what is lvalue required as left operand of assignment

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.

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?

How To Fix “error: lvalue required as left operand of assignment”

The message “error: lvalue required as left operand of assignment” can be shown quite frequently when you write your C/C++ programs. Check out the explanation below to understand why it happens.

Table of Contents

l-values And r-values

In C and C++, we can put expressions into many categories , including l-values and r-values

The history of these concepts can be traced back to Combined Programming Language. Their names are derived from the sides where they are typically located on an assignment statement.

Recent standards like C++17 actually define several categories like xvalue or prvalue. But the definitions of l-values and r-values are basically the same in all C and C++ standards.

In simple terms, l-values are memory addresses that C/C++ programs can access programmatically. Common examples include constants, variable names, class members, unions, bit-fields, and array elements.

In an assignment statement, the operand on the left-hand side should be a modifiable l-value because the operator will evaluate the right operand and assign its result to the left operand.

This example illustrates the common correct usage of l-values and r-values:

In the ‘x = 4’ statement, x is an l-value while the literal 4 is not. The increment operator also requires an l-value because it needs to read the operand value and modify it accordingly.

Similarly, dereferenced pointers like *p are also l-values. Notice that an l-value (like x) can be on the right side of the assignment statement as well.

Causes And Solutions For “error: lvalue required as left operand of assignment”

C/C++ compilers generates this error when you don’t provide a valid l-value to the left-hand side operand of an assignment statement. There are many cases you can make this mistake.

This code can’t be compiled successfully:

As we have mentioned, the number literal 4 isn’t an l-value, which is required for the left operand. You will need to write the assignment statement the other way around:

In the same manner, this program won’t compile either:

In C/C++, the ‘x + 1’ expression doesn’t evaluate to a l-value. You can fix it by switching the sides of the operands:

This is another scenario the compiler will complain about the left operand:

(-x) doesn’t evaluate to a l-value in C/C++, while ‘x’ does. You will need to change both operands to make the statement correct:

Many people also use an assignment operator when they need a comparison operator instead:

This leads to a compilation error:

The if statement above needs to check the output of a comparison statement:

if (strcmp (str1,str2) == 0)

C/C++ compilers will give you the message “ error: lvalue required as left operand of assignment ” when there is an assignment statement in which the left operand isn’t a modifiable l-value. This is usually the result of syntax misuse. Correct it, and the error should disappear.

Maybe you are interested :

Expression preceding parentheses of apparent call must have (pointer-to-) function type
ERROR: conditional jump or move depends on the uninitialized value(s)
Print a vector in C++

My name is Robert. I have a degree in information technology and two years of expertise in software development. I’ve come to offer my understanding on programming languages. I hope you find my articles interesting.

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Understanding lvalues and rvalues in C and C++

The terms lvalue and rvalue are not something one runs into often in C/C++ programming, but when one does, it's usually not immediately clear what they mean. The most common place to run into these terms are in compiler error & warning messages. For example, compiling the following with gcc :

True, this code is somewhat perverse and not something you'd write, but the error message mentions lvalue , which is not a term one usually finds in C/C++ tutorials. Another example is compiling this code with g++ :

Now the error is:

Here again, the error mentions some mysterious rvalue . So what do lvalue and rvalue mean in C and C++? This is what I intend to explore in this article.

A simple definition

This section presents an intentionally simplified definition of lvalues and rvalues . The rest of the article will elaborate on this definition.

An lvalue ( locator value ) represents an object that occupies some identifiable location in memory (i.e. has an address).

rvalues are defined by exclusion, by saying that every expression is either an lvalue or an rvalue . Therefore, from the above definition of lvalue , an rvalue is an expression that does not represent an object occupying some identifiable location in memory.

Basic examples

The terms as defined above may appear vague, which is why it's important to see some simple examples right away.

Let's assume we have an integer variable defined and assigned to:

An assignment expects an lvalue as its left operand, and var is an lvalue, because it is an object with an identifiable memory location. On the other hand, the following are invalid:

Neither the constant 4 , nor the expression var + 1 are lvalues (which makes them rvalues). They're not lvalues because both are temporary results of expressions, which don't have an identifiable memory location (i.e. they can just reside in some temporary register for the duration of the computation). Therefore, assigning to them makes no semantic sense - there's nowhere to assign to.

