assignment statement l'value

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Lvalue and Rvalue

Kenneth Leroy Busbee

Some programming languages use the idea of l-values and r-values , deriving from the typical mode of evaluation on the left and right hand side of an assignment statement. An lvalue refers to an object that persists beyond a single expression. An rvalue is a temporary value that does not persist beyond the expression that uses it. [1]

Lvalue and Rvalue refer to the left and right side of the assignment operator. The  Lvalue  (pronounced: L value) concept refers to the requirement that the operand on the left side of the assignment operator is modifiable, usually a variable.  Rvalue  concept pulls or fetches the value of the expression or operand on the right side of the assignment operator. Some examples:

The value 39 is pulled or fetched (Rvalue) and stored into the variable named age (Lvalue); destroying the value previously stored in that variable.

If the expression has a variable or named constant on the right side of the assignment operator, it would pull or fetch the value stored in the variable or constant. The value 18 is pulled or fetched from the variable named voting_age and stored into the variable named age.

If the expression is a test expression or Boolean expression, the concept is still an Rvalue one. The value in the identifier named age is pulled or fetched and used in the relational comparison of less than.

This is illegal because the identifier JACK_BENNYS_AGE does not have Lvalue properties. It is not a modifiable data object, because it is a constant.

Some uses of the Lvalue and Rvalue can be confusing in languages that support increment and decrement operators. Consider:

Postfix increment says to use my existing value then when you are done with the other operators; increment me. Thus, the first use of the oldest variable is an Rvalue context where the existing value of 55 is pulled or fetched and then assigned to the variable age; an Lvalue context. The second use of the oldest variable is an Lvalue context wherein the value of the oldest is incremented from 55 to 56.

• cnx.org: Programming Fundamentals – A Modular Structured Approach using C++
• Wikipedia: Value (computer science) ↵

Programming Fundamentals Copyright © 2018 by Kenneth Leroy Busbee is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License , except where otherwise noted.

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verilog, how to have a correct assignment

Please help me with this error here? I just want to write a simple fsm. Thank you!!!

all assignment of the output such as

they appears to have the same error as following:

• \$\begingroup\$ which program is producing that error? also, I think the error message describes precisely what the problem is. \$\endgroup\$ –  Marcus Müller Commented Apr 30, 2020 at 10:19
• \$\begingroup\$ Cadance Simulation on Linux \$\endgroup\$ –  David Commented May 1, 2020 at 12:23

I presume you doing that in an always block. However you've declared y_a / y_b / y_c as a wire (I can tell as the error says it's a net type).

To assign to a wire, you'll need to use the assign statement (continuous assignment).

To use the signals in an always block, you'll need to change them to a reg type.

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Verilog assign statement

Hardware schematic.

Signals of type wire or a similar wire like data type requires the continuous assignment of a value. For example, consider an electrical wire used to connect pieces on a breadboard. As long as the +5V battery is applied to one end of the wire, the component connected to the other end of the wire will get the required voltage.

In Verilog, this concept is realized by the assign statement where any wire or other similar wire like data-types can be driven continuously with a value. The value can either be a constant or an expression comprising of a group of signals.

Assign Syntax

The assignment syntax starts with the keyword assign followed by the signal name which can be either a single signal or a concatenation of different signal nets. The drive strength and delay are optional and are mostly used for dataflow modeling than synthesizing into real hardware. The expression or signal on the right hand side is evaluated and assigned to the net or expression of nets on the left hand side.

Delay values are useful for specifying delays for gates and are used to model timing behavior in real hardware because the value dictates when the net should be assigned with the evaluated value.

• LHS should always be a scalar or vector net or a concatenation of scalar or vector nets and never a scalar or vector register.
• RHS can contain scalar or vector registers and function calls.
• Whenever any operand on the RHS changes in value, LHS will be updated with the new value.
• assign statements are also called continuous assignments and are always active

In the following example, a net called out is driven continuously by an expression of signals. i1 and i2 with the logical AND & form the expression.

If the wires are instead converted into ports and synthesized, we will get an RTL schematic like the one shown below after synthesis.

Continuous assignment statement can be used to represent combinational gates in Verilog.

The module shown below takes two inputs and uses an assign statement to drive the output z using part-select and multiple bit concatenations. Treat each case as the only code in the module, else many assign statements on the same signal will definitely make the output become X.

