ide0054 c# use compound assignment

Thanks for all you work!

jnm2 2020-12-09

@HugCoder Perhaps this is fixed by the fix for #49294 which will be in 16.9.P2?

HugCoder 2020-12-09

@jnm2 I did find that issue as well, but it wasn't targetting the null operator mentioned here. It certainly could be addressing the issue of the null-coalescing assignment operator as well but I'm afraid it's above my league in this topic to say if it does. I'm not sure how specific the fixes for these kind of issues are and just wanted to make sure it wasn't overlooked and viewed as solved in VS 16.8.

This issue is about IDE0074 incorrectly recommending a ??= b to replace a ?? (a = b) for langversion < 8 and was fixed in 16.7.P4. The same concern can't apply to other kinds of compound assignment because they are legal in langversion < 8.

#49294 is about IDE0054 firing incorrectly in a C# 9 target-typed new initializer and will be fixed in 16.9.P2.

Is your concern that IDE0054 fired incorrectly in a C# 9 target-typed new initializer, specifically recommending ??= ? Or is there a new element involved such as with expressions, or am I misunderstanding? If I am, it would really help me to see a sample line of code.

Incidentally, 16.9.P2 just dropped in the preview channel, so we'll be able to test your concern about whether it was fixed for sure in that pull request in a minute or two :D

That's very close to the example I have at the top of this issue which IIRC didn't repro for 16.8.2 for me earlier today with <LangVersion>7.3</LangVersion> in my test csproj.

In any case, 16.8.3 doesn't repro this which matches your screenshot:

<PropertyGroup> <TargetFramework>net461</TargetFramework> </PropertyGroup>

< PropertyGroup > < TargetFramework >net461</ TargetFramework > </ PropertyGroup >

This makes me think there's something going on with your project that can't be seen from your description so far. E.g. are you multitargeting?

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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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Enable IDE0054 (Use compound assignment) #70564

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stephentoub commented Jun 10, 2022

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ghost commented Jun 10, 2022

tannergooding Jun 16, 2022

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Unrelated to this PR, but this entire class should be removed and we should just use System.Numerics.BitOperations

tannergooding left a comment

Again nice to see all this code get simpler and easier to read 👍

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複合代入を使用する (IDE0054 および IDE0074)

この記事では、 IDE0054 と IDE0074 という 2 つの関連するルールについて説明します。

これらの規則は、 複合代入 の使用に関するものです。 結合複合代入には IDE0074 が、その他の複合代入には IDE0054 がレポートされます。

オプション値によって、複合代入が必要かどうかを指定します。

オプションの構成の詳細については、「 オプションの書式 」を参照してください。

dotnet_style_prefer_compound_assignment

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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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