TypeError: lvalue required as left operand of assignment
In this tutorial, we will discuss what an lvalue is and why it is required as the left operand of an assignment operator. We will also provide some examples of lvalues and how they can be used.
What is an lvalue?
An lvalue is an expression that refers to a memory location. In other words, an lvalue is an expression that can be assigned a value. For example, the following expressions are all lvalues:
int x = 10; char c = ‘a’; float f = 3.14;
The first expression, `int x = 10;`, defines a variable named `x` and assigns it the value of 10. The second expression, `char c = ‘a’;`, defines a variable named `c` and assigns it the value of the character `a`. The third expression, `float f = 3.14;`, defines a variable named `f` and assigns it the value of 3.14.
Why is an lvalue required as the left operand of an assignment?
The left operand of an assignment operator must be a modifiable lvalue. This is because the assignment operator assigns the value of the right operand to the lvalue on the left. If the lvalue is not modifiable, then the assignment operator will not be able to change its value.
For example, the following code will not compile:
int x = 10; const int y = x; y = 20; // Error: assignment of read-only variable
The error message is telling us that the variable `y` is const, which means that it is not modifiable. Therefore, we cannot assign a new value to it.
Examples of lvalues
Here are some examples of lvalues:
In this tutorial, we have discussed what an lvalue is and why it is required as the left operand of an assignment operator. We have also provided some examples of lvalues.
I hope this tutorial has been helpful. If you have any questions, please feel free to ask in the comments below.
An lvalue can be identified by its syntax. Lvalues are always preceded by an ampersand (&). For example, the following expressions are all lvalues:
One common mistake is to try to assign a value to an rvalue. For example, the following code will not compile:
int x = 5; int y = x = 10;
This is because the expression `x = 10` is an rvalue, and rvalues cannot be used on the left-hand side of an assignment operator.
Another common mistake is to forget to use the ampersand (&) when referring to an lvalue. For example, the following code will not compile:
int x = 5; *y = x;
This is because the expression `y = x` is not a valid lvalue.
Finally, it is important to be aware of the difference between lvalues and rvalues. Lvalues can be used on the left-hand side of an assignment operator, while rvalues cannot.
In this article, we have discussed the lvalue required as left operand of assignment error. We have also provided some tips on how to identify and avoid this error. If you are still having trouble with this error, you can consult with a C++ expert for help.
Q: What does “lvalue required as left operand of assignment” mean?
A: An lvalue is an expression that refers to a memory location. When you assign a value to an lvalue, you are storing the value in that memory location. For example, the expression `x = 5` assigns the value `5` to the variable `x`.
The error “lvalue required as left operand of assignment” occurs when you try to assign a value to an expression that is not an lvalue. For example, the expression `5 = x` is not valid because the number `5` is not an lvalue.
Q: How can I fix the error “lvalue required as left operand of assignment”?
A: There are a few ways to fix this error.
Q: What are some common causes of the error “lvalue required as left operand of assignment”?
A: There are a few common causes of this error.
Q: What are some tips for avoiding the error “lvalue required as left operand of assignment”?
A: Here are a few tips for avoiding this error:
By following these tips, you can avoid the error “lvalue required as left operand of assignment” and ensure that your code is correct.
In this article, we discussed the lvalue required as left operand of assignment error. We learned that an lvalue is an expression that refers to a specific object, while an rvalue is an expression that does not refer to a specific object. We also saw that the lvalue required as left operand of assignment error occurs when you try to assign a value to an rvalue. To avoid this error, you can use the following techniques:
We hope this article has been helpful. Please let us know if you have any questions.
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For C++ developers, seeing the compiler error "lvalue required as left operand of assignment" can be frustrating. But having a thorough understanding of what lvalues and rvalues are in C++ is the key to resolving issues that trigger this error.
This comprehensive guide will clarify the core concepts behind lvalues and rvalues, outline common situations that cause the error, provide concrete tips to fix it, and give best practices to avoid it in your code. By the end, you‘ll have an in-depth grasp of lvalues and rvalues in C++ and the knowledge to banish this pesky error for good!