Assign reg variables

It is illegal to drive or assign reg type variables with an assign statement. This is because a reg variable is capable of storing data and does not require to be driven continuously. reg signals can only be driven in procedural blocks like initial and always .

Implicit Continuous Assignment

When an assign statement is used to assign the given net with some value, it is called explicit assignment. Verilog also allows an assignment to be done when the net is declared and is called implicit assignment.

Combinational Logic Design

Consider the following digital circuit made from combinational gates and the corresponding Verilog code.

Combinational logic requires the inputs to be continuously driven to maintain the output unlike sequential elements like flip flops where the value is captured and stored at the edge of a clock. So an assign statement fits the purpose the well because the output o is updated whenever any of the inputs on the right hand side change.

After design elaboration and synthesis, we do get to see a combinational circuit that would behave the same way as modeled by the assign statement.

See that the signal o becomes 1 whenever the combinational expression on the RHS becomes true. Similarly o becomes 0 when RHS is false. Output o is X from 0ns to 10ns because inputs are X during the same time.

Click here for a slideshow with simulation example !

• Table of Contents
• Course Home
• Assignments
• Peer Instruction (Instructor)
• Peer Instruction (Student)
• Change Course
• Instructor's Page
• Progress Page
• Edit Profile
• Change Password
• Scratch ActiveCode
• Scratch Activecode
• Instructors Guide
• About Runestone
• Report A Problem
• 1.1 Getting Started
• 1.1.1 Preface
• 1.1.2 About the AP CSA Exam
• 1.1.3 Transitioning from AP CSP to AP CSA
• 1.1.4 Java Development Environments
• 1.1.5 Growth Mindset and Pair Programming
• 1.1.6 Pretest for the AP CSA Exam
• 1.1.7 Survey
• 1.2 Why Programming? Why Java?
• 1.3 Variables and Data Types
• 1.4 Expressions and Assignment Statements
• 1.5 Compound Assignment Operators
• 1.6 Casting and Ranges of Values
• 1.7 Unit 1 Summary
• 1.8 Mixed Up Code Practice
• 1.9 Toggle Mixed Up or Write Code Practice
• 1.10 Coding Practice
• 1.11 Multiple Choice Exercises
• 1.12 Method Signatures (preview 2026 curriculum)
• 1.13 Calling Class Methods (preview 2026 curriculum)
• 1.3. Variables and Data Types" data-toggle="tooltip">
• 1.5. Compound Assignment Operators' data-toggle="tooltip" >

Time estimate: 90 min.

1.4. Expressions and Assignment Statements ¶

In this lesson, you will learn about assignment statements and expressions that contain math operators and variables.

1.4.1. Assignment Statements ¶

Assignment statements initialize or change the value stored in a variable using the assignment operator = . An assignment statement always has a single variable on the left hand side. The value of the expression (which can contain math operators and other variables) on the right of the = sign is stored in the variable on the left.

Figure 1: Assignment Statement (variable = expression;) ¶

Instead of saying equals for the = in an assignment statement, say “gets” or “is assigned” to remember that the variable gets or is assigned the value on the right. In the figure above score is assigned the value of the expression 10 times points (which is another variable) plus 5.

The following video by Dr. Colleen Lewis shows how variables can change values in memory using assignment statements.

As we saw in the video, we can set one variable’s value to a copy of the value of another variable like y = x; . This won’t change the value of the variable that you are copying from.

Let’s step through the following code in the Java visualizer to see the values in memory. Click on the Next button at the bottom of the code to see how the values of the variables change. You can run the visualizer on any Active Code in this e-book by just clicking on the Code Lens button at the top of each Active Code.

Activity: CodeLens 1.4.1.2 (asgnviz1)

1-4-3: What are the values of x, y, and z after the following code executes? You can step through this code by clicking on this Java visualizer link.

• x = 0, y = 1, z = 2
• These are the initial values in the variable, but the values are changed.
• x = 1, y = 2, z = 3
• x changes to y's initial value, y's value is doubled, and z is set to 3
• x = 2, y = 2, z = 3
• Remember that the equal sign doesn't mean that the two sides are equal. It sets the value for the variable on the left to the value from evaluating the right side.
• x = 0, y = 0, z = 3

The following has the correct code to ‘swap’ the values in x and y (so that x ends up with y’s initial value and y ends up with x’s initial value), but the code is mixed up and contains one extra block which is not needed in a correct solution. Drag the needed blocks from the left into the correct order on the right. Check your solution by clicking on the Check button. You will be told if any of the blocks are in the wrong order or if you need to remove one or more blocks. After three incorrect attempts you will be able to use the Help Me button to make the problem easier.

