#include <stdio.h>

int main(void)
{
   int i = 0;
   i = i++ + ++i;
   printf("%d\n", i); // 3

   i = 1;
   i = (i++);
   printf("%d\n", i); // 2 Should be 1, no ?

   volatile int u = 0;
   u = u++ + ++u;
   printf("%d\n", u); // 1

   u = 1;
   u = (u++);
   printf("%d\n", u); // 2 Should also be one, no ?

   register int v = 0;
   v = v++ + ++v;
   printf("%d\n", v); // 3 (Should be the same as u ?)

   int w = 0;
   printf("%d %d %d\n", w++, ++w, w); // shouldn't this print 0 2 2

   int x[2] = { 5, 8 }, y = 0;
   x[y] = y ++;
   printf("%d %d\n", x[0], x[1]); // shouldn't this print 0 8? or 5 0?
}
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Homework? Not trying to be a pain, but you should never write code with expressions like these. They are usually given as academic examples, sometimes showing that different compilers yield different output. – Jarrett Meyer
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@Jarett, nope, just needed some pointers to "sequence points". While working I found a piece of code with i = i++, I thougth "This isn't modifying the value of i". I tested and I wondered why. Since, i've removed this statment and replaced it by i++; – PiX
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Explain these undefined behaviors? Explain what about them? How they behave is undefined. – Jesse Millikan
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I think it's interesting that everyone ALWAYS assumes that questions like this are asked because the asker wants to USE the construct in question. My first assumption was that PiX knows that these are bad, but is curious why the behave they way the do on whataver compiler s/he was using... And yeah, what unWind said... it's undefined, it could do anything... including JCF (Jump and Catch Fire) – Brian Postow
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I'm curious: Why don't compilers seem to warn on constructs such as "u = u++ + ++u;" if the result is undefined? – Learn OpenGL ES
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Why would you expect i = (i++) be equal to 1? Parentheses override the natural order of evaluation precedence, so anything within them will happen first. So i++ will happen first (naturally, i++ would happen after assignment), making it 2. 2 would then be assigned back to i. i is 2. – swampf0etus
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(i++) still evaluates to 1, regardless of parentheses – Drew McGowen
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Whatever i = (i++); was intended to do, there is certainly a clearer way to write it. That would be true even if it were well defined. Even in Java, which defines the behavior of i = (i++);, it's still bad code. Just write i++; – Keith Thompson
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Just my cents: such statements are undefined behavior, because you read and write the same variable (memory spot). This allows compiler to do magic called "optimization" i.e. clean up your own mess. Naturally this comes with some limitations. Reading from memory is SLOOOW, therefore registers are used and then synced with real memory. Side effect is that compiler is now unsure which value to use: the one from memory or the one in register that has already been modified. (I like to explain it like that, makes most sense) – friendzis
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@LearnOpenGLES: They do. – Lightness Races in Orbit
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I have written an article about identifying undefined behaviour in expressions which covers many similar examples, but is defined in terms of the C++11 sequencing rules. Perhaps some of the readers here will find it useful. – Joseph Mansfield
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It reminds me of the interview questions asked by many software firms in India. Though the behaviour is undefined they still try to impose logic on the output. Also similar questions are mentioned in many C books of Yashwant Kanetkar. This kind of questions really make me sick :( – i_am_zero
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@LearnOpenGLES: My compiler(gcc 4.8.1) warns me on constructs like u=u++ & j=i++ + ++i; – Destructor
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Although this question is about C, it may be of interest to some aspects related to this is going to change with the next version of C++, with the voted in C++17 evaluation order guarantees (P0145R2) More: //allinonescript.com/questions/38501587/… – Johan Lundberg
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As mentioned in some comments, C / C++ don't have explicit rules on evaluation order. Some other languages do, in which case, this would not be an issue.The most unusual case is APL (A Programming Language) that evaluates expressions right to left (which allows for multiple assignments on a single line), with parenthesis used to override the order of evaluation. – rcgldr
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C compiler ouput different than Java compiler: int i=5; System.out.printf(",%d,%d,%d,%d,%d",i++,i--,++i,--i,i); gcc 5.3.0: Output: 4,5,5,5,5 Java1.8 Output: 5,6,6,5,5 – Akhilesh Dubey
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@i_am_zero: The fact that the Standard does not mandate a behavior in some situation does not mean that no implementations will specify how they process code in sufficient detail that only one possible behavior would be consistent with the spec. One problem with the Standard is that it has never attempted to catalog all the cases where an implementation would have to go out of its way not to behave in predictable fashion (e.g. using memcpy in cases where the source and destination might occasionally be equal, e.g. because the cost of an occasional redundant copy would be less than... – supercat
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...the cost of checking on every operation whether the copy was necessary). IMHO, the Standard would be better if it specified a basic execution model and then kinds of optimizations that programmers may enable. Given x=(*p)++ + (*q)++; b=*p; c=*p;, for example, it may be reasonable to say that with some optimizations enabled a compiler could at its option independently treat b and c as either holding the one plus the value that was read before the increment of *p, or as holding a value which is read from *p at any time between the increment and the assignment to b or c. – supercat
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@i_am_zero: Such rules would give compilers almost all of the useful flexibility they have under the present standard, but if combined with ways of converting indeterminate values to arbitrary values could allow some kinds of code to be written more efficiently than is currently possible. – supercat

