what is compound assignment c#

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Compound assignment operators in C#

A compound assignment operator has a shorter syntax to assign the result. The operation is performed on the two operands before the result is assigned to the first operand.

The following are the compound assignment operators in C#.

Let us see an example to learn how to work with compound assignment operators in C#.

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C# 8 - Null coalescing/compound assignment

There are a lot of cool new features in C# 8 and one of my favorites is the new Null coalescing assignment operator.

Stefán Jökull Sigurðarson

Hi friends.

There are a lot of cool new features in C# 8 and one of my favorites is the new Null coalescing assignment (or compound assignment, whichever you prefer) operator. If you have ever written code like this:

You can simplify this code A LOT!

That is the new ??= operator in action, which takes care of doing the null check and assignment for you in one sweet syntactic sugar-rush! I also find it more readable, but some people might disagree, and that fine. Just pick whatever you prefer :)

And the best part is that it actually compiles down to the same efficient code! Just take a look at this sharplab.io sample to see what I mean.

You might have solved this with a Lazy<string> or by using something like _someValue = _someValue ?? InitializeMyValue(); as well but that's still more code to write than the new operator and less efficient as well. The Lazy<T> approach has some overhead and the null-coalescing operator + assignment has the added inefficiency of always making the variable assignment even if there is no need to (you can take a closer look at the ASM part of the SharpLab example above) but there it is for reference:

Hope this helps! :)

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+= operator

• Add operation.
• Assignment of the result of add operation.
• Understanding += operator with a code -
• Statement i+=2 is equal to i=i+2 , hence 2 will be added to the value of i, which gives us 4.
• Finally, the result of addition, 4 is assigned back to i, updating its original value from 2 to 4.
• A special case scenario for all the compound assigned operators
• All compound assignment operators perform implicit casting.
• Casting the char value ( smaller data type ) to an int value( larger data type ), so it could be added to an int value, 2.
• Finally, the result of performing the addition resulted in an int value, which was casted to a char value before it could be assigned to a char variable, ch .

Example with += operator

-= operator.

• Subtraction operation.
• Assignment of the result of subtract operation.
• Statement i-=2 is equal to i=i-2 , hence, 2 will be subtracted from the value of i, which gives us 0.
• Finally, the result of subtraction i.e. 0 is assigned back to i, updating its value to 0.

Example with -= operator

*= operator.

• Multiplication operation.
• Assignment of the result of multiplication operation.
• Statement i*=2 is equal to i=i*2 , hence 2 will be multiplied with the value of i, which gives us 4.
• Finally, the result of multiplication, 4 is assigned back to i, updating its value to 4.

Example with *= operator

/= operator.

• Division operation.
• Assignment of the result of division operation.
• Statement i/=2 is equal to i=i/2 , hence 4 will be divided by the value of i, which gives us 2.
• Finally, the result of division i.e. 2 is assigned back to i, updating its value from 4 to 2.

Example with /= operator

%= operator.

• Modulus operation, which finds the remainder of a division operation.
• Assignment of the result of modulus operation.
• Statement i%=2 is equal to i=i%2 , hence 4 will be divided by the value of i and its remainder gives us 0.
• Finally, the result of this modulus operation i.e. 0 is assigned back to i, updating its value from 4 to 0.

Example with %= operator

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Posted on Jan 5, 2020

Coalescing operator and Compound assignment operator in C#

Hello! Today i want to ask you this: Have you ever used a Null Coalescing operator or even a Compound assignment operator in C# ?

Until today i had never heard about this things, so i want to share with you what i learned about and how it can be applied to your code.

The problem

Let's say you want to give a given variable the value of null.

Now, if we want to print the value on the screen it will accuse the following error:

Let's see how to get around this...

The old way 👎

The old and 'commom way' to check this is using if else operators like this:

We see that in this case we cannot place a default value to x and y operators. So we display on screen when it is null.

The Null Coalescing operator way 👌

First, a null coalescing operator (??) is used to define a default value for nullable value types or reference types. It returns the left-hand operand if the operand is not null, otherwise, it returns the right operand.

So it's mainly used to simplify checking for null values and also assign a default value to a variable when the value is null.

Using our example:

This way i can make a default value on x and y when one of them is null. And so, we can print on screen!

But can it be better?

The Compound assignment operator way 😎

The Compound assignment operator (??=) was introduced on C# 8.0 and has made our job easier. It simply reduces what we have to write and has the same result.

Instead of writing double x = x ?? 0.0; We can just write double x ??= 0.0;

Simple, right?

Hope you enjoyed this post, it's simple but it's something worth sharing for me.