1.4.2. Adding 1 to a Variable ¶

If you use a variable to keep score, you would probably increment it (add one to the current value) whenever score should go up. You can do this by setting the variable to the current value of the variable plus one ( score = score + 1 ) as shown below. The formula would look strange in math class, but it makes sense in coding because it is assigning a new value to the variable on the left that comes from evaluating the arithmetic expression on the right. So, the score variable is set to the previous value of score plus 1.

Try the code below to see how score is incremented by 1. Try substituting 2 instead of 1 to see what happens.

1.4.3. Input with Variables ¶

Variables are a powerful abstraction in programming because the same algorithm can be used with different input values saved in variables. The code below using the Scanner class will say hello to anyone who types in their name and will have different results for different name values. First, type in your name below the code and then click on run. Try again with a friend’s name. The code works for any name: behold, the power of variables!

The code below will say hello to anyone who types in their name. Type in your name below the code and then click on run. Try again with a friend’s name.

Although you will not be tested in the AP CSA exam on using the Java input or the Scanner or Console classes, learning how to do input in Java is very useful and fun. For more information on using the Scanner class, go to https://www.w3schools.com/java/java_user_input.asp , and for the newer Console class, https://howtodoinjava.com/java-examples/console-input-output/ . We are limited with the one way communication with the Java server in this Runestone ebook, but in most IDEs like replit, the input/output would be more interactive. Here are some examples in replit for Java Scanner Input Repl using the Scanner class and Java Console Input Repl using the Console class that you can try out.

1.4.4. Operators ¶

Java uses the standard mathematical operators for addition ( + ), subtraction ( - ), and division ( / ). The multiplication operator is written as * , as it is in most programming languages, since the character sets used until relatively recently didn’t have a character for a real multiplication sign, × , and keyboards still don’t have a key for it. Likewise no ÷ .

You may be used to using ^ for exponentiation, either from a graphing calculator or tools like Desmos. Confusingly ^ is an operator in Java, but it has a completely different meaning than exponentiation and isn’t even exactly an arithmetic operator. You will learn how to use the Math.pow method to do exponents in Unit 2.

Arithmetic expressions can be of type int or double . An arithmetic expression consisting only of int values will evaluate to an int value. An arithmetic expression that uses at least one double value will evaluate to a double value. (You may have noticed that + was also used to combine String and other values into new String s. More on this when we talk about String s more fully in Unit 2.)

Java uses the operator == to test if the value on the left is equal to the value on the right and != to test if two items are not equal. Don’t get one equal sign = confused with two equal signs == . They mean very different things in Java. One equal sign is used to assign a value to a variable. Two equal signs are used to test a variable to see if it is a certain value and that returns true or false as you’ll see below. Also note that using == and != with double values can produce surprising results. Because double values are only an approximation of the real numbers even things that should be mathematically equivalent might not be represented by the exactly same double value and thus will not be == . To see this for yourself, write a line of code below to print the value of the expression 0.3 == 0.1 + 0.2 ; it will be false !

Run the code below to see all the operators in action. Do all of those operators do what you expected? What about 2 / 3? Isn’t it surprising that it prints 0? See the note below about truncating division with integers. Change the code to make it print the decimal part of the division too. You can do this by making at least one of the numbers a double like 2.0.

When Java sees you doing integer division (or any operation with integers) it assumes you want an integer result so it throws away anything after the decimal point in the answer. This is called truncating division . If you need a double answer, you should make at least one of the values in the expression a double like 2.0.

With division, another thing to watch out for is dividing by 0. An attempt to divide an integer by zero will result in an ArithmeticException error message. Try it in one of the active code windows above.

Operators can be used to create compound expressions with more than one operator. You can either use a literal value which is a fixed value like 2, or variables in them. When compound expressions are evaluated, operator precedence rules are used, just like when we do math (remember PEMDAS?), so that * , / , and % are done before + and - . However, anything in parentheses is done first. It doesn’t hurt to put in extra parentheses if you are unsure as to what will be done first or just to make it more clear.

In the example below, try to guess what it will print out and then run it to see if you are right. Remember to consider operator precedence . How do the parentheses change the precedence?