So it should now be clear what the error message in the first code snippet means. foo returns a temporary value which is an rvalue. Attempting to assign to it is an error, so when seeing foo() = 2; the compiler complains that it expected to see an lvalue on the left-hand-side of the assignment statement.

Not all assignments to results of function calls are invalid, however. For example, C++ references make this possible:

Here foo returns a reference, which is an lvalue , so it can be assigned to. Actually, the ability of C++ to return lvalues from functions is important for implementing some overloaded operators. One common example is overloading the brackets operator [] in classes that implement some kind of lookup access. std::map does this:

The assignment mymap[10] works because the non-const overload of std::map::operator[] returns a reference that can be assigned to.

Modifiable lvalues

Initially when lvalues were defined for C, it literally meant "values suitable for left-hand-side of assignment". Later, however, when ISO C added the const keyword, this definition had to be refined. After all:

So a further refinement had to be added. Not all lvalues can be assigned to. Those that can are called modifiable lvalues . Formally, the C99 standard defines modifiable lvalues as:

[...] an lvalue that does not have array type, does not have an incomplete type, does not have a const-qualified type, and if it is a structure or union, does not have any member (including, recursively, any member or element of all contained aggregates or unions) with a const-qualified type.

Conversions between lvalues and rvalues

Generally speaking, language constructs operating on object values require rvalues as arguments. For example, the binary addition operator '+' takes two rvalues as arguments and returns an rvalue:

As we've seen earlier, a and b are both lvalues. Therefore, in the third line, they undergo an implicit lvalue-to-rvalue conversion . All lvalues that aren't arrays, functions or of incomplete types can be converted thus to rvalues.

What about the other direction? Can rvalues be converted to lvalues? Of course not! This would violate the very nature of an lvalue according to its definition [1] .

This doesn't mean that lvalues can't be produced from rvalues by more explicit means. For example, the unary '*' (dereference) operator takes an rvalue argument but produces an lvalue as a result. Consider this valid code:

Conversely, the unary address-of operator '&' takes an lvalue argument and produces an rvalue:

The ampersand plays another role in C++ - it allows to define reference types. These are called "lvalue references". Non-const lvalue references cannot be assigned rvalues, since that would require an invalid rvalue-to-lvalue conversion:

Constant lvalue references can be assigned rvalues. Since they're constant, the value can't be modified through the reference and hence there's no problem of modifying an rvalue. This makes possible the very common C++ idiom of accepting values by constant references into functions, which avoids unnecessary copying and construction of temporary objects.

CV-qualified rvalues

If we read carefully the portion of the C++ standard discussing lvalue-to-rvalue conversions [2] , we notice it says:

An lvalue (3.10) of a non-function, non-array type T can be converted to an rvalue. [...] If T is a non-class type, the type of the rvalue is the cv-unqualified version of T. Otherwise, the type of the rvalue is T.

What is this "cv-unqualified" thing? CV-qualifier is a term used to describe const and volatile type qualifiers.

From section 3.9.3:

Each type which is a cv-unqualified complete or incomplete object type or is void (3.9) has three corresponding cv-qualified versions of its type: a const-qualified version, a volatile-qualified version, and a const-volatile-qualified version. [...] The cv-qualified or cv-unqualified versions of a type are distinct types; however, they shall have the same representation and alignment requirements (3.9)

But what has this got to do with rvalues? Well, in C, rvalues never have cv-qualified types. Only lvalues do. In C++, on the other hand, class rvalues can have cv-qualified types, but built-in types (like int ) can't. Consider this example:

The second call in main actually calls the foo () const method of A , because the type returned by cbar is const A , which is distinct from A . This is exactly what's meant by the last sentence in the quote mentioned earlier. Note also that the return value from cbar is an rvalue. So this is an example of a cv-qualified rvalue in action.