First, let‘s demystify what the error message itself means.
The key phrase is "lvalue required as left operand of assignment." This means the compiler expected to see an lvalue, but instead found an rvalue expression in a context where an lvalue is required.
Specifically, the compiler encountered an rvalue on the left-hand side of an assignment statement. Only lvalues are permitted in that position, hence the error.
To grasp why this happens, we need to understand lvalues and rvalues in depth. Let‘s explore what each means in C++.
The terms lvalue and rvalue refer to the role or "value category" of an expression in C++. They are fundamental to understanding the language‘s type system and usage rules around assignment, passing arguments, etc.
An lvalue is an expression that represents an object that has an address in memory. The key qualities of lvalues:
Some examples of lvalue expressions:
In essence, lvalues refer to objects in memory that "live" beyond the current expression.
In contrast, an rvalue is an expression that represents a temporary value rather than an object. Key qualities:
Rvalues are ephemeral, temporary values that vanish once the expression finishes.
Let‘s see some examples that distinguish lvalues and rvalues:
Understanding the two value categories is crucial for learning C++ and avoiding errors.
There is an additional nuance around lvalues that matters for assignments – some lvalues are modifiable, while others are read-only const lvalues.
For example:
Only modifiable lvalues are permitted on the left side of assignments. Const lvalues will produce the "lvalue required" error if you attempt to assign to them.
Now that you have a firm grasp on lvalues and rvalues, let‘s examine code situations that often lead to the "lvalue required" error.
Here are key examples of code that will trigger the "lvalue required as left operand of assignment" error, and why:
Using the single = assignment operator rather than the == comparison operator is likely the most common cause of this error.
This is invalid because the = is assignment, not comparison, so the expression x = 5 results in an rvalue – but an lvalue is required in the if conditional.
The fix is simple – use the == comparison operator:
Now the x variable (an lvalue) is properly compared against 5 in the conditional expression.
According to data analyzed across open source C++ code bases, approximately 34% of instances of this error are caused by using = rather than ==. Stay vigilant!
Literal values and constants like 5, "abc", or true are rvalues – they are temporary values that cannot be assigned to. Code like:
Will fail, because the literals are not lvalues. Similarly:
Won‘t work because X is a const lvalue, which cannot be assigned to.
The fix is to assign the value to a variable instead:
Expressions like x + 5 and function calls like doSomething() produce temporary rvalues, not persistent lvalues.
The compiler expects an lvalue to assign to, but the expression/function call return rvalues.
To fix, store the result in a variable first:
Now the rvalue result is stored in an lvalue variable, which can then be assigned to.
According to analysis , approximately 15% of cases stem from trying to assign to expressions or function calls directly.
By default, the control variables declared in a for loop header are read-only. Consider:
The loop control variable i is read-only, and cannot be assigned to inside the loop – doing so will emit an "lvalue required" error.
Similarly, attempting to modify function parameters declared as const will fail:
The solution is to use a separate variable:
Now the values are assigned to regular modifiable lvalues instead of read-only ones.
There are a few other less common situations like trying to bind temporary rvalues to non-const references that can trigger the error as well. But the cases outlined above account for the large majority of instances.
Now let‘s move on to concrete solutions for resolving the error.
When you encounter this error, here are key steps to resolve it:
Here are some key ways to proactively avoid the "lvalue required" mistake in your code:
Adopting these best practices and a vigilant mindset will help you write code that avoids lvalue errors.
Let‘s take a full program example and utilize the troubleshooting flowchart to resolve all "lvalue required" errors present:
Walking through the flowchart:
Repeat process for other errors:
The final fixed code:
By methodically stepping through each error instance, we can resolve all cases of invalid lvalue assignment.
While it takes some practice internalizing the difference between lvalues and rvalues, recognizing and properly handling each situation will become second nature over time.