1.4.5. The Remainder Operator ¶

The operator % in Java is the remainder operator. Like the other arithmetic operators is takes two operands. Mathematically it returns the remainder after dividing the first number by the second, using truncating integer division. For instance, 5 % 2 evaluates to 1 since 2 goes into 5 two times with a remainder of 1.

While you may not have heard of remainder as an operator, think back to elementary school math. Remember when you first learned long division, before they taught you about decimals, how when you did a long division that didn’t divide evenly, you gave the answer as the number of even divisions and the remainder. That remainder is what is returned by this operator. In the figures below, the remainders are the same values that would be returned by 2 % 3 and 5 % 2 .

Figure 1: Long division showing the integer result and the remainder ¶

Sometimes people—including Professor Lewis in the next video—will call % the modulo , or mod , operator. That is not actually correct though the difference between remainder and modulo, which uses Euclidean division instead of truncating integer division, only matters when negative operands are involved and the signs of the operands differ. With positive operands, remainder and mod give the same results. Java does have a method Math.floorMod in the Math class if you need to use modulo instead of remainder, but % is all you need in the AP exam.

Here’s the video .

In the example below, try to guess what it will print out and then run it to see if you are right.

The result of x % y when x is smaller than y is always x. The value y can’t go into x at all (goes in 0 times), since x is smaller than y, so the result is just x. So if you see 2 % 3 the result is 2.

1-4-11: What is the result of 158 % 10?

• This would be the result of 158 divided by 10. % gives you the remainder.
• % gives you the remainder after the division.
• When you divide 158 by 10 you get a remainder of 8.

1-4-12: What is the result of 3 % 8?

• 8 goes into 3 no times so the remainder is 3. The remainder of a smaller number divided by a larger number is always the smaller number!
• This would be the remainder if the question was 8 % 3 but here we are asking for the reminder after we divide 3 by 8.
• What is the remainder after you divide 3 by 8?

1.4.6. Programming Challenge : Dog Years ¶

In this programming challenge, you will calculate your age, and your pet’s age from your birthdates, and your pet’s age in dog years. In the code below, type in the current year, the year you were born, the year your dog or cat was born (if you don’t have one, make one up!) in the variables below. Then write formulas in assignment statements to calculate how old you are, how old your dog or cat is, and how old they are in dog years which is 7 times a human year. Finally, print it all out. If you are pair programming, switch drivers (who has control of the keyboard in pair programming) after every line of code.

Calculate your age and your pet’s age from the birthdates, and then your pet’s age in dog years.

Your teacher may suggest that you use a Java IDE like replit.com for this challenge so that you can use input to get these values using the Scanner class . Here is a repl template that you can use to get started if you want to try the challenge with input.

1.4.7. Summary ¶

Arithmetic expressions include expressions of type int and double .

The arithmetic operators consist of + , - , * , / , and % also known as addition, subtraction, multiplication, division, and remainder.

An arithmetic operation that uses two int values will evaluate to an int value. With integer division, any decimal part in the result will be thrown away.

An arithmetic operation that uses at least one double value will evaluate to a double value.

Operators can be used to construct compound expressions.

During evaluation, operands are associated with operators according to operator precedence to determine how they are grouped. ( * , / , % have precedence over + and - , unless parentheses are used to group those.)

An attempt to divide an integer by zero will result in an ArithmeticException .

The assignment operator ( = ) allows a program to initialize or change the value stored in a variable. The value of the expression on the right is stored in the variable on the left.

During execution, expressions are evaluated to produce a single value.

The value of an expression has a type based on the types of the values and operators used in the expression.

1.4.8. AP Practice ¶

The following is a 2019 AP CSA sample question.

1-4-14: Consider the following code segment.

What is printed when the code segment is executed?

• 0.666666666666667
• Don't forget that division and multiplication will be done first due to operator precedence.
• Yes, this is equivalent to (5 + ((a/b)*c) - 1).
• Don't forget that division and multiplication will be done first due to operator precedence, and that an int/int gives an int truncated result where everything to the right of the decimal point is dropped.

• Intro to Programming Logic  > 

Assignment Statements

Assignment statements in a program come in two forms – with and without mutations. Assignments without mutation are those that give a value to a variable without using the old value of that variable. Assignments with mutation are variable assignments that use the old value of a variable to calculate a value for the variable.

For example, an increment statement like x = x + 1 MUTATES the value of x by updating its value to be one bigger than it was before. In order to make sense of such a statement, we need to know the previous value of x .

