5 ms·

UB can not travel back in time in C. Although it is true that it can affect previous instructions, but that code is reordered or transformed in complicated ways

by uecker 2y ago

UB can not travel back in time in C. Although it is true that it can affect previous instructions, but that code is reordered or transformed in complicated ways is true even without UB.

kazinator 2y ago
If a compiler can determine that some statement is UB, it can treat that as an assertion that the code is unreachable. All other statements which reach only that code and no other are also unreachable. A compiler's analysis can go backward in time. That is to say, the compiler can build a model of what happens in some section of code over time, and analyze it whichever way it wants. You cannot go back in time from execution time to translation time, but the translator can follow the code as if it were executing it at translation time.
emmericp 2y ago
The time-travelling UB interpretation was popularized by this blog post about 10 years ago [1]. I'm not enough of a specification lawyer to say that this is definitely true, but the reasoning and example given there seems sound to me. [1] https://devblogs.microsoft.com/oldnewthing/20140627-00/?p=633 https://devblogs.microsoft.com/oldnewthing/20140627-00/?p=63...
- uecker 2y ago
  Yes, random blog posts did a lot of damage here. Also broken compilers [1]. Note that blog post is correct about C++ but incorrectly assumes this is true for C as well. [1]. https://developercommunity.visualstudio.com/t/Invalid-optimization-in-CC/10337428?q=muecker https://developercommunity.visualstudio.com/t/Invalid-optimi...
  - kibwen 2y ago
    I'm inclined to trust Raymond Chen and John Regehr on these matters, so if you assert that they're incorrect here then a source to back up your assertion would help your argument.
    - uecker 2y ago
      I am a member of WG14. You should check the C standard. I do not see how "time-travel" is a possible reading of the definition of UB in C. We added another footnote to C23 to counter this idea: https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf "Any other behavior during execution of a program is only affected as a direct consequence of the concrete behavior that occurs when encountering the erroneous or non portable program construct or data. In particular, all observable behavior (5.1.2.4) appears as specified in this document when it happens before an operation with undefined behavior in the execution of the program." I should point out that compilers also generally do not do true time-travel: Consider this example: https://godbolt.org/z/rPG14rrbj https://godbolt.org/z/rPG14rrbj
      - lmm 2y ago
        While interactions with volatile and interactive streams cannot time-travel, anything else is free to - the standard only imposes requirements on a conforming implementation in terms of the contents of files at program termination, and programs with undefined behaviour are not required to terminate, so there are approximately no requirements on a program that invokes undefined behaviour.
      - grumbelbart 2y ago
        So maybe we have different definitions of "time travel". But I recall that - if a compiler finds that condition A would lead to UB, it can assume that A is never true - that fact can "backpropagate" to, for example, eliminate comparisons long before the UB. Here is an older discussion: https://softwareengineering.stackexchange.com/q/291548 https://softwareengineering.stackexchange.com/q/291548 Is that / will that no longer be true for C23? Or does "time-travel" mean something else in this context?
        uecker 2y ago
        There may be different definitions, but also a lot of incorrect information. Nothing changes with C23 except that we added a note that clarifies that UB can not time-travel. The semantic model in C only requires that observable effects are preserved. Everything else can be changed by the optimizer as long as it does not change those observable effects (known as the "as if" principle). This is generally the basis of most optimizations. Thus, I call time-travel only when it would affect previous observable effects, and this what is allowed for UB in C++ but not in C. Earlier non-observable effects can be changed in any case and is nothing speicifc to UB. So if you call time-travel also certain optimization that do not affect earlier observable behavior, then this was and is still allowed. But the often repeated statement that a compiler can assume that "A is never true" does not follow (or only in very limited sense) from the definition of UB in ISO C (and never did), so one has to be more careful here. In particular it is not possible to remove I/O before UB. The following code has to print 0 when called with zero and a compiler which would remove the I/O would not be conforming. int foo(int x) { printf("%d\n", x); fflush(stdout); return 1 / x; } In the following example int foo(int x) { if (x) bar(x); return 1 / x; } the compiler could indeed remove the "if" but not because it were allowed to assume that x can never be zero, but because 1 / 0 can have arbitrary behavior, so could also call "bar()" and then it is called for zero and non-zero x and the if condition could be removed (not that compilers would do this)
  - quietbritishjim 2y ago