13 Answers 11

up vote 491 down vote accepted

C has the concept of undefined behavior, i.e. some language constructs are syntactically valid but you can't predict the behavior when the code is run.

As far as I know, the standard doesn't explicitly say why the concept of undefined behavior exists. In my mind, it's simply because the language designers wanted there to be some leeway in the semantics, instead of i.e. requiring that all implementations handle integer overflow in the exact same way, which would very likely impose serious performance costs, they just left the behavior undefined so that if you write code that causes integer overflow, anything can happen.

So, with that in mind, why are these "issues"? The language clearly says that certain things lead to undefined behavior. There is no problem, there is no "should" involved. If the undefined behavior changes when one of the involved variables is declared volatile, that doesn't prove or change anything. It is undefined; you cannot reason about the behavior.

Your most interesting-looking example, the one with

u = (u++);

is a text-book example of undefined behavior (see Wikipedia's entry on sequence points).

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I knew it was undefined, (The idea of seing this code in production frighten me :)) but I tried to understand what was the reason for these results. Especially why u = u++ incremented u. In java for example: u = u++ returns 0 as (my brain) expected :) Thanks for the sequence points links BTW. – PiX
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Obviously because of the brackets around the u++ the compiler has decided to incerement u and then return it. As it is undefined behaviuor in C this is ligitimate. A different compiler or even a different machine and the same one may give a different answer. I do not know java, but perhaps the behaviour is clearly defined. – ChrisBD
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@PiX: Things are undefined for a number of possible reasons. These include: there is no clear "right result", different machine architectures would strongly favour different results, existing practice is not consistent, or beyond the scope of the standard (e.g. what filenames are valid). – Richard
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@PiX Java goes out of its way to have defined behaviors for many things that are undefined in C. – Laurence Gonsalves
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@PaulManta, If you see this, editing answers is not intended for the addition of irrelevant information to already-accepted answers. This is a C question and the answer was fine as it was to describe the situation in C standards from C90 to C11. Editing is for fixing syntax and style. – Pascal Cuoq
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The spirit of C: Trust the programmer... no matter how insane he is. – Fiddling Bits
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unwind you called it undefined behaviour, but is there any explanation for why it is so? – user3124504
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@rusty Not sure what you mean. The term "undefined behavior" is used in the C standard. It means that even though some constructs are syntactically valid and will typically compile, they lead to undefine behavior i.e. they do not make sense and should be avoided since your program is broken if it has undefined behavior. – unwind
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Just to confuse everyone, some such examples are now well-defined in C11, e.g. i = ++i + 1; . – M.M
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@MattMcNabb that is only well defined in C++11 not in C11. – Shafik Yaghmour
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I've edited the question to add the UB in evaluation of function arguments, as this question is often used as a duplicate for that. (The last example) – Antti Haapala
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Reading the Standard and the published rationale, It's clear why the concept of UB exists. The Standard was never intended to fully describe everything a C implementation must do to be suitable for any particular purpose (see the discussion of the "One Program" rule), but instead relies upon implementors' judgment and desire to produce useful quality implementations. A quality implementation suitable for low-level systems programming will need to define the behavior of actions that wouldn't be needed in high-end number crunching.applications. Rather than try to complicate the Standard... – supercat
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...by getting into extreme detail about which corner cases are or are not defined, the authors of the Standard recognized that implementors should be better paced to judge which kinds of behaviors will be needed by the kinds of programs they're expected to support. Hyper-modernist compilers pretend that making certain actions UB was intended to imply that no quality program should need them, but the Standard and rationale are inconsistent with such a supposed intent. – supercat
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@supercat Well put, I'd recommend adding that to your answer. – jrh
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@jrh: I wrote that answer before I'd realized how out of hand the hyper-modernist philosophy had gotten. What irks me is the progression from "We don't need to officially recognize this behavior because the platforms where it's needed can support it anyway" to "We can remove this behavior without providing a usable replacement because it was never recognized and thus any code needing it was broken". Many behaviors should have been deprecated long ago in favor of replacements that were in every way better, but that would have required acknowledging their legitimacy. – supercat