In contrast, a statement like y = x + 1 assigns to y one more than the value in x . We do not need to know the previous value of y , as we are not using it in the assignment statement. (We do need to know the value of x ).

Assignments without mutation

We have already seen the steps necessary to process assignment statements that do not involve variable mutation. Recall that we can declare as a Premise any assignment statement or claim from a previous proof block involving variables that have not since changed.

For example, suppose we want to verify the following program so the assert statement at the end will hold (this example again eliminates the Logika mode notation and the necessary import statements, which we will continue to do in subsequent examples):

Since none of the statements involve variable mutation, we can do the verification in a single proof block:

Note that we did need to do AndI so that the last claim was y == z ∧ y == 6 , even though we had previously established the claims y == z and y == 6 . In order for an assert to hold (at least until we switch Logika modes in chapter 10), we need to have established EXACTLY the claim in the assert in a previous proof block.

Assignments with mutation

Assignments with mutation are trickier – we need to know the old value of a variable in order to reason about its new value. For example, if we have the following program:

Then we might try to add the following proof blocks:

…but then we get stuck in the second proof block. There, x is supposed to refer to the CURRENT value of x (after being incremented), but both our attempted claims are untrue. The current value of x is not one more than itself (this makes no sense!), and we can tell from reading the code that x is now 5, not 4.

To help reason about changing variables, Logika has a special Old(varName) function that refers to the OLD value of a variable called varName , just before the latest update. In the example above, we can use Old(x) in the second proof block to refer to x ’s value just before it was incremented. We can now change our premises and finish the verification as follows:

By the end of the proof block following a variable mutation, we need to express everything we know about the variable’s current value WITHOUT using the Old terminology, as its scope will end when the proof block ends. Moreover, we only ever have one Old value available in a proof block – the variable that was most recently changed. This means we will need proof blocks after each variable mutation to process the changes to any related facts.

Variable swap example

Suppose we have the following program:

We can see that this program gets two user input values, x and y , and then swaps their values. So if x was originally 4 and y was originally 6, then at the end of the program x would be 6 and y would be 4.

We would like to be able to assert what we did – that x now has the original value from y , and that y now has the original value from x . To do this, we might invent dummy constants called xOrig and yOrig that represent the original values of those variables. Then we can add our assert:

We can complete the verification by adding proof blocks after assignment statements, being careful to update all we know (without using the Old value) by the end of each block:

Notice that in each proof block, we express as much as we can about all variables/values in the program. In the first proof block, even though xOrig and yOrig were not used in the previous assignment statement, we still expressed how the current values our other variables compared to xOrig and yOrig . It helps to think about what you are trying to claim in the final assert – since our assert involved xOrig and yOrig , we needed to relate the current values of our variables to those values as we progressed through the program.

Last modified by: Julie Thornton Apr 25, 2024

Solve error: lvalue required as left operand of assignment

In this tutorial you will know about one of the most occurred error in C and C++ programming, i.e.  lvalue required as left operand of assignment.

lvalue means left side value. Particularly it is left side value of an assignment operator.

rvalue means right side value. Particularly it is right side value or expression of an assignment operator.

In above example  a  is lvalue and b + 5  is rvalue.

In C language lvalue appears mainly at four cases as mentioned below:

• Left of assignment operator.
• Left of member access (dot) operator (for structure and unions).
• Right of address-of operator (except for register and bit field lvalue).
• As operand to pre/post increment or decrement for integer lvalues including Boolean and enums.

Now let see some cases where this error occur with code.

When you will try to run above code, you will get following error.

Solution: In if condition change assignment operator to comparison operator, as shown below.

Above code will show the error: lvalue required as left operand of assignment operator.

Here problem occurred due to wrong handling of short hand operator (*=) in findFact() function.

Solution : Just by changing the line ans*i=ans to ans*=i we can avoid that error. Here short hand operator expands like this,  ans=ans*i. Here left side some variable is there to store result. But in our program ans*i is at left hand side. It’s an expression which produces some result. While using assignment operator we can’t use an expression as lvalue.

The correct code is shown below.

Above code will show the same lvalue required error.

Reason and Solution: Ternary operator produces some result, it never assign values inside operation. It is same as a function which has return type. So there should be something to be assigned but unlike inside operator.

The correct code is given below.

Some Precautions To Avoid This Error

There are no particular precautions for this. Just look into your code where problem occurred, like some above cases and modify the code according to that.