Rvalue references (C++11)

Rvalue references and the related concept of move semantics is one of the most powerful new features the C++11 standard introduces to the language. A full discussion of the feature is way beyond the scope of this humble article [3] , but I still want to provide a simple example, because I think it's a good place to demonstrate how an understanding of what lvalues and rvalues are aids our ability to reason about non-trivial language concepts.

I've just spent a good part of this article explaining that one of the main differences between lvalues and rvalues is that lvalues can be modified, and rvalues can't. Well, C++11 adds a crucial twist to this distinction, by allowing us to have references to rvalues and thus modify them, in some special circumstances.

As an example, consider a simplistic implementation of a dynamic "integer vector". I'm showing just the relevant methods here:

So, we have the usual constructor, destructor, copy constructor and copy assignment operator [4] defined, all using a logging function to let us know when they're actually called.

Let's run some simple code, which copies the contents of v1 into v2 :

What this prints is:

Makes sense - this faithfully represents what's going on inside operator= . But suppose that we want to assign some rvalue to v2 :

Although here I just assign a freshly constructed vector, it's just a demonstration of a more general case where some temporary rvalue is being built and then assigned to v2 (this can happen for some function returning a vector, for example). What gets printed now is this:

Ouch, this looks like a lot of work. In particular, it has one extra pair of constructor/destructor calls to create and then destroy the temporary object. And this is a shame, because inside the copy assignment operator, another temporary copy is being created and destroyed. That's extra work, for nothing.

Well, no more. C++11 gives us rvalue references with which we can implement "move semantics", and in particular a "move assignment operator" [5] . Let's add another operator= to Intvec :

The && syntax is the new rvalue reference . It does exactly what it sounds it does - gives us a reference to an rvalue, which is going to be destroyed after the call. We can use this fact to just "steal" the internals of the rvalue - it won't need them anyway! This prints:

What happens here is that our new move assignment operator is invoked since an rvalue gets assigned to v2 . The constructor and destructor calls are still needed for the temporary object that's created by Intvec(33) , but another temporary inside the assignment operator is no longer needed. The operator simply switches the rvalue's internal buffer with its own, arranging it so the rvalue's destructor will release our object's own buffer, which is no longer used. Neat.

I'll just mention once again that this example is only the tip of the iceberg on move semantics and rvalue references. As you can probably guess, it's a complex subject with a lot of special cases and gotchas to consider. My point here was to demonstrate a very interesting application of the difference between lvalues and rvalues in C++. The compiler obviously knows when some entity is an rvalue, and can arrange to invoke the correct constructor at compile time.

One can write a lot of C++ code without being concerned with the issue of rvalues vs. lvalues, dismissing them as weird compiler jargon in certain error messages. However, as this article aimed to show, getting a better grasp of this topic can aid in a deeper understanding of certain C++ code constructs, and make parts of the C++ spec and discussions between language experts more intelligible.

Also, in the new C++ spec this topic becomes even more important, because C++11's introduction of rvalue references and move semantics. To really grok this new feature of the language, a solid understanding of what rvalues and lvalues are becomes crucial.

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Understanding the meaning of lvalues and rvalues in C++

A lightweight introduction to a couple of basic C++ features that act as a foundation for bigger structures.

I have been struggling with the concepts of lvalue and rvalue in C++ since forever. I think that now is the right time to understand them for good, as they are getting more and more important with the evolution of the language.

Once the meaning of lvalues and rvalues is grasped, you can dive deeper into advanced C++ features like move semantics and rvalue references (more on that in future articles).

Lvalues and rvalues: a friendly definition

First of all, let's keep our heads away from any formal definition. In C++ an lvalue is something that points to a specific memory location. On the other hand, a rvalue is something that doesn't point anywhere. In general, rvalues are temporary and short lived, while lvalues live a longer life since they exist as variables. It's also fun to think of lvalues as containers and rvalues as things contained in the containers . Without a container, they would expire.

Let me show you some examples right away.

Here 666 is an rvalue; a number (technically a literal constant ) has no specific memory address, except for some temporary register while the program is running. That number is assigned to x , which is a variable. A variable has a specific memory location, so its an lvalue. C++ states that an assignment requires an lvalue as its left operand: this is perfectly legal.