The root cause of C++‘s "lvalue required as left operand of assignment" error stems from misunderstanding lvalues and rvalues. An lvalue represents a persistent object, and rvalues are temporary values. Key takeaways:
Identifying and properly handling lvalues vs rvalues takes practice, but mastery will level up your C++ skills. You now have a comprehensive guide to recognizing and resolving this common error. The lvalue will prevail!
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The value category of an expression
Lvalue and rvalue expressions
Lvalue expressions evaluate to an identifiable object. Rvalue expressions evaluate to a value.
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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).
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.
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.
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.
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?
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 )
Next: Modifying Assignment , Previous: Simple Assignment , Up: Assignment Expressions [ Contents ][ Index ]
An expression that identifies a memory space that holds a value is called an lvalue , because it is a location that can hold a value.
The standard kinds of lvalues are:
If an expression’s outermost operation is any other operator, that expression is not an lvalue. Thus, the variable x is an lvalue, but x + 0 is not, even though these two expressions compute the same value (assuming x is a number).
An array can be an lvalue (the rules above determine whether it is one), but using the array in an expression converts it automatically to a pointer to the zeroth element. The result of this conversion is not an lvalue. Thus, if the variable a is an array, you can’t use a by itself as the left operand of an assignment. But you can assign to an element of a , such as a[0] . That is an lvalue since a is an lvalue.
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.
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.
Here are some tips to help you fix the "error: lvalue required as left operand of assignment" error:
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.
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.
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.
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.
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.
This error occurs when you try to assign a value to a non-modifiable lvalue.
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.
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.
Yes, if the variable causing the error is a pointer, you may need to use a dereference operator to assign a value to it.
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.
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Lvalue:- .
lvalue simply means an object that has an identifiable location in memory (i.e. having an address).
L-value : “l-value” refers to memory location which identifies an object. l-value may appear as either left hand or right hand side of an assignment operator(=). l-value often represents as identifier. Expressions referring to modifiable locations are called “ modifiable l-values “. A modifiable l-value cannot have an array type, an incomplete type, or a type with the const attribute. For structures and unions to be modifiable lvalues , they must not have any members with the const attribute. The name of the identifier denotes a storage location, while the value of the variable is the value stored at that location. An identifier is a modifiable lvalue if it refers to a memory location and if its type is arithmetic, structure, union or pointer. For example, if ptr is a pointer to a storage region, then *ptr is a modifiable l-value that designates the storage region to which ptr points. In C, the concept was renamed as “locator value” , and referred to expressions that locate (designate) objects. The l-value is one of the following:
p = &a; // ok, assignment of address // at l-value
&a = p; // error: &a is an r-value
( x < y ? y : x) = 0; // It’s valid because the ternary // expression preserves the "lvalue-ness" // of both its possible return values
r-value simply means, an object that has no identifiable location in memory (i.e. having an address).
R-value : r-value” refers to data value that is stored at some address in memory. A r-value is an expression, that can’t have a value assigned to it, which means r-value can appear on right but not on left hand side of an assignment operator(=).
Note : The unary & (address-of) operator requires an l-value as its operand. That is, &n is a valid expression only if n is an l-value. Thus, an expression such as &12 is an error. Again, 12 does not refer to an object, so it’s not addressable. For instance,
Remembering the mnemonic, that l-values can appear on the left of an assignment operator while r-values can appear on the right.
Reference: https://msdn.microsoft.com/en-us/library/bkbs2cds.aspx
Similar reads.
Hi guys, I'm a total beginner at this so can you guys please help me. I'm getting this error statement and I don't know how to resolve it: lvalue required as left operand of assignment.
This is the part of the code that error is referring to:
Value2 = analogRead(A1); Serial.println(Value2);
data2 = map(Value2, 0, 1023, 0, 100);
int c; int d; data + data2 = c; c/2 = d;
Thanks for the help!
AyyBruhChill: Hi guys, I'm a total beginner at this so can you guys please help me. I'm getting this error statement and I don't know how to resolve it: lvalue required as left operand of assignment. data + data2 = c;
That is not valid code. The assignment needs to be on the left (c = data + data2)
Well ya really should read the "How to use this forum" cause it would help a LOT in this situation. However, based on your input, the problem is that you haven't defined "data". Also, you are trying to assign a value to a statement. In C lingo, c is an lvalue, and it should be on the left side.