I think the relevant parts of the C99 standard are 6.5 Expressions, §2

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

and 6.5.16 Assignment operators, §4:

The order of evaluation of the operands is unspecified. If an attempt is made to modify the result of an assignment operator or to access it after the next sequence point, the behavior is undefined.

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Would the above imply that 'i=i=5;" would be Undefined Behavior? – supercat
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@supercat as far as I know i=i=5 is also undefined behavior – dhein
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@Zaibis: The rationale I like to use for most places rule applies that in theory a mutli-processor platform could implement something like A=B=5; as "Write-lock A; Write-Lock B; Store 5 to A; store 5 to B; Unlock B; Unock A;", and a statement like C=A+B; as "Read-lock A; Read-lock B; Compute A+B; Unlock A and B; Write-lock C; Store result; Unlock C;". That would ensure that if one thread did A=B=5; while another did C=A+B; the latter thread would either see both writes as having taken place or neither. Potentially a useful guarantee. If one thread did I=I=5;, however, ... – supercat
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... and the compiler didn't notice that both writes were to the same location (if one or both lvalues involve pointers, that may be hard to determine), the generated code could deadlock. I don't think any real-world implementations implement such locking as part of their normal behavior, but it would be permissible under the standard, and if hardware could implement such behaviors cheaply it might be useful. On today's hardware such behavior would be way too expensive to implement as a default, but that doesn't mean it would always be thus. – supercat
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@supercat but wouldn't the sequence point access rule of c99 alone be enough to declare it as undefined behavior? So it doesn't matter what technically the hardware could implement? – dhein
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@Zaibis: Rules which characterize actions as Undefined Behavior aren't supposed to exist merely to allow implementations to behave in hostile fashion. They're supposed to exist to allow implementers to either do something more efficiently or more usefully than would be possible in their absence. To understand why the specs characterize something as UB, it's helpful to identify something useful the rule would allow implementations to do which they otherwise could not. – supercat
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@supercat I absolutly agree to that what you say about the behavior of undefined behavior(^^). But this doesn't change the point that if something is in the standard listed as UB you can expect, it is well defined just because it would be easy to implement as well defined construct. If the standard says it is UB, then the answer to the question is it UB? is "Yes!", and not "It could... [...]". – dhein
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@Zaibis: The answer to almost any question of the form "Why is X in language/framework Y Undefined Behavior" is "Because that's what the standard for Y says", but that's hardly enlightening. In most cases, however, what someone asking such a question really wants to know is "Why did the makers of the standard specify that". In most cases, things are specified as UB (rather than partially-specified behaviors) to allow for the possibility of an implementation which might do something unexpected. For example, the spec could have said that p1=malloc(4); p2=malloc(5); r=p1>p2;... – supercat
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...may result in r arbitrarily holding 1 or 0, with no guarantee that the value will relate in any way to future comparisons among the same or different operands. Such a spec (returning an arbitrary 0 or 1) would have allowed an efficient memmove to be written in portable fashion [if dest > src, apply a top-down copy, else bottom-up; if the regions don't overlap, either will work so the comparison result wouldn't matter]. I believe the standard says such comparison is UB, however; if every machine could easily--at worst--arbitrarily yield a 0 or 1, there'd be no reason not to say so. – supercat

Just compile and disassemble your line of code, if you are so inclined to know how exactly it is you get what you are getting.

This is what I get on my machine, together with what I think is going on:

$ cat evil.c
void evil(){
  int i = 0;
  i+= i++ + ++i;
}
$ gcc evil.c -c -o evil.bin
$ gdb evil.bin
(gdb) disassemble evil
Dump of assembler code for function evil:
   0x00000000 <+0>:   push   %ebp
   0x00000001 <+1>:   mov    %esp,%ebp
   0x00000003 <+3>:   sub    $0x10,%esp
   0x00000006 <+6>:   movl   $0x0,-0x4(%ebp)  // i = 0   i = 0
   0x0000000d <+13>:  addl   $0x1,-0x4(%ebp)  // i++     i = 1
   0x00000011 <+17>:  mov    -0x4(%ebp),%eax  // j = i   i = 1  j = 1
   0x00000014 <+20>:  add    %eax,%eax        // j += j  i = 1  j = 2
   0x00000016 <+22>:  add    %eax,-0x4(%ebp)  // i += j  i = 3
   0x00000019 <+25>:  addl   $0x1,-0x4(%ebp)  // i++     i = 4
   0x0000001d <+29>:  leave  
   0x0000001e <+30>:  ret
End of assembler dump.