Then with x , which is an lvalue, you can do stuff like that:

Here I'm grabbing the the memory address of x and putting it into y , through the address-of operator & . It takes an lvalue argument and produces an rvalue. This is another perfectly legal operation: on the left side of the assignment we have an lvalue (a variable), on the right side an rvalue produced by the address-of operator.

However, I can't do the following:

Yeah, that's obvious. But the technical reason is that 666 , being a literal constant — so an rvalue, doesn't have a specific memory location. I am assigning y to nowhere.

This is what GCC tells me if I run the program above:

He is damn right; the left operand of an assigment always require an lvalue, and in my program I'm using an rvalue ( 666 ).

I can't do that either:

He is right again. The & operator wants an lvalue in input, because only an lvalue has an address that & can process.

Functions returning lvalues and rvalues

We know that the left operand of an assigment must be an lvalue. Hence a function like the following one will surely throw the lvalue required as left operand of assignment error:

Crystal clear: setValue() returns an rvalue (the temporary number 6 ), which cannot be a left operand of assignment. Now, what happens if a function returns an lvalue instead? Look closely at the following snippet:

It works because here setGlobal returns a reference, unlike setValue() above. A reference is something that points to an existing memory location (the global variable) thus is an lvalue, so it can be assigned to. Watch out for & here: it's not the address-of operator, it defines the type of what's returned (a reference).

The ability to return lvalues from functions looks pretty obscure, yet it is useful when you are doing advanced stuff like implementing some overloaded operators. More on that in future chapters.

Lvalue to rvalue conversion

An lvalue may get converted to an rvalue: that's something perfectly legit and it happens quite often. Let's think of the addition + operator for example. According to the C++ specifications, it takes two rvalues as arguments and returns an rvalue.

Let's look at the following snippet:

Wait a minute: x and y are lvalues, but the addition operator wants rvalues: how come? The answer is quite simple: x and y have undergone an implicit lvalue-to-rvalue conversion . Many other operators perform such conversion — subtraction, addition and division to name a few.

Lvalue references

What about the opposite? Can an rvalue be converted to lvalue? Nope. It's not a technical limitation, though: it's the programming language that has been designed that way.

In C++, when you do stuff like

you are declarying yref as of type int& : a reference to y . It's called an lvalue reference . Now you can happily change the value of y through its reference yref .

We know that a reference must point to an existing object in a specific memory location, i.e. an lvalue. Here y indeed exists, so the code runs flawlessly.

Now, what if I shortcut the whole thing and try to assign 10 directly to my reference, without the object that holds it?

On the right side we have a temporary thing, an rvalue that needs to be stored somewhere in an lvalue.

On the left side we have the reference (an lvalue) that should point to an existing object. But being 10 a numeric constant, i.e. without a specific memory address, i.e. an rvalue, the expression clashes with the very spirit of the reference.

If you think about it, that's the forbidden conversion from rvalue to lvalue. A volatile numeric constant (rvalue) should become an lvalue in order to be referenced to. If that would be allowed, you could alter the value of the numeric constant through its reference. Pretty meaningless, isn't it? Most importantly, what would the reference point to once the numeric value is gone?

The following snippet will fail for the very same reason:

I'm passing a temporary rvalue ( 10 ) to a function that takes a reference as argument. Invalid rvalue to lvalue conversion. There's a workaround: create a temporary variable where to store the rvalue and then pass it to the function (as in the commented out code). Quite inconvenient when you just want to pass a number to a function, isn't it?

Const lvalue reference to the rescue

That's what GCC would say about the last two code snippets:

GCC complains about the reference not being const , namely a constant . According to the language specifications, you are allowed to bind a const lvalue to an rvalue . So the following snippet works like a charm:

And of course also the following one:

The idea behind is quite straightforward. The literal constant 10 is volatile and would expire in no time, so a reference to it is just meaningless. Let's make the reference itself a constant instead, so that the value it points to can't be modified. Now the problem of modifying an rvalue is solved for good. Again, that's not a technical limitation but a choice made by the C++ folks to avoid silly troubles.

