is equivalent to x = (int) (x + 4.5) , where the result of the addition is explicitly cast to an int . The fractional part of the double value is truncated, and the resulting int value is assigned back to x . It is advisable to use Method 2 ( x += 4.5 ) to avoid runtime errors and to obtain the desired output.

Same automatic narrowing primitive conversion is applicable for other compound assignment operators as well, including -= , *= , /= , and %= .

3. (-=) operator: 

This operator is a compound of ‘-‘ and ‘=’ operators. It operates by subtracting the variable’s value on the right from the current value of the variable on the left and then assigning the result to the operand on the left. 

4. (*=) operator:

 This operator is a compound of ‘*’ and ‘=’ operators. It operates by multiplying the current value of the variable on the left to the value on the right and then assigning the result to the operand on the left. 

5. (/=) operator: 

This operator is a compound of ‘/’ and ‘=’ operators. It operates by dividing the current value of the variable on the left by the value on the right and then assigning the quotient to the operand on the left. 

6. (%=) operator: 

This operator is a compound of ‘%’ and ‘=’ operators. It operates by dividing the current value of the variable on the left by the value on the right and then assigning the remainder to the operand on the left. 

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