(I... suppose that the 0x00000014 instruction was some kind of compiler optimization?)

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how do i get the machine code? I use Dev C++, and i played around with 'Code Generation' option in compiler settings, but go no extra file output or any console output – bad_keypoints
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@ronnieaka gcc evil.c -c -o evil.bin and gdb evil.bindisassemble evil, or whatever the Windows equivalents of those are :) – badp
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is -0x4(%ebp) = 4 at the end? – kchoi
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This answer does not really address the question of Why are these constructs undefined behavior?. – Shafik Yaghmour
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@ShafikYaghmour I'm addressing the questions in the question body ("why am I not getting the results I am getting?"), see the comments in the code. Given that this is undefined behaviour, I can only show how to get the actual assembly he's compiled. – badp
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Perhaps the answer is in there but I think most would not be able to figure it out without some elaboration. Just add some explanatory text and it becomes an answer. – Shafik Yaghmour
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@ShafikYaghmour I must admit that the assembly is kinda baffling me; especially the instruction at +20. But why am I trying to make sense of it? – badp
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As an aside, it'll be easier to compile to assembly (with gcc -S evil.c), which is all that's needed here. Assembling then disassembling it is just a roundabout way of doing it. – Kat
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For the record, if for whatever reason you're wondering what a given construct does -- and especially if there's any suspicion that it might be undefined behavior -- the age-old advice of "just try it with your compiler and see" is potentially quite perilous. You will learn, at best, what it does under this version of your compiler, under these circumstances, today. You will not learn much if anything about what it's guaranteed to do. In general, "just try it with your compiler" leads to nonportable programs that work only with your compiler. – Steve Summit

While it is unlikely that any compilers and processors would actually do so, it would be legal, under the C standard, for the compiler to implement "i++" with the sequence:

In a single operation, read `i` and lock it to prevent access until further notice
Compute (1+read_value)
In a single operation, unlock `i` and store the computed value

While I don't think any processors support the hardware to allow such a thing to be done efficiently, one can easily imagine situations where such behavior would make multi-threaded code easier (e.g. it would guarantee that if two threads try to perform the above sequence simultaneously, i would get incremented by two) and it's not totally inconceivable that some future processor might provide a feature something like that.

If the compiler were to write i++ as indicated above (legal under the standard) and were to intersperse the above instructions throughout the evaluation of the overall expression (also legal), and if it didn't happen to notice that one of the other instructions happened to access i, it would be possible (and legal) for the compiler to generate a sequence of instructions that would deadlock. To be sure, a compiler would almost certainly detect the problem in the case where the same variable i is used in both places, but if a routine accepts references to two pointers p and q, and uses (*p) and (*q) in the above expression (rather than using i twice) the compiler would not be required to recognize or avoid the deadlock that would occur if the same object's address were passed for both p and q.

The behavior can't really be explained because it invokes both unspecified behavior and undefined behavior, so we can not make any general predictions about this code, although if you read Olve Maudal's work such as Deep C and Unspecified and Undefined sometimes you can make good guesses in very specific cases with a specific compiler and environment but please don't do that anywhere near production.

So moving on to unspecified behavior, in draft c99 standard section6.5 paragraph 3 says(emphasis mine):

The grouping of operators and operands is indicated by the syntax.74) Except as specified later (for the function-call (), &&, ||, ?:, and comma operators), the order of evaluation of subexpressions and the order in which side effects take place are both unspecified.

So when we have a line like this:

i = i++ + ++i;

we do not know whether i++ or ++i will be evaluated first. This is mainly to give the compiler better options for optimization.

