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 .
Lvalue references must be bound to a modifiable lvalue.
As C++ developers, we‘ve all likely encountered the dreaded "expression must be a modifiable lvalue" error at some point. This confusing compile-time error can be a roadblock, but if we take the time to truly understand it, it can help us write safer and more robust C++ code overall.
In this comprehensive guide, we‘ll demystify this error completely by looking at what it means, why it happens, how to diagnose it, and best practices to avoid it in the future. Let‘s get started!
When the compiler throws this error, it‘s telling us that we tried to assign something to an expression that cannot be assigned to. Let‘s break it down word-by-word:
Put simply, the error occurs because we tried to assign a value to something that cannot be assigned to. The expression on the left hand side of the assignment operator = needs to be a modifiable lvalue.
This is because the assignment operator stores the rvalue (right hand side expression) at the location in memory represented by the lvalue (left hand side). For this modification of memory to work correctly, the lvalue must refer to a valid modifiable location.
To really understand this error, we need to be clear on the difference between lvalues and rvalues in C++.
Lvalues are expressions that represent identifiable memory locations that can be accessed/modified in some way. Some key properties of lvalues:
Some examples of lvalues:
Rvalues are expressions that do not represent identifiable memory locations. They cannot be assigned to. Some examples:
So in summary, lvalues have identifiable memory locations and can appear on the left side of assignments whereas rvalues cannot.
Understanding lvalues vs rvalues, we can now look at some of the most common scenarios that cause this error:
When we declare a variable with const , it becomes read-only and its value cannot be changed. Trying to assign to it results in our error:
Similarly, assigning to a constant literal like a string also causes an error:
Condition statements like if and while expect a boolean expression, not an assignment.
Accidentally using = instead of == leads to an assignment inside the condition, causing invalid code:
The condition x = 5 assigns 5 to x, when we meant to check if x == 5.
The assignment operator = has lower precedence than other operators like + , * . So code like this:
Is treated as
This essentially tries to assign 10 to the temporary rvalue x + 5 , causing an error.
When declaring multiple variables in one statement, we must take care to assign to the correct identifier on the left side:
This assigns 10 to y instead of x due to ordering.
Assignment operators can be overloaded in classes. If the overloaded assignment operator is not implemented correctly, it can lead to this error.
For example, overloaded assignment should return a modifiable lvalue but sometimes programmers mistakenly return a temporary rvalue instead.
Temporary rvalues that get created in expressions cannot be assigned to:
Some other examples are trying to assign to literals directly or dereferencing an incorrect pointer location.
Only modifiable lvalues pointing to valid memory can be assigned to.
Trying to assign to an out of bounds array index is invalid:
Now that we‘ve seen common causes of the error, let‘s look at how to diagnose it when it shows up in our code:
With some careful analysis of the code and error messages, identifying the root cause becomes much easier.
Once we diagnose the issue, it can be resolved by using a valid modifiable lvalue on the left hand side of the assignment:
Let‘s see a couple of examples fixing code with this error:
1. Fixing Const Assignment
2. Fixing Conditional Assignment
3. Fixing Invalid Array Index
By properly understanding lvalues vs rvalues in C++, we can find and fix the root causes of this error.
Learning from the examples and causes above, we can establish coding best practices that help avoid these kinds of errors:
Adopting these best practices proactively will help us write code that avoids lvalue errors.
The cryptic "expression must be a modifiable lvalue" error in C++ is trying to tell us that we improperly tried assigning to something that cannot be assigned to in the language.
By learning lvalue/rvalue concepts deeply and following best practices around assignments, we can eliminate these kinds of bugs in our code.
Some key lessons:
Persistently applying these lessons will ensure we write assignment-bug-free C++ code!
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Prerequisites: lvalue and rvalue in C++ , References in C++ “l-value” refers to a memory location that identifies an object. “r-value” refers to the data value that is stored at some address in memory. References in C++ are nothing but the alternative to the already existing variable. They are declared using the ‘&’ before the name of the variable.
Example:
Below is the implementation for lvalue and rvalue:
Explanation : The following code will print True as both the variable are pointing to the same memory location. b is just an alternative name to the memory assigned to the variable a. The reference declared in the above code is lvalue reference (i.e., referring to variable in the lvalue) similarly the references for the values can also be declared.
rvalue references have two properties that are useful :
Important: lvalue references can be assigned with the rvalues but rvalue references cannot be assigned to the lvalue .
Uses of the lvalue references :
Note: When the function return lvalue reference the expression becomes lvalue expression.