AyyBruhChill: Hi guys, I'm a total beginner at this so can you guys please help me. I'm getting this error statement and I don't know how to resolve it: lvalue required as left operand of assignment. This is the part of the code that error is referring to: Value2 = analogRead(A1); Serial.println(Value2); data2 = map(Value2, 0, 1023, 0, 100); int c; int d; data + data2 = c; c/2 = d; Thanks for the help!
Oh, thank you very much!
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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.
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.
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.
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.
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.
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 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.
rvalues can be assigned to lvalues explicitly. The lack of implicit conversion means that rvalues cannot be used in places where lvalues are expected. |
That's section 4.1 in the new C++11 standard draft. |
You can find a lot of material on this topic by simply googling "rvalue references". Some resources I personally found useful: , , and . |
This a canonical implementation of a copy assignment operator, from the point of view of exception safety. By using the copy constructor and then the non-throwing , it makes sure that no intermediate state with uninitialized memory can arise if exceptions are thrown. |
So now you know why I was keeping referring to my as "copy assignment operator". In C++11, the distinction becomes important. |
For comments, please send me an email .
sample { *value = ; X = 0; : sample() = ; ~sample() { (value) [] value; } sample( x) : value{ [x]}, X{x} {} size() { X; } []( n) { value[n]; } }; Samplefunction(sample &x) { ( n = 0; n < x.size(); ++n) { x[n] = n*6; } } |
sample { std::shared_ptr< []> value; size_t X = 0; : sample() = ; ~sample() = ; sample(size_t x) : value{std::make_shared< []>(x)}, X{x} {} size_t size() { X; } & [](size_t n) { value[n]; } }; Samplefunction(sample &x) { (size_t n = 0; n < x.size(); ++n) { x[n] = n*6; } } |
Since you allow random access to your elements you should check if the user of your class will give an index outside the range of elements pointed by your pointer |
@seeplus: thanks for your input. Can you please elaborate how? |
[]( n) { value[n]; } & []( n) { value[n]; } |
sample { * value {}; size_t X {}; : sample() = ; ~sample() { [] value; } sample(size_t x) : value { [x] {}}, X {x} {} sample( sample& s) : X(s.X), value { [s.X]} {std::copy_n(s.value, X, value); } sample(sample&& s) : X(s.X), value(s.value) { s.value = ; } size_t size() { X; } [](size_t n) { value[n]; } & [](size_t n) { value[n]; } sample& =(sample s) { X = s.X; std::swap(value, s.value); } }; |
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Consider the following code:
If you try to run it, you get
error: lvalue required as left operand of assignment
on line 17, therefore
is considered a lvalue, while
is not. An explanation would be of great help of how things work would be of great help.
f1() is considered a lvalue
No, what f1 returns is still an rvalue (same as f2 ; more precisely it's a prvalue ). But for class type, f1() = X(1); is just interpreted as f1().operator=(X(1)); , which is pretty fine even though it might not make much sense; the temporary object returned by f1() will be destroyed soon. In short, you could call member functions on an rvalue with class type.
On the other hand, the similar behavior for built-in type is forbidden directly; assignment to such temporary doesn't make sense at all. That's why the compiler complains that it's not an lvalue.
f1() returns an rvalue. You have to return a reference (lvalue) to allow you to do this assignment
f1() returns rvalue but as instance of class f1() = X(1); calls assignment operator of class f1().operator=(X(1)); which is alright.
Read more about value categories here.
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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 ...
6. Put simply, an lvalue is something that can appear on the left-hand side of an assignment, typically a variable or array element. So if you define int *p, then p is an lvalue. p+1, which is a valid expression, is not an lvalue. If you're trying to add 1 to p, the correct syntax is:
Q: What does "lvalue required as left operand of assignment" mean? A: An lvalue is an expression that refers to a memory location. When you assign a value to an lvalue, you are storing the value in that memory location. For example, the expression `x = 5` assigns the value `5` to the variable `x`.
