Pre & post increment operator behavior in C, C++, Java, & C#

c# java c++ c

16,813

Solution 1

Java and C# evaluate expressions from left to right, and the side-effects are visible immediately.

In C++, the order of evaluation of subexpressions is unspecified, and modifying the same object twice without an intervening sequence point is undefined behavior.

Solution 2

I don't have the time to write up a detailed description of the differences between C++, C, C# and Java. I will merely say that the C# behaviour of the pre and post increment operators is fully specified (in single-threaded scenarios; if you want to know about its atomicity, guarantees about observations of read and write orders in multi-processor weak memory models and so on, you're on your own to do that research.) It is not fully specified in C and C++; a compiler has broad lattitude to do whatever it pleases with re-ordering side effects. I have never used Java so I'm not going to hazard a guess as to what Java does.

For more information on what C# does you should read the C# specification. For a short take on it, read my answer to this question:

What is the difference between i++ and ++i?

For an even shorter take:

Subexpressions in a C# expression are logically grouped by precedence and associativity, and then evaluated from left to right regardless. (So for example, A() + B() * C() evaluates A(), then B(), then C(). The fact that the multiplication "comes before" the addition is irrelevant; the subexpressions are always evaluated left to right.)

If the evaluation of a subexpression causes a side effect because of a pre or post increment subexpression then the side effect happens immediately before the result is produced.

Solution 3

In C++, this is undefined behaviour, so any answer would be correct. See Undefined behavior and sequence points for further details.

Not sure about other languages, but I would expect this code to be incorrect there, too.

EDIT:
See Eric Lippert's answer about C#. He disproves my assumption about C#'s behaviour.

Solution 4

In C++ at least this undefined behaviour. Quoting the C++ standard:

Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression.

Solution 5

The Java memory model ensures the order the loads and stores are done, so it should come out the same on any JVM (I believe).

It looks like C++ has the same order of operations, but once you use it twice on a line you start running into other stuff (Vlad is right there). If you try other C++ compilers, you may find they come up with different answers.

I'm sure C# has the same order of operations, but my guess is they have a memory model (like Java) that ensures consistency, but I don't have a lot of knowledge here.

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Author by

Nick

please delete me

Updated on July 05, 2022

Comments

Nick almost 2 years

DISCLAIMER: This is not a real-world example. It is just a theoretical question of how these languages work.

What exactly are the differences between C/C++, C#, and Java when it comes to post & pre increment operators?

This is what I get with VC++10, Java 1.6, and C# 4

int a = 2;
int b = a++ + a++;
int c = ++a + a++ + a++;

      +-----+------+------+----+
      |  C  | C++  | Java | C# |
+-----+-----+------+------+----+
| a   |  7  |  7   |   7  |  7 |
+-----+-----+------+------+----+
| b   |  4  |  4   |   5  | 5  |
+-----+-----+------+------+----+
| c   | 15  |  15  |  16  | 16 |
+-----+-----+------+------+----+

Kirill V. Lyadvinsky almost 13 years

I'd say: any answer would be incorrect.
Martin York almost 13 years

Its well defined in the other two languages. Just not defined in C++
Vlad almost 13 years

@Kirill: as I understand this, the undefined behaviour means that anything the program does is correct (including formatting the hard disk and launching nasal demons), from the compiler point of view. The fact that such a behaviour is usually not what the code author expects is a completely unrelated issue and not compiler's fault.
Vlad almost 13 years

@Martin: could you please point to some documentation defining the behaviour in Java and C#? Just curious.
Blindy almost 13 years

Exactly, the people saying it's undefined in C# and Java are plain wrong.
Nick almost 13 years

Sounds like you're the go-to man when it comes to pre & post increment operators :)
fredoverflow almost 13 years

The Java memory model is concerned with concurrency. I'm pretty sure it does not apply here.
Martin York almost 13 years

@Vlad: stackoverflow.com/questions/6457130/… I could be wrong about Java. But both languages designers saw all the silly question from C++ noob and said lets not get distracted by stupid questions and just define the behavior. :-)
fredoverflow almost 13 years

As soon as I my eyes saw "Eric Lippert", my finger clicked on "upvote" :)
Nick almost 13 years

+1 Well -- this sound like a great short answer. Unless it is a hassle, I would appreciate some references
Jedihomer Townend almost 13 years

That's what I get for typing out of memory. You're right, it's specified elsewhere.
Eric Lippert almost 13 years

What does memory model have to do with it? This isn't a question about multithreading.
Vlad almost 13 years

@Martin: I've already discovered the answer about C# (see my modified answer). Maybe you could point to the explanation about Java, too?
Eric Lippert almost 13 years

@Vlad: subexpressions are evaluated left to right, period, end of story. (Full disclosure: C# 4 has some bizarre bugs where sometimes we accidentally evaluate subexpressions in the wrong order when the named-and-optional-parameter call analyzer gets confused; those bugs will be fixed in later releases.)
Nick almost 13 years

References or no references, this sums it all up and is the accepted answer
Vlad almost 13 years

@Eric: Actually, I removed my question, because the grouping would logically not affect anyway the evaluation order, so the question doesn't make sense. In any case, thanks for the answer!
John Bode almost 13 years

@Nick: For C, see N1256 (PDF), sect 6.5, para 2; sect 6.5.2.4, para 2. I can't get to the C++ standard currently, but the language is largely the same.
Rasmus Faber almost 13 years

@Nick: Java specifies this in JLS §15.7 "Evaluation Order": "The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right."
Nick almost 13 years

@John Thanks a bunch! C behavior was the missing link :)
Nick almost 13 years

@Ramus this is great, right on the money!
matth almost 13 years

For C++, it is ISO/IEC 14882:2003(E) §5, ¶4.
configurator almost 13 years

You've never used Java? You really should try it sometime; you learn so much about how not to design a programming language...
Eric Lippert almost 13 years

@configurator: We have two former Java language guys on the C# design team, so I think we've got that covered. :-)
Daniel Fischer about 11 years

int b= a++ + a++; is undefined behaviour in C and C++, since a is modified twice without intervening sequence point (ditto for the int c = ... thing).
commit about 11 years

Yes you are right, I had also read that but I am just confuse and want to know how its behavior is undefined? Because in C/C++ if you also evaluate int a=2; and int b = a++ + a++ + a++ + a++; b's output will be same 8, it will take value and then do post increment of that operand after finishing of that statement. I hope you will get what I am trying to say.
Daniel Fischer about 11 years

Undefined means there's no restriction on what happens by the standard. It would not violate the standard if the compiler inserted an abort() call for the undefined behaviour. But of course no serious implementation does such things. Variations closer to the code include different orders of evaluation (right-to-left, middle-to-outer, invent-your-own) and the question of when the incremented value is stored, whether a is read only once or several times. So it would be legal for the value of a to be read three times, and the incremented value of a stored after evaluating the expression.
Daniel Fischer about 11 years

That could lead to int c = ++a + a++ + a++; being evaluated as 5 + 4 + 4, and then three times storing 4+1 into a, thus c == 13 and a == 5 afterward. Or the middle term evaluated first, and the incremented value not stored yet, then the ++a with immediate storing of the incremented value, then the right a++ with a new reading of the value, leading to c = 5 + 4 + 5 and a could be 5 or 6, depending on which pending store from the two a++ is done last.