We also have undefined behavior here as well since the program is modifying variables(i, u, etc..) more than once between sequence points. From draft standard section 6.5 paragraph 2(emphasis mine):

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

it cites the following code examples as being undefined:

i = ++i + 1;
a[i++] = i; 

In all these examples the code is attempting to modify an object more than once in the same sequence point, which will end with the ; in each one of these cases:

i = i++ + ++i;
^   ^       ^

i = (i++);
^    ^

u = u++ + ++u;
^   ^       ^

u = (u++);
^    ^

v = v++ + ++v;
^   ^       ^

Unspecified behavior is defined in the draft c99 standard in section 3.4.4 as:

use of an unspecified value, or other behavior where this International Standard provides two or more possibilities and imposes no further requirements on which is chosen in any instance

and undefined behavior is defined in section 3.4.3 as:

behavior, upon use of a nonportable or erroneous program construct or of erroneous data, for which this International Standard imposes no requirements

and notes that:

Possible undefined behavior ranges from ignoring the situation completely with unpredictable results, to behaving during translation or program execution in a documented manner characteristic of the environment (with or without the issuance of a diagnostic message), to terminating a translation or execution (with the issuance of a diagnostic message).

The C standard says that a variable should only be assigned at most once between two sequence points. A semi-colon for instance is a sequence point.
So every statement of the form:

i = i++;
i = i++ + ++i;

and so on violate that rule. The standard also says that behavior is undefined and not unspecified. Some compilers do detect these and produce some result but this is not per standard.

However, two different variables can be incremented between two sequence points.

while(*src++ = *dst++);

The above is a common coding practice while copying/analysing strings.

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Of course it doesn't apply to different variables within one expression. It would be a total design failure if it did! All you need in the 2nd example is for both to be incremented between the statement ending and the next one beginning, and that's guaranteed, precisely because of the concept of sequence points at the centre of all this. – underscore_d

In //allinonescript.com/questions/29505280/incrementing-array-index-in-c someone asked about a statement like:

int k[] = {0,1,2,3,4,5,6,7,8,9,10};
int i = 0;
int num;
num = k[++i+k[++i]] + k[++i];
printf("%d", num);

which prints 7... the OP expected it to print 6.

The ++i increments aren't guaranteed to all complete before the rest of the calculations. In fact, different compilers will get different results here. In the example you provided, the first 2 ++i executed, then the values of k[] were read, then the last ++i then k[].

num = k[i+1]+k[i+2] + k[i+3];
i += 3

Modern compilers will optimize this very well. In fact, possibly better than the code you originally wrote (assuming it had worked the way you had hoped).

Another way of answering this, rather than getting bogged down in arcane details of sequence points and undefined behavior, is simply to ask, what are they supposed to mean? What was the programmer trying to do?

The first fragment asked about, i = i++ + ++i, is pretty clearly insane in my book. No one would ever write it in a real program, it's not obvious what it does, there's no conceivable algorithm someone could have been trying to code that would have resulted in this particular contrived sequence of operations. And since it's not obvious to you and me what it's supposed to do, it's fine in my book if the compiler can't figure out what it's supposed to do, either.

The second fragment, i = i++, is a little easier to understand. Someone is clearly trying to increment i, and assign the result back to i. But there are a couple ways of doing this in C. The most basic way to add 1 to i, and assign the result back to i, is the same in almost any programming language:

i = i + 1

C, of course, has a handy shortcut:

i++

This means, "add 1 to i, and assign the result back to i". So if we construct a hodgepodge of the two, by writing

i = i++

what we're really saying is "add 1 to i, and assign the result back to i, and assign the result back to i". We're confused, so it doesn't bother me too much if the compiler gets confused, too.

Realistically, the only time these crazy expressions get written is when people are using them as artificial examples of how ++ is supposed to work. And of course it is important to understand how ++ works. But one practical rule for using ++ is, "If it's not obvious what an expression using ++ means, don't write it."

We used to spend countless hours on comp.lang.c discussing expressions like these and why they're undefined. Two of my longer answers, that try to really explain why, are on the web at http://www.eskimo.com/~scs/readings/undef.950321.html and http://www.eskimo.com/~scs/readings/precvsooe.960725.html .

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A rather nasty gotcha with regard to Undefined Behavior is that while it used to be safe on 99.9% of compilers to use *p=(*q)++; to mean if (p!=q) *p=(*q)++; else *p= __ARBITRARY_VALUE; that is no longer the case. Hyper-modern C would require writing something like the latter formulation (though there's no standard way of indicating code doesn't care what's in *p) to achieve the level of efficiency compilers used to provide with the former (the else clause is necessary in order to let the compiler optimize out the if which some newer compilers would require). – supercat
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I've seen at least 5 similar questions about these ++ and -- madness last week or so. These seem to be some professors' favorite topic to puzzle their students.. – artm

Most of the answers here quoted from C standard emphasizing that the behavior of these constructs are undefined. To understand why the behavior of these constructs are undefined, let's understand these terms first in the light of C11 standard:

Sequenced: (5.1.2.3)

Given any two evaluations A and B, if A is sequenced before B, then the execution of A shall precede the execution of B.

