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My understanding of the whole sequence points thing is basic. All I have is some crude intuitive idea that "once a sequence point is encountered, we can be sure all side effects of previous evaluations are complete". I also read that in statements like printf("%d",a++,++a,a++) the behavior is undefined as a comma doesn't signify a sequence point while a semi-colon does. So instead of making guesses and going by intuition, I feel a very rigorous and conclusive answer to this will help me a lot.

So are the following kind of statements safe & certain in C:

int a=4,*ptr=&a;  //Statement 1

x+=4,y=x*2;  //Statement 2  (Assume x and y are integer variables)

If yes, how? Especially in the second case, if a comma is not a sequence point, how can we be sure x has been incremented by 4 before we use it in the assignment y=x*2? And for the first statement, how can I be sure a has been initialized and allocated memory before I assign its address to ptr? Should I play safe and use the following for the above:

int a=4,*ptr;
ptr=&a;

and 

x+=4;
y=x*2;

Edit My understanding of the comma operator tells me that those statements are safe. But after reading about sequence points and how something like printf("%d",a++,++a,a++) is undefined, I am having second thoughts.

Mat
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Rüppell's Vulture
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  • `,` is not a sequence point in `printf("%d",a++,++a,a++)` but it is one in `x+=4,y=x*2;`. And `int a=4,*ptr=&a;` is not a statement. – Pascal Cuoq Apr 28 '13 at 12:47
  • @PascalCuoq What is `int a=4,*ptr=&a;` then? Doesn't it declare and initialize `a` and `ptr`? – Rüppell's Vulture Apr 28 '13 at 12:48
  • @PascalCuoq `int a=4,*ptr=&a;` is correct after all, right? – Rüppell's Vulture Apr 28 '13 at 12:48
  • `int a=4,*ptr=&a;` is not a statement because it is a declaration. See C99 clauses 6.7:1 and 6.8.2:1. – Pascal Cuoq Apr 28 '13 at 12:50
  • Variable `a` does not need to be initialized to take its address, but nevertheless there is a question in that part of your question. It has already been asked here: http://stackoverflow.com/questions/16219082/does-int-a-1-b-a-invoke-undefined-behavior – Pascal Cuoq Apr 28 '13 at 12:52
  • I know that probably this is just an exercise, but nonetheless: don't live dangerously. If you aren't sure if there's a sequence point, just split the damn thing in two statements. There's no need to be extra clever in these things (also, *you* may have checked the standard, but the next programmer will be puzzled over your code). This applies in general to potentially UB-prone code. – Matteo Italia Apr 28 '13 at 13:10

2 Answers2

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The comma that separates function arguments in a function call is not a comma operator - it's just punctuation that happens to be spelled in the same way as the comma operator. There's no sequence point between the evaluation of different function arguments, so that's why you get UB in that case.

On the other hand, in your expression:

x+=4,y=x*2;

the comma here is a comma operator, which introduces a sequence point; there's no UB.

In a declaration, the comma between declarators is also not a comma operator; however, the end of a full declarator (a declarator not part of another declarator) does introduce a sequence point, so a declaration like:

int a = 2, b = a + 1;

is not UB.

caf
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3

There is no UB here.
As per C99 standard:

C. Annex C (informative): Sequence points
#1 The following are the sequence points described in 5.1.2.3:
-- The call to a function, after the arguments have been evaluated (6.5.2.2).
-- The end of the first operand of the following operators: logical AND && (6.5.13); logical OR || (6.5.14); conditional ? (6.5.15); comma , (6.5.17).

Thus there is no comma operator involved upon function call. Despite presence of comma.

alexrider
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  • @Mat Can you please give a simpler answer?This one is a little tough for me. – Rüppell's Vulture Apr 28 '13 at 12:58
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    @SheerFish: the `,` that separates function arguments isn't the comma operator. The comma operator introduces a sequence point. The one between function arguments doesn't, it is only syntax. – Mat Apr 28 '13 at 13:00
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    @alexrider: please stop expanding on C++ rules answering a C question. – Mat Apr 28 '13 at 13:06
  • @Mat Thanks, that concise comment told me a lot.But if alexrider is wrong how come he got so many upvotes?He is right or not? – Rüppell's Vulture Apr 28 '13 at 13:13
  • @Mat: I don't see why he shouldn't do that, this is a common question also in C++ (about a "language zone" that is similar but not exactly the same in the two languages), so adding such standard references doesn't hurt now and may help future readers. – Matteo Italia Apr 28 '13 at 13:15
  • @SheerFish: he's right; C and C++ mostly have the same rules for sequence points and the memory model. – Mat Apr 28 '13 at 13:15
  • @Matteo: then the corresponding quotes from the C standard would help just as much (since they're very similar). And if there are subtle differences between C and C++, answering a question about C with C++ legalese is dangerous for future readers. – Mat Apr 28 '13 at 13:17
  • Even C11 and C99 express sequence point rules differently. If C++ rules were “mostly the same” to those of C, should they be mostly the same to C99 or mostly the same to C11? – Pascal Cuoq Apr 28 '13 at 17:49