    > Also broken compilers [1]. The issue you linked to is not a counter example because, as the poster said, g may terminate the program in which case that snippet does not have undefined behaviour even if b is zero. The fact that they bothered to mention that g may terminate the program seems like an acknowledgement that it would be valid to do that time travelling if it didn't. > Note that blog post is correct about C++ but incorrectly assumes this is true for C as well. Presumably you're referring to this line of the C++ standard, which does not appear in the C standard: > However, if any such execution contains an undefined operation, this International Standard places no requirement on the implementation executing that program with that input (not even with regard to operations preceding the first undefined operation). I looked at every instance of the word "undefined" in the C standard and, granted, it definitely didn't have anything quite so clear about time travel as that. But it also didn't make any counter claims that operations before are valid. It pretty much just said that undefined behaviour causes behaviour that is undefined! So, without strong evidence, it seem presumptuous to assume that operations provably before undefined behaviour are well defined.
    - uecker 2y ago
      The poster is me. You are right that this is not an example for time-travel. There aren't really good examples for true time travel because compilers generally do not do this. But my point is that with compilers behaving like this, people might confuse this for time-traveling UB. I have certainly met some who did and the blog posts seems to have similar examples (but I haven't looked closely now). Note that I am a member of WG14. We added more clarification to C23 to make clear that this is not a valid interpretation of UB, see here: https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf
      - quietbritishjim 2y ago
        Ok, fair enough. I must admit I was looking at C99 as I thought that was most generally relevant, I don't follow recent C standards (as much as I do those for C++) and C23 hasn't been ratified yet. I've found your new snippet: > In particular, all observable behavior (5.1.2.4) appears as specified in this document when it happens before an operation with undefined behavior in the execution of the program. I consider that a change in the standard but, of course, that's allowed, especially as it's backwards compatible for well defined programs. The wording is a little odd: it makes it sound a like you need some undefined behaviour in order to make the operations beforehand work, and, taken very literally, that operations between two undefined behaviours will work (because they're still "before an operation with undefined behavior"). But I suppose the intention is clear.
ant6n 2y ago
> but that code is reordered or transformed in complicated ways is true even without UB. Without undefined behavior, the compiler emits code that has the behavior defined by the code —- the ordering may be altered, but not the behavior.
- uecker 2y ago
  Yes, and with undefined behavior, the compiler has to emit code that has the behavior defined by the code up to the operation that has undefined behavior.
  - kazinator 2y ago
    That is false. If a compiler determines that some statement has undefined behavior, it can treat it as unreachable, and, transitively, other code before it as unreachable. printf("hello\n"); // this doesn't have to print x = x / 0; // because this is effectively a notreached() assertion
    - jstimpfle 2y ago
      This is in direct contradiction to what uecker says. Can you back up your claim -- for both C and C++? Putting your code in godbolt with -O3 did not remove the print statement for me in either C or C++. But I didn't experiment with different compilers or compiler flags, or more complicated program constructions. https://godbolt.org/z/8nbbd3jPW https://godbolt.org/z/8nbbd3jPW I've often said that I've never noticed any surprising consequences from UB personally. I know I'm on thin ice here and running risk of looking very ignorant. There are a lot of blogposts and comments that spread what seems like FUD from my tiny personal lookout. It just seems hard to come across measureable evidence of actual miscompilations happening in the wild that show crazy unpredictable behaviour -- I would really like to have some of it to even be able to start tallying the practical impact. And disregarding whatever formulations there are in the standard -- I think we can all agree that insofar compilers don't already do this, they should be fixed to reject programs with an error message should they be able to prove UB statically -- instead of silently producing something else or acting like the code wouldn't exist. Is there an error in my logic -- is there a reason why this shouldn't be practically possible for compilers to do, just based on how UB is defined? With all the flaws that C has, UB seems like a relatively minor one to me in practice. Another example: https://godbolt.org/z/b5j99enTn https://godbolt.org/z/b5j99enTn This is an adaption from the Raymond Chen post, and it seems to actually compile to a "return 1" when compiling with C++ (not with C), at least with the settings I tried. And even the "return 1" for me is understandable given that we actually hit a bug and there are no observeable side-effects before the UB happens. (But again, the compiler should instead be so friendly and emit a diagnostic about what it's doing here, or better return an error). Un-comment the printf statement and you'll see that the code totally changes. The printf actually happens now. So again, what uecker says about observable effects seems to apply.