Unsequenced:

If A is not sequenced before or after B, then A and B are unsequenced.

Evaluations can be one of two things:

  • value computations, which work out the result of an expression; and
  • side effects, which are modifications of objects.

Sequence Point:

The presence of a sequence point between the evaluation of expressions A and B implies that every value computation and side effect associated with A is sequenced before every value computation and side effect associated with B.

Now coming to the question, for the expressions like

int i = 1;
i = i++;

standard says that:

6.5 Expressions:

If a side effect on a scalar object is unsequenced relative to either a different side effect on the same scalar object or a value computation using the value of the same scalar object, the behavior is undefined. [...]

Therefore, the above expression invokes UB because two side effects on the same object i is unsequenced relative to each other. That means it is not sequenced whether the side effect by assignment to i will be done before or after the side effect by ++.
Depending on whether assignment occurs before or after the increment, different results will be produced and that's the one of the case of undefined behavior.

Lets rename the i at left of assignment be il and at the right of assignment (in the expression i++) be ir, then the expression be like

il = ir++     // Note that suffix l and r are used for the sake of clarity.
              // Both il and ir represents the same object.  

An important point regarding Postfix ++ operator is that:

just because the ++ comes after the variable does not mean that the increment happens late. The increment can happen as early as the compiler likes as long as the compiler ensures that the original value is used.

It means the expression il = ir++ could be evaluated either as

temp = ir;      // i = 1
ir = ir + 1;    // i = 2   side effect by ++ before assignment
il = temp;      // i = 1   result is 1  

or

temp = ir;      // i = 1
il = temp;      // i = 1   side effect by assignment before ++
ir = ir + 1;    // i = 2   result is 2  

resulting in two different results 1 and 2 which depends on the sequence of side effects by assignment and ++ and hence invokes UB.

Often this question is linked as a duplicate of questions related to code like

printf("%d %d\n", i, i++);

or

printf("%d %d\n", ++i, i++);

or similar variants.

While this is also undefined behaviour as stated already, there are subtle differences when printf() is involved when comparing to a statement such as:

   x = i++ + i++;

In the following statement:

printf("%d %d\n", ++i, i++);

the order of evaluation of arguments in printf() is unspecified. That means, expressions i++ and ++i could be evaluated in any order. C11 standard has some relevant descriptions on this:

Annex J, unspecified behaviours

The order in which the function designator, arguments, and subexpressions within the arguments are evaluated in a function call (6.5.2.2).

3.4.4, unspecified behavior

Use of an unspecified value, or other behavior where this International Standard provides two or more possibilities and imposes no further requirements on which is chosen in any instance.

EXAMPLE An example of unspecified behavior is the order in which the arguments to a function are evaluated.

The unspecified behaviour itself is NOT an issue. Consider this example:

printf("%d %d\n", ++x, y++);

This too has unspecified behaviour because the order of evaluation of ++x and y++ is unspecified. But it's perfectly legal and valid statement. There's no undefined behaviour in this statement. Because the modifications (++x and y++) are done to distinct objects.

What renders the following statement

printf("%d %d\n", ++i, i++);

as undefined behaviour is the fact that these two expressions modify the same object i without an intervening sequence point.


Another detail is that the comma involved in the printf() call is a separator, not the comma operator.

This is an important distinction because the comma operator does introduce a sequence point between the evaluation of their operands, which makes the following legal:

int i = 5;
int j;

j = (++i, i++);  // No undefined behaviour here because the comma operator 
                 // introduces a sequence point between '++i' and 'i++'

printf("i=%d j=%d\n",i, j); // prints: i=7 j=6

The comma operator evaluates its operands left-to-right and yields only the value of the last operand. So in j = (++i, i++);, ++i increments i to 6 and i++ yields old value of i (6) which is assigned to j. Then i becomes 7 due to post-increment.

So if the comma in the function call were to be a comma operator then

printf("%d %d\n", ++i, i++);

will not be a problem. But it invokes undefined behaviour because the comma here is a separator.


For those who are new to undefined behaviour would benefit from reading What Every C Programmer Should Know About Undefined Behavior to understand the concept and many other variants of undefined behaviour in C.

This post: Undefined, unspecified and implementation-defined behavior is also relevant.