This makes possible the very common C++ idiom of accepting values by constant references into functions, as I did in the previous snipped above, which avoids unnecessary copying and construction of temporary objects.

Under the hood the compiler creates an hidden variable for you (i.e. an lvalue) where to store the original literal constant, and then bounds that hidden variable to your reference. That's basically the same thing I did manually in a couple of snippets above. For example:

Now your reference points to something that exists for real (until it goes out of scope) and you can use it as usual, except for modifying the value it points to:

Understanding the meaning of lvalues and rvalues has given me the chance to figure out several of the C++'s inner workings. C++11 pushes the limits of rvalues even further, by introducing the concept of rvalue references and move semantics , where — surprise! — rvalues too are modifiable. I will restlessly dive into that minefield in one of my next articles.

Thomas Becker's Homepage - C++ Rvalue References Explained ( link ) Eli Bendersky's website - Understanding lvalues and rvalues in C and C++ ( link ) StackOverflow - Rvalue Reference is Treated as an Lvalue? ( link ) StackOverflow - Const reference and lvalue ( link ) CppReference.com - Reference declaration ( link )

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Else without IF and L-Value Required Error in C

Else without if.

This error is shown if we write anything in between if and else clause. Example:

L-value required

This error occurs when we put constants on left hand side of = operator and variables on right hand side of it. Example:

Example 2: At line number 12, it will show an error L-value because arr++ means arr=arr+1.Now that is what there is difference in normal variable and array. If we write a=a+1 (where a is normal variable), compiler will know its job and there will be no error but when you write arr=arr+1 (where arr is name of an array) then, compiler will think arr contain address and how we can change address. Therefore it will take arr as address and left side will be constant, hence it will show error as L-value required.

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Assignment operators

Simple assignment operator =, compound assignment operators.

The result of an assignment expression is not an lvalue.

All assignment operators have the same precedence and have right-to-left associativity.

The simple assignment operator has the following form:

lvalue = expr

The operator stores the value of the right operand expr in the object designated by the left operand lvalue .

The left operand must be a modifiable lvalue. The type of an assignment operation is the type of the left operand.

If the left operand is not a class type or a vector type , the right operand is implicitly converted to the type of the left operand. This converted type will not be qualified by const or volatile .

If the left operand is a class type, that type must be complete. The copy assignment operator of the left operand will be called.

If the left operand is an object of reference type, the compiler will assign the value of the right operand to the object denoted by the reference.

The compound assignment operators consist of a binary operator and the simple assignment operator. They perform the operation of the binary operator on both operands and store the result of that operation into the left operand, which must be a modifiable lvalue.

The following table shows the operand types of compound assignment expressions:

The following table lists the compound assignment operators and shows an expression using each operator:

Although the equivalent expression column shows the left operands (from the example column) twice, it is in effect evaluated only once.

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Troubleshooting 'error: lvalue required as left operand of assignment': Tips to Fix Assignment Errors in Your Code

Are you struggling with the "error: lvalue required as left operand of assignment" error in your code? Don't worry; this error is common among developers and can be fixed with a few simple tips. In this guide, we will walk you through the steps to troubleshoot and fix this error.

Understanding the Error

The "error: lvalue required as left operand of assignment" error occurs when you try to assign a value to a non-modifiable lvalue. An lvalue refers to an expression that can appear on the left-hand side of an assignment operator, whereas an rvalue can only appear on the right-hand side.

Tips to Fix Assignment Errors

Here are some tips to help you fix the "error: lvalue required as left operand of assignment" error:

1. Check for Typographical Errors

The error may occur due to typographical errors in your code. Make sure that you have spelled the variable name correctly and used the correct syntax for the assignment operator.

2. Check the Scope of Your Variables

The error may occur if you try to assign a value to a variable that is out of scope. Make sure that the variable is declared and initialized before you try to assign a value to it.

3. Check the Type of Your Variables

The error may occur if you try to assign a value of a different data type to a variable. Make sure that the data type of the value matches the data type of the variable.