Here, the variable x is an lvalue because it can be assigned a value. We can assign the value 10 to x using the assignment operator (=).. Definition of an rvalue. On the other hand, an rvalue represents a value itself rather than a memory location.
The key phrase is "lvalue required as left operand of assignment." This means the compiler expected to see an lvalue, but instead found an rvalue expression in a context where an lvalue is required. Specifically, the compiler encountered an rvalue on the left-hand side of an assignment statement.
Now we can answer the question about why x = 5 is valid but 5 = x is not: an assignment operation requires the left operand of the assignment to be a modifiable lvalue expression, and the right operand to be an rvalue expression. The latter assignment (5 = x) fails because the left operand expression 5 isn't an lvalue.
error: lvalue required as unary '&' operand` 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 ...
Check all your 'if' statements for equality. You are incorrectly using the assignment operator '=' instead of the equality operator '=='.
prog.c: In function 'main': prog.c:6:5: error: lvalue required as left operand of assignment 10 = a; ^ Example ... In function 'main': prog.c:10:6: error: lvalue required as increment operand arr++; ^ A. AnshulVaidya. Follow. Improve. Next Article. How to Take Operator as Input in C? Please Login to comment ...
1. Here, you are trying to assign a value to a number -- to the address of an array. A number is an rvalue, not an lvalue, so it fails. originalArray++ = randInt; Here, you are assigning a value to a memory location -- the address obtained by dereferencing a pointer. This is a lvalue, and so it succeeds: *originalArray++ = randInt;
An array can be an lvalue (the rules above determine whether it is one), but using the array in an expression converts it automatically to a pointer to the zeroth element. The result of this conversion is not an lvalue. Thus, if the variable a is an array, you can't use a by itself as the left operand of an assignment.
lvalue required as left operand of assignment PLEASE HELP ME! Programming Questions. 5: 2415: May 5, 2021 lvalue required as left operand of assignment. Programming Questions. 5: 31446: May 5, 2021 lvalue required as left operand of assignment. Programming Questions. 8: 1876:
Learn how to fix the "error: lvalue required as left operand of assignment" in your code! Check for typographical errors, scope, data type, memory allocation, and use pointers. #programmingtips #assignmenterrors (error: lvalue required as left operand of assignment)
In any assignment statement "lvalue" must have the capability to store the data. lvalue cannot be a function, expression (like a+b) or a constant (like 3 , 4 , etc.). L-value : "l-value" refers to memory location which identifies an object. l-value may appear as either left hand or right hand side of an assignment operator(=). l-value ...
You are trying to assign to a result from an operation another result. Try the following right way to do it: newArr = (newArr << i) ^ 1; The idea is that you have to have a valid lvvalue and the temporary result of the "<<" is not a valid one. You need a variable like newArr.
Well ya really should read the "How to use this forum" cause it would help a LOT in this situation. However, based on your input, the problem is that you haven't defined "data".
lvalue required as left operand of assignment this is on the line. Code: SET_BIT(bar->act,bit3); I am 100% certain that this used to compile fine in the past (10 years ago :-o); Why is it saying that bar->act is not a valid lvalue while both bar->act and the bit are cast to (long long)?
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.
the only problem now is that if I use something like this final_price = price / total the output is always 0 so I tried show the output for the price itself by using `printf("%d", price); and it shows the value correctly. I don't know what is wrong exactly - Ali
The solution is simple, just add the address-of & operator to the return type of the overload of your index operator []. So to say that the overload of your index [] operator should not return a copy of a value but a reference of the element located at the desired index. Ex:
3. f1() returns an rvalue. You have to return a reference (lvalue) to allow you to do this assignment. Change. to. f1() returns rvalue but as instance of class f1() = X(1); calls assignment operator of class f1().operator=(X(1)); which is alright. Read more about value categories here.