While the syntax of C allows such constructs, the behaviour of the program is undefined, because a shall in C standard is not obeyed. C99 6.5p2:

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. [72] Furthermore, the prior value shall be read only to determine the value to be stored [73]

With footnote 73 further clarifying that

This paragraph renders undefined statement expressions such as

i = ++i + 1;
a[i++] = i;

while allowing

i = i + 1;
a[i] = i;

You can detect such errors in a program by for example using a recent version of GCC with -Wall and -Werror, and then GCC will outright refuse to compile your program. The following is the output of gcc (Ubuntu 6.2.0-5ubuntu12) 6.2.0 20161005:

% gcc plusplus.c -Wall -Werror -pedantic
plusplus.c: In function ‘main’:
plusplus.c:6:6: error: operation on ‘i’ may be undefined [-Werror=sequence-point]
    i = i++ + ++i;
    ~~^~~~~~~~~~~
plusplus.c:6:6: error: operation on ‘i’ may be undefined [-Werror=sequence-point]
plusplus.c:10:6: error: operation on ‘i’ may be undefined [-Werror=sequence-point]
    i = (i++);
    ~~^~~~~~~
plusplus.c:14:6: error: operation on ‘u’ may be undefined [-Werror=sequence-point]
    u = u++ + ++u;
    ~~^~~~~~~~~~~
plusplus.c:14:6: error: operation on ‘u’ may be undefined [-Werror=sequence-point]
plusplus.c:18:6: error: operation on ‘u’ may be undefined [-Werror=sequence-point]
    u = (u++);
    ~~^~~~~~~
plusplus.c:22:6: error: operation on ‘v’ may be undefined [-Werror=sequence-point]
    v = v++ + ++v;
    ~~^~~~~~~~~~~
plusplus.c:22:6: error: operation on ‘v’ may be undefined [-Werror=sequence-point]
cc1: all warnings being treated as errors

The important part is to know what a sequence point is -- and what is a sequence point and what isn't. For example the comma operator is a sequence point, so

j = (i ++, ++ i);

is well-defined, and will increment i by one yield the old value, discard that value; then at comma operator, settle the side effects; and then increment i by one, and the resulting value becomes the value of the expression - i.e. this is just a contrived way to write j = (i += 2) which is yet again a "clever" way to write

i += 2;
j = i;

However, the , in function argument lists is not a comma operator, and there is no sequence point between evaluations of distinct arguments; instead they're unsequenced with regard to each other; so the function call

int i = 0;
printf("%d %d\n", i++, ++i, i);

has undefined behaviour because there is no sequence point between the evaluations of i++ and ++i in function arguments, and the value of i is therefore modified twice, by both i++ and ++i, between the previous and the next sequence point.

The reason is that the program is running undefined behavior. The problem lies in the evaluation order, because there is no sequence points required according to C++98 standard ( no operations is sequenced before or after another according to C++11 terminology).

However if you stick to one compiler, you will find the behavior persistent, as long as you don't add function calls or pointers, which would make the behavior more messy.

  • So first the GCC: Using Nuwen MinGW 15 GCC 7.1 you will get:

    #include<stdio.h>
    int main(int argc, char ** argv)
    {
    int i = 0;
    i = i++ + ++i;
    printf("%d\n", i); // 2
    
    i = 1;
    i = (i++);
    printf("%d\n", i); //1
    
    volatile int u = 0;
    u = u++ + ++u;
    printf("%d\n", u); // 2
    
    u = 1;
    u = (u++);
    printf("%d\n", u); //1
    
    register int v = 0;
    v = v++ + ++v;
    printf("%d\n", v); //2
    

    }

How does GCC work? it evaluates sub expressions at a left to right order for the right hand side (RHS) , then assigns the value to the left hand side (LHS) . This is exactly how Java and C# behave and define their standards. (Yes, the equivalent software in Java and C# has defined behaviors). It evaluate each sub expression one by one in the RHS Statement in a left to right order; for each sub expression: the ++c (pre-increment) is evaluated first then the value c is used for the operation, then the post increment c++).

according to GCC C++: Operators

In GCC C++, the precedence of the operators controls the order in which the individual operators are evaluated

the equivalent code in defined behavior C++ as GCC understands:

#include<stdio.h>
int main(int argc, char ** argv)
{
    int i = 0;
    //i = i++ + ++i;
    int r;
    r=i;
    i++;
    ++i;
    r+=i;
    i=r;
    printf("%d\n", i); // 2

    i = 1;
    //i = (i++);
    r=i;
    i++;
    i=r;
    printf("%d\n", i); // 1

    volatile int u = 0;
    //u = u++ + ++u;
    r=u;
    u++;
    ++u;
    r+=u;
    u=r;
    printf("%d\n", u); // 2

    u = 1;
    //u = (u++);
    r=u;
    u++;
    u=r;
    printf("%d\n", u); // 1

    register int v = 0;
    //v = v++ + ++v;
    r=v;
    v++;
    ++v;
    r+=v;
    v=r;
    printf("%d\n", v); //2
}

Then we go to Visual Studio. Visual Studio 2015, you get:

#include<stdio.h>
int main(int argc, char ** argv)
{
    int i = 0;
    i = i++ + ++i;
    printf("%d\n", i); // 3

    i = 1;
    i = (i++);
    printf("%d\n", i); // 2 

    volatile int u = 0;
    u = u++ + ++u;
    printf("%d\n", u); // 3

    u = 1;
    u = (u++);
    printf("%d\n", u); // 2 

    register int v = 0;
    v = v++ + ++v;
    printf("%d\n", v); // 3 
}

How does visual studio work, it takes another approach, it evaluates all pre-increments expressions in first pass, then uses variables values in the operations in second pass, assign from RHS to LHS in third pass, then at last pass it evaluates all the post-increment expressions in one pass.

So the equivalent in defined behavior C++ as Visual C++ understands:

#include<stdio.h>
int main(int argc, char ** argv)
{
    int r;
    int i = 0;
    //i = i++ + ++i;
    ++i;
    r = i + i;
    i = r;
    i++;
    printf("%d\n", i); // 3

    i = 1;
    //i = (i++);
    r = i;
    i = r;
    i++;
    printf("%d\n", i); // 2 

    volatile int u = 0;
    //u = u++ + ++u;
    ++u;
    r = u + u;
    u = r;
    u++;
    printf("%d\n", u); // 3

    u = 1;
    //u = (u++);
    r = u;
    u = r;
    u++;
    printf("%d\n", u); // 2 

    register int v = 0;
    //v = v++ + ++v;
    ++v;
    r = v + v;
    v = r;
    v++;
    printf("%d\n", v); // 3 
}

as Visual Studio documentation states at Precedence and Order of Evaluation:

Where several operators appear together, they have equal precedence and are evaluated according to their associativity. The operators in the table are described in the sections beginning with Postfix Operators.

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I've edited the question to add the UB in evaluation of function arguments, as this question is often used as a duplicate for that. (The last example) – Antti Haapala
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Also the question is about c now, not C++ – Antti Haapala

A good explanation about what happens in this kind of computation is provided in the document n1188 from the ISO W14 site.

I explain the ideas.

The main rule from the standard ISO 9899 that applies in this situation is 6.5p2.

Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression. Furthermore, the prior value shall be read only to determine the value to be stored.

The sequence points in an expression like i=i++ are before i= and after i++.

In the paper that I quoted above it is explained that you can figure out the program as being formed by small boxes, each box containing the instructions between 2 consecutive sequence points. The sequence points are defined in annex C of the standard, in the case of i=i++ there are 2 sequence points that delimit a full-expression. Such an expression is syntactically equivalent with an entry of expression-statement in the Backus-Naur form of the grammar (a grammar is provided in annex A of the Standard).

So the order of instructions inside a box has no clear order.

i=i++

can be interpreted as

tmp = i
i=i+1
i = tmp

or as

tmp = i
i = tmp
i=i+1

because both all these forms to interpret the code i=i++ are valid and because both generate different answers, the behavior is undefined.

So a sequence point can be seen by the beginning and the end of each box that composes the program [the boxes are atomic units in C] and inside a box the order of instructions is not defined in all cases. Changing that order one can change the result sometimes.

EDIT:

Other good source for explaining such ambiguities are the entries from c-faq site (also published as a book) , namely here and here and here .

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I've edited the question to add the UB in evaluation of function arguments, as this question is often used as a duplicate for that. (The last example) – Antti Haapala
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@AnttiHaapala Other question connected with that is a[i]=i++ -- which index of a is modified ? – alinsoar
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I've just added that too, thanks. – Antti Haapala
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@AnttiHaapala thanks, I also completed a few more comments. – alinsoar
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How this answer added new to the existing answers? Also the explanations for i=i++ is very similar to this answer. – haccks
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@haccks I did not read the other answers. I wanted to explain in my own language what I learned from the mentioned document from the official site of ISO 9899 open-std.org/jtc1/sc22/wg14/www/docs/n1188.pdf – alinsoar

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