Thanks for your time!😊

Links: https://dzone.com/articles/nullable-types-and-null-coalescing-operator-in-c

https://dev.to/mpetrinidev/the-null-coalescing-operator-in-c-8-0-4ib4

https://docs.microsoft.com/pt-br/dotnet/csharp/language-reference/operators/null-coalescing-operator

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Good article. Thank you.

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Awesome Dude, simple but unknown by most programmers

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Compound Assignment in

About compound assignment.

Many operators can also be used as compound assignments , which are a shorthand notation where x = op y can be written as x op= y :

C# Compound Assignments

The core statement of the solution is as follows:

Compound Assignments

In this code, you use the compound assignment operator (+=), which is a shortcut that does the same thing as the previous solution.

You will see compound assignments in all C-family programming languages.

A program that illustrates compound assignment in connection with subtraction, multiplication, and division.

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?? and ??= operators - the null-coalescing operators

• 9 contributors

The null-coalescing operator ?? returns the value of its left-hand operand if it isn't null ; otherwise, it evaluates the right-hand operand and returns its result. The ?? operator doesn't evaluate its right-hand operand if the left-hand operand evaluates to non-null. The null-coalescing assignment operator ??= assigns the value of its right-hand operand to its left-hand operand only if the left-hand operand evaluates to null . The ??= operator doesn't evaluate its right-hand operand if the left-hand operand evaluates to non-null.

The left-hand operand of the ??= operator must be a variable, a property , or an indexer element.

The type of the left-hand operand of the ?? and ??= operators can't be a non-nullable value type. In particular, you can use the null-coalescing operators with unconstrained type parameters:

The null-coalescing operators are right-associative. That is, expressions of the form

are evaluated as

The ?? and ??= operators can be useful in the following scenarios:

In expressions with the null-conditional operators ?. and ?[] , you can use the ?? operator to provide an alternative expression to evaluate in case the result of the expression with null-conditional operations is null :

When you work with nullable value types and need to provide a value of an underlying value type, use the ?? operator to specify the value to provide in case a nullable type value is null :

Use the Nullable<T>.GetValueOrDefault() method if the value to be used when a nullable type value is null should be the default value of the underlying value type.

You can use a throw expression as the right-hand operand of the ?? operator to make the argument-checking code more concise:

The preceding example also demonstrates how to use expression-bodied members to define a property.

You can use the ??= operator to replace the code of the form

with the following code:

Operator overloadability

The operators ?? and ??= can't be overloaded.

C# language specification

For more information about the ?? operator, see The null coalescing operator section of the C# language specification .

For more information about the ??= operator, see the feature proposal note .

• Null check can be simplified (IDE0029, IDE0030, and IDE0270)
• C# operators and expressions
• ?. and ?[] operators
• ?: operator

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Why it is not possible to overload compound assignment operator in C#?

The title is be misleading, so please read entire question :-) .

By "compound assignment operator" I have in mind a construct like this op= , for example += . Pure assignment operator ( = ) does not belong to my question.

By "why" I don't mean an opinion, but resource (book, article, etc) when some of the designers, or their coworkers, etc. express their reasoning (i.e. the source of the design choice).

I am puzzled by asymmetry found in C++ and C# ( yes, I know C# is not C++ 2.0 ) -- in C++ you would overload operator += and then almost automatically write appropriate + operator relying on previously defined operator. In C# it is in reverse -- you overload + and += is synthesised for you.

If I am not mistaken the later approach kills a chance for optimization in case of actual += , because new object has to be created. So there has to be some big advantage of such approach, but MSDN is too shy to tell about it.

And I wonder what that advantage is -- so if you spotted explanation in some C# book, tech-talk video, blog entry, I will be grateful for the reference.

The closest thing I found is a comment on Eric Lippert blog Why are overloaded operators always static in C#? by Tom Brown. If the static overloading was decided first it simply dictates which operators can be overloaded for structs. This further dictates what can be overloaded for classes.

• language-design

• 3 Do you have a link for the new C++ style? Naively, overloading += first seems absurd; why would you overload a combined operation rather than the parts of it? –  Telastyn Commented May 30, 2015 at 16:18
• 1 I believe the term for these is "compound assignment operators". –  Ixrec Commented May 30, 2015 at 16:30
• 2 @greenoldman Telastyn was probably implying that implementing += in terms of + seems more natural than the other way around, since semantically += is a composition of + and =. To the best of my knowledge , having + call += is largely an optimization trick. –  Ixrec Commented May 30, 2015 at 16:33
• 1 @Ixrec: Sometimes, a+=b makes sense, while a=a+b doesn't: Consider that identity might be important, an A cannot be duplicated, whatever. Sometimes, a=a+b makes sense, while a+=b doesn't: Consider immutable strings. So, one actually needs the ability to decide which one to overload separately. Of course, auto-generating the missing one, if all neccessary building-blocks exist and it's not explicitly disables, is a good idea. Not that C# allows that atm, afaik. –  Deduplicator Commented May 30, 2015 at 16:40
• 4 @greenoldman - I understand that motivation, but personally, I would find *= mutating a reference type to be semantically incorrect. –  Telastyn Commented May 30, 2015 at 17:52

I can't find the reference for this, but my understanding was that the C# team wanted to provide a sane subset of operator overloading. At the time, operator overloading had a bad rap. People asserted that it obfuscated code, and could only be used for evil. By the time C# was being designed, Java had shown us that no operator overloading was kind of annoying.