Uses of rvalue references :
Program 1:
Program 2:
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Greetings everyone, I am thankful for any help that you may give me. I am working on a program that takes a trips start time and end time, then shaves 25% off the time it takes to complete the trip. However, while doing a static cast, I get the error of 'non-lvalue in assignment'. Everything else seems to work, at least from what I can tell. There are no other errors or warnings during an attempt to compile. Here is my source code: ------------------------------------------------------------------------------------------------------------------------------------ #include <iostream> #include <iomanip> using namespace std; int main () { int starttime = 0; cout << "\nPlease enter the start time for the old trip in 24 hour format (hhmm): "; cin >> starttime; cin.get(); int endtime = 0; cout << "\nPlease enter the end time for the old trip in 24 hour format (hhmm) :"; cin >> endtime; cin.get(); // Declare integers c and pt for static casting later. int newtime = 0; int pt = 0; //Get hours and minutes from the start time float hours = endtime / 100 - starttime / 100; float mins = endtime % 100 - starttime % 100; float starthours = starttime/100; float startmins = starttime%100; // Take Minutes, divide by sixty to get hours(from minutes) float minToHours = mins/60.0; // Same as above, but with the starting minutes float startMin = startmins/60.0; // Get the total hours (from minutes to hours, and regular hours) float added = minToHours + hours; // Same as above, but with starting minutes and hours float added1 = startMin + starthours; float newhours = added - (added / 4); float pretime = added1 + newhours; static_cast<int>(pt) = added1 + newhours; float fmins = pretime - pt; // Declare final numbers for the newtime calculation float newHours = pt * 100; float newMinutes = fmins * 60; static_cast<int>(newtime) = newHours + newMinutes; cout.fill ('0'); cout << "Your new ending time is: " << setw(4) << newtime << endl; system("PAUSE"); return 0; } ------------------------------------------------------------------------------------------------------------------------------------------------- These are the two static casts I am having problems with: static_cast<int>(pt) = added1 + newhours; static_cast<int>(newtime) = newHours + newMinutes; Again, thank you for the help!
Okay, I figured it out on my own after a good nights sleep and a little more research. I am posting the solution simply for anyone who may in the future be searching for similar information. In my original source code I did this: static_cast<int>(pt) = added1 + newhours; My problem is that the static_cast operator uses the parenthetical statement, and doesn't really listen to anything after that. So, the correct way to static cast the about red line is this way: static_cast<int>(pt = added1 + newhours); So, all in all a simple mistake. Gotta love that about programming, BE VERY carefull about what you type! It makes a difference! DISCLAIMER: I am no programming expert by any stretch of the imagination. This information worked for me, with the MinGW Developer Studio. It may not work with a different compiler, or you may not even have the problem I did. I dunno!
As far as I know, static_cast take a value of any type and typecast it to a given type. It's similar to regular C typecast operator, but stricter. The way to use the static_cast is as follows: cast_value = static_cast<type>(value); For example: double d = 15.4; int i = static_cast<int>(d); In your original code, instead of typecasting the value, you typecast the left hand value contained in the variable, which of course would do nothing value-wise because it is already an integer. It also throws the compiler off, because instead of treating the left hand statement as a variable in where the right hand value would be assigned, the static_cast actually convert the _value_ of the variable to an int and we all know that we can't assign the right hand value into another value. This is why the compiler came up with the non l-value error in assignment. Your subsequent typecasting also not entirely accurate, instead of typecasting the r-value, you actually typecast the _entire_ statement, and do nothing with the typecasting result. However, the value contained in variable pt is correct, because the assignment actually cast the value automatically (when you assigned a float to an int variable). Depending on the compiler and warning/error settings, this should, however, throws a warning on assignment of incorrect type. See the example above for correct usage.
n9352527, Thank you very much for the information! I appreciate the response. That does make more sense, good to learn. Thanks again.
excuse me but i received the same thing (Non-lvalue in assignment) when i was doing my homework. here is the code : #include <iostream> #include <cstdlib> #include <cstdio> using namespace std; int main () { double a = 0.0; double b = 0.0; double c = 0.0; double d1; double d2; double d3; double x = 0; cout << "Please input the 3 remainders.\n"; cin >> a; cin >> b; cin >> c; 3*x + a = d1; 5*x+b=d2; 7*x+c=d3; while ( d1 != d2 != d3) { x++; } cout << ("Your number is", d1 ,"."); system ("PAUSE"); return 0; } Please help!!!!
EEMajor said: These are the two static casts I am having problems with: static_cast<int>(pt) = added1 + newhours; static_cast<int>(newtime) = newHours + newMinutes; Click to expand...
alphaboy95 said: excuse me but i received the same thing (Non-lvalue in assignment) when i was doing my homework. here is the code : #include <iostream> #include <cstdlib> #include <cstdio> using namespace std; int main () { double a = 0.0; double b = 0.0; double c = 0.0; double d1; double d2; double d3; double x = 0; cout << "Please input the 3 remainders.\n"; cin >> a; cin >> b; cin >> c; 3*x + a = d1; // <=== error here 5*x+b=d2; // <=== error here 7*x+c=d3; // <=== error here while ( d1 != d2 != d3) // <=== possible error here { x++; } cout << ("Your number is", d1 ,"."); // I doubt that this will work system ("PAUSE"); // I'm not sure, but I don't think this will work return 0; } Please help!!!! Click to expand...