4. Check the Memory Allocation of Your Variables

The error may occur if you try to assign a value to a variable that has not been allocated memory. Make sure that you have allocated memory for the variable before you try to assign a value to it.

5. Use Pointers

If the variable causing the error is a pointer, you may need to use a dereference operator to assign a value to it. Make sure that you use the correct syntax for the dereference operator.

Q1. What does "lvalue required as left operand of assignment" mean?

This error occurs when you try to assign a value to a non-modifiable lvalue.

Q2. How do I fix the "lvalue required as left operand of assignment" error?

You can fix this error by checking for typographical errors, checking the scope of your variables, checking the type of your variables, checking the memory allocation of your variables, and using pointers.

Q3. Why does the "lvalue required as left operand of assignment" error occur?

This error occurs when you try to assign a value to a non-modifiable lvalue, or if you try to assign a value of a different data type to a variable.

Q4. Can I use the dereference operator to fix the "lvalue required as left operand of assignment" error?

Yes, if the variable causing the error is a pointer, you may need to use a dereference operator to assign a value to it.

Q5. How can I prevent the "lvalue required as left operand of assignment" error?

You can prevent this error by declaring and initializing your variables before you try to assign a value to them, making sure that the data type of the value matches the data type of the variable, and allocating memory for the variable before you try to assign a value to it.

Understanding and Resolving the 'lvalue Required: Left Operand Assignment' Error in C++

Abstract: In C++ programming, the 'lvalue Required: Left Operator Assignment' error occurs when assigning a value to an rvalue. In this article, we'll discuss the error in detail, provide examples, and discuss possible solutions.

Understanding and Resolving the "lvalue Required Left Operand Assignment" Error in C++

In C++ programming, one of the most common errors that beginners encounter is the "lvalue required as left operand of assignment" error. This error occurs when the programmer tries to assign a value to an rvalue, which is not allowed in C++. In this article, we will discuss the concept of lvalues and rvalues, the causes of this error, and how to resolve it.

Lvalues and Rvalues

In C++, expressions can be classified as lvalues or rvalues. An lvalue (short for "left-value") is an expression that refers to a memory location and can appear on the left side of an assignment. An rvalue (short for "right-value") is an expression that does not refer to a memory location and cannot appear on the left side of an assignment.

For example, consider the following code:

In this code, x is an lvalue because it refers to a memory location that stores the value 5. The expression x = 10 is also an lvalue because it assigns the value 10 to the memory location referred to by x . However, the expression 5 is an rvalue because it does not refer to a memory location.

Causes of the Error

The "lvalue required as left operand of assignment" error occurs when the programmer tries to assign a value to an rvalue. This is not allowed in C++ because rvalues do not have a memory location that can be modified. Here are some examples of code that would cause this error:

In each of these examples, the programmer is trying to assign a value to an rvalue, which is not allowed. The error message indicates that an lvalue is required as the left operand of the assignment operator ( = ).

Resolving the Error

To resolve the "lvalue required as left operand of assignment" error, the programmer must ensure that the left operand of the assignment operator is an lvalue. Here are some examples of how to fix the code that we saw earlier:

In each of these examples, we have ensured that the left operand of the assignment operator is an lvalue. This resolves the error and allows the program to compile and run correctly.

The "lvalue required as left operand of assignment" error is a common mistake that beginners make when learning C++. To avoid this error, it is important to understand the difference between lvalues and rvalues and to ensure that the left operand of the assignment operator is always an lvalue. By following these guidelines, you can write correct and efficient C++ code.

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C++ Programming: From Problem Analysis to Program Design (5th Edition) by D.S. Malik
"lvalue required as left operand of assignment" on cppreference.com

Learn how to resolve 'lvalue Required: Left Operand Assignment' error in C++ by understanding the concept of lvalues and rvalues and applying the appropriate solutions.

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Tags: : C++ Programming Error Debugging

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err: lvalue required as left operand of assignment

I get this error (error: lvalue required as left operand of assignment) with the following code:

= is an assignment operator. == is a comparison operator.

This code is trying to assign the value 1 to Serial.read(), which it can't do.

Thank you. I forgot.