So C# wanted to balance operator overloading so that it was more difficult to do evil, but you could make nice things too. Changing the semantics of assignment was one of those things that was deemed always evil. By allowing += and its kin to be overloaded, it would allow that sort of thing. For example, if += mutated the reference rather than making a new one, it wouldn't follow the expected semantics, leading to bugs.

• Thank you, if you don't mind I will keep this question open bit longer, and if nobody beat you with the reasoning of actual design, I will close it with accepting yours. Once again -- thank you. –  greenoldman Commented May 30, 2015 at 18:29
• @greenoldman - as well you should. I too hope someone come with a answer with more concrete references. –  Telastyn Commented May 30, 2015 at 20:04
• So, this is a place where the impossibility of talking both about the pointer and the pointee comfortably and unambiguously rears it's head? Damn shame. –  Deduplicator Commented May 30, 2015 at 21:15
• "People asserted that it obfuscated code" - it's still true though, have you seen some of the F# libraries? Operator noise. E.g. quanttec.com/fparsec –  Den Commented Jun 3, 2015 at 9:39

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C# Tutorial

C# examples, c# assignment operators, assignment operators.

Assignment operators are used to assign values to variables.

In the example below, we use the assignment operator ( = ) to assign the value 10 to a variable called x :

Try it Yourself »

The addition assignment operator ( += ) adds a value to a variable:

A list of all assignment operators:

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Advantage of using compound assignment

What is the real advantage of using compound assignment in C/C++ (or may be applicable to many other programming languages as well)?

I looked at few links like microsoft site , SO post1 , SO Post2 . But the advantage says exp1 is evaluated only once in case of compound statement. How exp1 is really evaluated twice in first case? I understand that current value of exp1 is read first and then new value is added. Updated value is written back to the same location. How this really happens at lower level in case of compound statement? I tried to compare assembly code of two cases, but I did not see any difference between them.

• compound-assignment

• There's a good chance that your compiler optimised it. Anyway, assembly language usually has increment opcodes. In fact, they don't have anything else. If you do something like 1 + 2 in C, it would be compiled to something like move 1,a and add 2,a . –  SeverityOne Commented Apr 19, 2018 at 13:22
• @SeverityOne's response should be the answer... –  Frederick Commented Apr 19, 2018 at 13:27

6 Answers 6

For simple expressions involving ordinary variables, the difference between

is syntactical only. The two expressions will behave exactly the same, and might well generate identical assembly code. (You're right; in this case it doesn't even make much sense to ask whether a is evaluated once or twice.)

Where it gets interesting is when the left-hand side of the assignment is an expression involving side effects. So if you have something like

it makes much more of a difference! The former tries to increment p twice (and is therefore undefined). But the latter evaluates p++ precisely once, and is well-defined.

As others have mentioned, there are also advantages of notational convenience and readability. If you have

it can be hard to spot the bug. But if you use

it's impossible to even have that bug, and a later reader doesn't have to scrutinize the terms to rule out the possibility.

A minor advantage is that it can save you a pair of parentheses (or from a bug if you leave those parentheses out). Consider:

Finally (thanks to Jens Gustedt for reminding me of this), we have to go back and think a little more carefully about what we meant when we said "Where it gets interesting is when the left-hand side of the assignment is an expression involving side effects." Normally, we think of modifications as being side effects, and accesses as being "free". But for variables qualified as volatile (or, in C11, as _Atomic ), an access counts as an interesting side effect, too. So if variable a has one of those qualifiers, a = a + b is not a "simple expression involving ordinary variables", and it may not be so identical to a += b , after all.