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Hi all, I know this is a trivial question but I really got stuck. I keep getting "non-1value in assignment" when trying to access the pointer variable "next" using the getNext() function. It didn't show any error if I change the last 2 lines to
The problem arises when I try using getNext(), as shown in the code below(located in the last 2 lines). This is my school assignments and I'm not allowed to set the pointer "next" to public.. kindly need your help as I've already stuck in this for so long.
Pass your after pointer to your temp_ptr's setNext method, something like temp_ptr->setNext(after->getNext());
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missang() { (A == 0 && B != 0 && C != 0) { A = 180 - C - B; } (A != 0 && B == 0 && C !=0) { B = 180 - A - C; } (A != 0 && B != 0 && C == 0) { C = 180 - A - B; } } |
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Hi i would like to make an external single linked list. I have a problem with "Non-Ivalue in assignment" and its occuring on line "this = currP->next" i tried making it currP.next but its also producing an error
An lvalue is a variable that can reside at the left side of the equal operator. This means its value can be changed. You can't change the value of this , it's just not permitted, therefore the error.
You could re-write your function as follows:
deleteNode() will now return a pointer to the start of the rest of the list and the recursive algorithm will concatenate the first part with the last part. It's not tested so a few adjustments may be necessary but I hope you get the point.
A lvalue is a semantic rule. It means "left-value".
Examples of lvalues are:
this is not a lvalue. You just can't assign anything to it. It's the reference to the method caller.
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Errors with c++ pointers to arrays: invalid type argument of unary * (have 'int), lvalue required as left operand of assignment 3 lvalue required as left operand of assignment c++ arrays
Since this is an assignment I would appreciate not getting code for an answer, but rather if someone can tell me where to look for the issue? Here is the set function I am trying to write. The private data in this class is an array, Hand TheHand[3] and int numCards(the number of cards in TheHand) in the .h
An L-Value (Left-Value) is an operand* that can appear on both the left-hand & right-hand side of an assignment. L-Values generally have an address in memory, so the compiler knows where to store the value on the right-hand side. An R-Value is also an operand that generally doesn't have an address in memory. These are usually magic constants or ...
Zhuge (4664) The assignment (=) operator makes the left side equal to the right. The right side is the constant 1, which can't be changed to be equal to whatever is at ary [j]. In other words, the constant 1 is not a proper l-value for the operator. You should just switch the order of the operands.
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 ...
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.
An lvalue reference (commonly just called a reference since prior to C++11 there was only one type of reference) acts as an alias for an existing lvalue (such as a variable).. To declare an lvalue reference type, we use an ampersand (&) in the type declaration: int // a normal int type int& // an lvalue reference to an int object double& // an lvalue reference to a double object
Call functions that return non-const lvalue references like int& func() Use array element indices that are in bounds ... Let's see a couple of examples fixing code with this error: 1. Fixing Const Assignment // Broken const float PI = 3.14; PI = 2.5; // Fixed float PI = 3.14; PI = 2.5; 2. Fixing Conditional Assignment
Note: When the function return lvalue reference the expression becomes lvalue expression. Uses of rvalue references: They are used in working with the move constructor and move assignment. cannot bind non-const lvalue reference of type 'int&' to an rvalue of type 'int'.
The errors you mention about non lvalues come from the first three lines I marked as errors. An assignment statement has to have a variable (an lvalue) on the left side, and some value on the right side. The l in lvalue stands for left. You need to rewrite these three lines following this model: Rich (BB code):
You can assign something to a variable, but not to "meter2 - meter". You probably meant meter2 = remainder + meter;
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Hi all, I know this is a trivial question but I really got stuck. I keep getting "non-1value in assignment" when trying to access the ...
r=(5&&c)=10; which is equivalent to either. r= 0 =10; or. r= 1 =10; The reason behind either expression is that 5 && c will return either true or false depending upon the value of c (whether it is 0 or not). As per the comment by OP. How to wire r=5 and c=10. You can do this as.
when i compile this code i get "error: non-lvalue in assignment" why? Just so you know, this program soves for the missing parts of a triangle, los and loc are law of sines and law of cosines respectively. The errors are commented and are in the last appx 30 lines of code. /* Note, a capital letter represents an angle measure and
error: non-lvalue in assignment. C / C++ Forums on Bytes. >>Thanks for the response.... Point Taken but this is not the case. Thus, if a person writes a text file on her or his computer and does not use UNICODE to save it, the current code page is used.
The name lvalue comes originally from the assignment expression E1 = E2, in which the left operand E1 is required to be a (modifiable) lvalue. It is perhaps better considered as representing an object "locator value". What is sometimes called rvalue is in this International Standard described as the "value of an expression".
1. To elaborate a little on this answer, instead of assigning to this, you could copy the data which for a simple class like this you can do with *this = *currP->next. If you don't know what the difference is, read more about pointer dereferencing. - Some programmer dude.
Thus, if a person writes a text file on her or his computer and does not use UNICODE to save it, the current code page is used. If this file is given to someone with some other current codepage, the file is not displayed correctly. Simply converting the file to Unicode will make the data display properly. When performing the encoding process ...