Mostly 90% of this error occurs when we do mistake in comparison and assignment operations. When using pointers also we should careful about this error. And there are some rare reasons like short hand operators and ternary operators like above mentioned. We can easily rectify this error by finding the line number in compiler, where it shows error: lvalue required as left operand of assignment.

Programming Assignment Help on Assigncode.com, that provides homework ecxellence in every technical assignment.

Comment below if you have any queries related to above tutorial.

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Basic structure of c program, introduction to c programming language, variables, constants and keywords in c, first c program – print hello world message, 6 thoughts on “solve error: lvalue required as left operand of assignment”.

hi sir , i am andalib can you plz send compiler of c++.

i want the solution by char data type for this error

#include #include #include using namespace std; #define pi 3.14 int main() { float a; float r=4.5,h=1.5; {

a=2*pi*r*h=1.5 + 2*pi*pow(r,2); } cout<<" area="<<a<<endl; return 0; } what's the problem over here

#include using namespace std; #define pi 3.14 int main() { float a,p; float r=4.5,h=1.5; p=2*pi*r*h; a=1.5 + 2*pi*pow(r,2);

cout<<" area="<<a<<endl; cout<<" perimeter="<<p<<endl; return 0; }

You can't assign two values at a single place. Instead solve them differetly

Hi. I am trying to get a double as a string as efficiently as possible. I get that error for the final line on this code. double x = 145.6; int size = sizeof(x); char str[size]; &str = &x; Is there a possible way of getting the string pointing at the same part of the RAM as the double?

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Assignment Operator in C Language | L- value Error in C programming

Introduction:.

In our previous articles, we discussed the Arithmetic Operators in C Language . In today’s article, We will learn about the Assignment Operator in C Programming. We also discuss the L-Value error in programming.

Assignment Operator in C Langauge ( = ) :

We represent the Assignment operator using the equal to( = ) symbol in C programming.

Syntax Of Assignment Operator:

The assignment operator is a binary operator so it must need two operands.

The Assignment Operation will copy the R-Value to L-Value .

The left-side value ( L-Value ) of the Assignment Operator must be a Variable .

Example Program to understand Assignment Operator:

Here x is the left-side value or L-Value and The value 10 is the right-side value or R-Value, By using the assignment operator we assigned the value 10 to the variable x .

Program Output:

Example 2 : L – Value Error in C programming :

📢 The Left-side or L-Value of the Assignment Operator must be a Variable.

Program Output

Complex Assignment Operations in C Programming:

For example,

Simple Assginment Operation:

Compound addition ( add then assigne += ):.

Here first x + 20 is calculated then x is updated with the resulting value of x + 20 operation.

Conclusion:

In this article, We discussed the Assignment Operators and Compound Assignment operators in the C language. We also looked at a few example code snippets and L-Value Error.

Related Tutorials:

Related programs:, share this:, you may also like..., array programs in c language, sum of n natural numbers in c program, pointers programs in c programming language, 2 responses, leave a reply cancel reply.

• Assignment Statement

An Assignment statement is a statement that is used to set a value to the variable name in a program .

Assignment statement allows a variable to hold different types of values during its program lifespan. Another way of understanding an assignment statement is, it stores a value in the memory location which is denoted by a variable name.

The symbol used in an assignment statement is called as an operator . The symbol is ‘=’ .

Note: The Assignment Operator should never be used for Equality purpose which is double equal sign ‘==’.

The Basic Syntax of Assignment Statement in a programming language is :

variable = expression ;

variable = variable name

expression = it could be either a direct value or a math expression/formula or a function call

Few programming languages such as Java, C, C++ require data type to be specified for the variable, so that it is easy to allocate memory space and store those values during program execution.

data_type variable_name = value ;

In the above-given examples, Variable ‘a’ is assigned a value in the same statement as per its defined data type. A data type is only declared for Variable ‘b’. In the 3 rd line of code, Variable ‘a’ is reassigned the value 25. The 4 th line of code assigns the value for Variable ‘b’.

Assignment Statement Forms

This is one of the most common forms of Assignment Statements. Here the Variable name is defined, initialized, and assigned a value in the same statement. This form is generally used when we want to use the Variable quite a few times and we do not want to change its value very frequently.

Tuple Assignment

Generally, we use this form when we want to define and assign values for more than 1 variable at the same time. This saves time and is an easy method. Note that here every individual variable has a different value assigned to it.

(Code In Python)

Sequence Assignment

(Code in Python)

Multiple-target Assignment or Chain Assignment

In this format, a single value is assigned to two or more variables.