by uecker 2y ago

UB can not travel back in time in C. Although it is true that it can affect previous instructions, but that code is reordered or transformed in complicated ways is true even without UB.

kazinator 2y ago

If a compiler can determine that some statement is UB, it can treat that as an assertion that the code is unreachable. All other statements which reach only that code and no other are also unreachable. A compiler's analysis can go backward in time. That is to say, the compiler can build a model of what happens in some section of code over time, and analyze it whichever way it wants. You cannot go back in time from execution time to translation time, but the translator can follow the code as if it were executing it at translation time.

emmericp 2y ago

The time-travelling UB interpretation was popularized by this blog post about 10 years ago [1]. I'm not enough of a specification lawyer to say that this is definitely true, but the reasoning and example given there seems sound to me. [1] https://devblogs.microsoft.com/oldnewthing/20140627-00/?p=633 https://devblogs.microsoft.com/oldnewthing/20140627-00/?p=63...

uecker 2y ago
Yes, random blog posts did a lot of damage here. Also broken compilers [1]. Note that blog post is correct about C++ but incorrectly assumes this is true for C as well. [1]. https://developercommunity.visualstudio.com/t/Invalid-optimization-in-CC/10337428?q=muecker https://developercommunity.visualstudio.com/t/Invalid-optimi...
- kibwen 2y ago
  I'm inclined to trust Raymond Chen and John Regehr on these matters, so if you assert that they're incorrect here then a source to back up your assertion would help your argument.
  - uecker 2y ago
    I am a member of WG14. You should check the C standard. I do not see how "time-travel" is a possible reading of the definition of UB in C. We added another footnote to C23 to counter this idea: https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf "Any other behavior during execution of a program is only affected as a direct consequence of the concrete behavior that occurs when encountering the erroneous or non portable program construct or data. In particular, all observable behavior (5.1.2.4) appears as specified in this document when it happens before an operation with undefined behavior in the execution of the program." I should point out that compilers also generally do not do true time-travel: Consider this example: https://godbolt.org/z/rPG14rrbj https://godbolt.org/z/rPG14rrbj
    - lmm 2y ago
      While interactions with volatile and interactive streams cannot time-travel, anything else is free to - the standard only imposes requirements on a conforming implementation in terms of the contents of files at program termination, and programs with undefined behaviour are not required to terminate, so there are approximately no requirements on a program that invokes undefined behaviour.
    - grumbelbart 2y ago
      So maybe we have different definitions of "time travel". But I recall that - if a compiler finds that condition A would lead to UB, it can assume that A is never true - that fact can "backpropagate" to, for example, eliminate comparisons long before the UB. Here is an older discussion: https://softwareengineering.stackexchange.com/q/291548 https://softwareengineering.stackexchange.com/q/291548 Is that / will that no longer be true for C23? Or does "time-travel" mean something else in this context?
      - uecker 2y ago
        There may be different definitions, but also a lot of incorrect information. Nothing changes with C23 except that we added a note that clarifies that UB can not time-travel. The semantic model in C only requires that observable effects are preserved. Everything else can be changed by the optimizer as long as it does not change those observable effects (known as the "as if" principle). This is generally the basis of most optimizations. Thus, I call time-travel only when it would affect previous observable effects, and this what is allowed for UB in C++ but not in C. Earlier non-observable effects can be changed in any case and is nothing speicifc to UB. So if you call time-travel also certain optimization that do not affect earlier observable behavior, then this was and is still allowed. But the often repeated statement that a compiler can assume that "A is never true" does not follow (or only in very limited sense) from the definition of UB in ISO C (and never did), so one has to be more careful here. In particular it is not possible to remove I/O before UB. The following code has to print 0 when called with zero and a compiler which would remove the I/O would not be conforming. int foo(int x) { printf("%d\n", x); fflush(stdout); return 1 / x; } In the following example int foo(int x) { if (x) bar(x); return 1 / x; } the compiler could indeed remove the "if" but not because it were allowed to assume that x can never be zero, but because 1 / 0 can have arbitrary behavior, so could also call "bar()" and then it is called for zero and non-zero x and the if condition could be removed (not that compilers would do this)
- quietbritishjim 2y ago
  > Also broken compilers [1]. The issue you linked to is not a counter example because, as the poster said, g may terminate the program in which case that snippet does not have undefined behaviour even if b is zero. The fact that they bothered to mention that g may terminate the program seems like an acknowledgement that it would be valid to do that time travelling if it didn't. > Note that blog post is correct about C++ but incorrectly assumes this is true for C as well. Presumably you're referring to this line of the C++ standard, which does not appear in the C standard: > However, if any such execution contains an undefined operation, this International Standard places no requirement on the implementation executing that program with that input (not even with regard to operations preceding the first undefined operation). I looked at every instance of the word "undefined" in the C standard and, granted, it definitely didn't have anything quite so clear about time travel as that. But it also didn't make any counter claims that operations before are valid. It pretty much just said that undefined behaviour causes behaviour that is undefined! So, without strong evidence, it seem presumptuous to assume that operations provably before undefined behaviour are well defined.
  - uecker 2y ago
    The poster is me. You are right that this is not an example for time-travel. There aren't really good examples for true time travel because compilers generally do not do this. But my point is that with compilers behaving like this, people might confuse this for time-traveling UB. I have certainly met some who did and the blog posts seems to have similar examples (but I haven't looked closely now). Note that I am a member of WG14. We added more clarification to C23 to make clear that this is not a valid interpretation of UB, see here: https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf https://www.open-std.org/jtc1/sc22/wg14/www/docs/n3220.pdf
    - quietbritishjim 2y ago
      Ok, fair enough. I must admit I was looking at C99 as I thought that was most generally relevant, I don't follow recent C standards (as much as I do those for C++) and C23 hasn't been ratified yet. I've found your new snippet: > In particular, all observable behavior (5.1.2.4) appears as specified in this document when it happens before an operation with undefined behavior in the execution of the program. I consider that a change in the standard but, of course, that's allowed, especially as it's backwards compatible for well defined programs. The wording is a little odd: it makes it sound a like you need some undefined behaviour in order to make the operations beforehand work, and, taken very literally, that operations between two undefined behaviours will work (because they're still "before an operation with undefined behavior"). But I suppose the intention is clear.