• Got the real issue now. I think most books does not highlight this concept of two time evaluation by giving good example. In fact this is the most important difference than the readability. @dbush also gave another good example where we can easily see the result of two time evaluation. Disadvantage indicated by chux also give us better clarity. –  Rajesh Commented Apr 19, 2018 at 14:58
• No, they are not always the same. It depends on what a is. If it is volatile or _Atomic the result is fundamentally different. –  Jens Gustedt Commented Apr 19, 2018 at 18:54
• How would tmp = *XYZ; be interpreted and why? my compiler interprets this as: Copy *XYZ to temp, and why not tmp = tmp * XYZ ? Note: XYZ is pointer. –  M Sharath Hegde Commented Aug 30, 2018 at 6:58
• 1 @MSharathHegde If the compound assignment operator were spelled " =* ", there might be some ambiguity here, but since it is in fact spelled " *= ", it's perfectly clear that tmp=*XYZ must involve a pointer access, not a multiplication. (Now, in the earliest days of C, the compound assignment operator was spelled " =* ", and there was ambiguity, which is why the compound assignment operators were rejiggered to their modern form.) –  Steve Summit Commented Aug 30, 2018 at 9:27

Evaluating the left side once can save you a lot if it's more than a simple variable name. For example:

In this case some_long_running_function() is only called once. This differs from:

Which calls the function twice.

There is a disadvantage too. Consider the effect of types.

10 + b addition below will use int math and overflow ( undefined behavior )

• even clang -Weverything don't produce warning... it's should ! –  Stargateur Commented Apr 19, 2018 at 14:42

This is what the standard 6.5.16.2 says:

A compound assignment of the form E1 op = E2 is equivalent to the simple assignment expression E1 = E1 op ( E2 ), except that the lvalue E1 is evaluated only once

So the "evaluated once" is the difference. This mostly matters in embedded systems where you have volatile qualifiers and don't want to read a hardware register several times, as that could cause unwanted side-effects.

That's not really possible to reproduce here on SO, so instead here's an artificial example to demonstrate why multiple evaluations could lead to different program behavior:

The simple assignment version did not only give a different result, it also introduced unspecified behavior in the code, so that two different results are possible depending on the compiler.

• correct. in addition to volatile it can also make a difference for atomic qualified types. –  Jens Gustedt Commented Apr 19, 2018 at 18:57

Not sure what you're after. Compound assignment is shorter, and therefore simpler (less complex) than using regular operations.

Consider this:

Which one is easier to read and understand, and verify?

This, to me, is a very very real advantage, and is just as true regardless of semantic details like how many times something is evaluated.

• Arguably the main issue with your example is that you don't follow the law of Demeter. The difference between x += dt * speed and x =x + dt * speed is less, but still worth having. –  Pete Kirkham Commented Apr 19, 2018 at 13:38
• I agree with this answer. Modern compilers usually do not need any "help" by programmers using specific language constructs. In fact today better readability of a code fragment usually also means better optimization by the compiler. –  Blue Commented Apr 19, 2018 at 13:41
• @unwind I am mainly talking about performance related issues rather than readability. According to Microsoft, "However, the compound-assignment expression is not equivalent to the expanded version because the compound-assignment expression evaluates expression1 only once, while the expanded version evaluates expression1 twice: in the addition operation and in the assignment operation". Here is what I am expecting some kind of explanation. –  Rajesh Commented Apr 19, 2018 at 13:49

A language like C is always going to be an abstraction of the underlying machine opcodes. In the case of addition, the compiler would first move the left operand into the accumulator, and add the right operand to it. Something like this (pseudo-assembler code):

This is what 1+2 would compile to in assembler. Obviously, this is perhaps over-simplified, but you get the idea.

Also, compiler tend to optimise your code, so exp1=exp1+b would very likely compile to the same opcodes as exp1+=b .

And, as @unwind remarked, the compound statement is a lot more readable.

• I am using TDM-GCC-64 compiler. Optimization is turned off. Here is the log 'gcc.exe -Wall -s -pedantic -Wextra -Wall -g -c "C:\sample_Project_Only_Main\main.c" -o Debug\main.o g++.exe -o Debug\sample_Project_Only_Main.exe Debug\main.o " –  Rajesh Commented Apr 19, 2018 at 13:54
• Fair enough. Still, assembly always works along the lines of "add something to a register or memory location", which is what the += operator does. –  SeverityOne Commented Apr 19, 2018 at 17:03

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• Why is Legion helping you whilst geth are fighting you?
• Why did the Fallschirmjäger have such terrible parachutes?
• Reusing own code at work without losing licence
• Can Shatter damage Manifest Mind?
• Motion of the COM of 2-body system
• Why does a halfing's racial trait lucky specify you must use the next roll?
• How does the summoned monster know who is my enemy?
• Cannot open and HTML file stored on RAM-disk with a browser
• How can moral disagreements be resolved when the conflicting parties are guided by fundamentally different value systems?
• Is it possible to accurately describe something without describing the rest of the universe?
• Which hash algorithms support binary input of arbitrary bit length?
• Which version of Netscape is this, on which OS?