Augmented Assignment

In this format, we use the combination of mathematical expressions and values for the Variable. Other augmented Assignment forms are: &=, -=, **=, etc.

Browse more Topics Under Data Types, Variables and Constants

• Concept of Data types
• Built-in Data Types
• Constants in Programing Language 
• Access Modifier
• Variables of Built-in-Datatypes
• Declaration/Initialization of Variables
• Type Modifier

Few Rules for Assignment Statement

Few Rules to be followed while writing the Assignment Statements are:

• Variable names must begin with a letter, underscore, non-number character. Each language has its own conventions.
• The Data type defined and the variable value must match.
• A variable name once defined can only be used once in the program. You cannot define it again to store other types of value.
• If you assign a new value to an existing variable, it will overwrite the previous value and assign the new value.

FAQs on Assignment Statement

Q1. Which of the following shows the syntax of an  assignment statement ?

• variablename = expression ;
• expression = variable ;
• datatype = variablename ;
• expression = datatype variable ;

Answer – Option A.

Q2. What is an expression ?

• Same as statement
• List of statements that make up a program
• Combination of literals, operators, variables, math formulas used to calculate a value
• Numbers expressed in digits

Answer – Option C.

Q3. What are the two steps that take place when an  assignment statement  is executed?

• Evaluate the expression, store the value in the variable
• Reserve memory, fill it with value
• Evaluate variable, store the result
• Store the value in the variable, evaluate the expression.

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Data Types, Variables and Constants

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Different Forms of Assignment Statements in Python

We use Python assignment statements to assign objects to names. The target of an assignment statement is written on the left side of the equal sign (=), and the object on the right can be an arbitrary expression that computes an object.

There are some important properties of assignment in Python :-

• Assignment creates object references instead of copying the objects.
• Python creates a variable name the first time when they are assigned a value.
• Names must be assigned before being referenced.
• There are some operations that perform assignments implicitly.

Assignment statement forms :-

1. Basic form:

This form is the most common form.

2. Tuple assignment:

When we code a tuple on the left side of the =, Python pairs objects on the right side with targets on the left by position and assigns them from left to right. Therefore, the values of x and y are 50 and 100 respectively.

3. List assignment:

This works in the same way as the tuple assignment.

4. Sequence assignment:

In recent version of Python, tuple and list assignment have been generalized into instances of what we now call sequence assignment – any sequence of names can be assigned to any sequence of values, and Python assigns the items one at a time by position.

5. Extended Sequence unpacking:

It allows us to be more flexible in how we select portions of a sequence to assign.

Here, p is matched with the first character in the string on the right and q with the rest. The starred name (*q) is assigned a list, which collects all items in the sequence not assigned to other names.

This is especially handy for a common coding pattern such as splitting a sequence and accessing its front and rest part.

6. Multiple- target assignment:

In this form, Python assigns a reference to the same object (the object which is rightmost) to all the target on the left.

7. Augmented assignment :

The augmented assignment is a shorthand assignment that combines an expression and an assignment.

There are several other augmented assignment forms:

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Error in task: syntax error in assignment statement l-value

I have written code for a synchronous FIFO that contains tasks. I am getting an error in task t_write that says

syntax error in assignment statement l-value

However, I do not understand the cause of the error. Any help would be appreciated.

Source code:

Verilog is likely interoperating FIFO(WRITE_PTR) as a function (which cannot be a l-value) because you used parentheses. For array indexing you must use square brackets [ ] .

You have the same problem in your read task. Either your compiler errors out before detecting it or it is lower in the log file.

FYI, your codes may simulate but it definitely will not synthesize. Your pointer, counters, and FIFO are being assigned in multiple always blocks which is not legal for synthesis. You need to combined them into one always block. Use else-if or case statements. Example:

Also you should assign synchronous logic with non-blocking ( <= ). This will prevent race conditions in the simulator’s scheduler with other always blocks that sample the output values. It will not impacted synthesis.

I will also recommended you change @(COUNT) to @* . It will not change your logic. @* is auto sensitivity and is the preferred way of doing combination logic since it was added in 2001. You should only hard code the sensitivity list of combination logic if required to use Verilog-1995 or special case analog behavioral models.

• thank you so much. that solved my problem. As for you suggestion on using multiple always blocks, I will try merging it into one as you suggested and give a try –  badri18 Commented Jan 21, 2018 at 10:59

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