ant6n 2y ago

> but that code is reordered or transformed in complicated ways is true even without UB. Without undefined behavior, the compiler emits code that has the behavior defined by the code —- the ordering may be altered, but not the behavior.

uecker 2y ago
Yes, and with undefined behavior, the compiler has to emit code that has the behavior defined by the code up to the operation that has undefined behavior.
- kazinator 2y ago
  That is false. If a compiler determines that some statement has undefined behavior, it can treat it as unreachable, and, transitively, other code before it as unreachable. printf("hello\n"); // this doesn't have to print x = x / 0; // because this is effectively a notreached() assertion
  - jstimpfle 2y ago
    This is in direct contradiction to what uecker says. Can you back up your claim -- for both C and C++? Putting your code in godbolt with -O3 did not remove the print statement for me in either C or C++. But I didn't experiment with different compilers or compiler flags, or more complicated program constructions. https://godbolt.org/z/8nbbd3jPW https://godbolt.org/z/8nbbd3jPW I've often said that I've never noticed any surprising consequences from UB personally. I know I'm on thin ice here and running risk of looking very ignorant. There are a lot of blogposts and comments that spread what seems like FUD from my tiny personal lookout. It just seems hard to come across measureable evidence of actual miscompilations happening in the wild that show crazy unpredictable behaviour -- I would really like to have some of it to even be able to start tallying the practical impact. And disregarding whatever formulations there are in the standard -- I think we can all agree that insofar compilers don't already do this, they should be fixed to reject programs with an error message should they be able to prove UB statically -- instead of silently producing something else or acting like the code wouldn't exist. Is there an error in my logic -- is there a reason why this shouldn't be practically possible for compilers to do, just based on how UB is defined? With all the flaws that C has, UB seems like a relatively minor one to me in practice. Another example: https://godbolt.org/z/b5j99enTn https://godbolt.org/z/b5j99enTn This is an adaption from the Raymond Chen post, and it seems to actually compile to a "return 1" when compiling with C++ (not with C), at least with the settings I tried. And even the "return 1" for me is understandable given that we actually hit a bug and there are no observeable side-effects before the UB happens. (But again, the compiler should instead be so friendly and emit a diagnostic about what it's doing here, or better return an error). Un-comment the printf statement and you'll see that the code totally changes. The printf actually happens now. So again, what uecker says about observable effects seems to apply.