3 ms·

If you truly believe so, then can you give an example of input-conditional UB causing unexpected observable behavior, before the input is actually read? This sh

If you truly believe so, then can you give an example of input-conditional UB causing unexpected observable behavior, before the input is actually read? This should be impossible, since otherwise the program would have incorrect behavior if a non-UB-producing input is given.

saagarjha 4y ago
How this might happen is that one branch of your program may have unconditional undefined behavior, which can be detected at the check itself. This would let a compiler elide the entire branch, even side effects that would typically run.
- tpush 4y ago
  That example doesn't contradict LegionMammal978's point though, if I understood correctly. He's saying that the 'time-travel' wouldn't extend to before checking the conditional.
- LegionMammal978 4y ago
  The compiler can elide the unconditional-UB branch and its side effects, and it can elide the check itself. But it cannot elide the input operation that produces the value which is checked, nor can it elide any side effects before that input operation, unless it can statically prove that no input values can possibly result in the non-UB branch.
lmm 4y ago
If it's provably input-conditional then of course it's impossible. But the C implementation does not have to obey the sequence point rules or perform observable effects in the correct order for invocations that contain UB, and it doesn't have to implement "possible" non-UB-containing invocations if you can't find them. E.g. if you write a program to search for a counterexample for something like the Collantz Conjecture, that loops trying successively higher numbers until it finds one and then exits, GCC may compile that into a program that exits immediately (since looping forever is, arguably, undefined behaviour) - there's a real example of a program that does this for Fermat's Last Theorem.
- LegionMammal978 4y ago
  > If it's provably input-conditional then of course it's impossible. My entire point pertains to programs with input-conditional UB: that is, programs for which there exists an input that makes it result in UB, and there also exists an input that makes it not result in UB. Arguably, it would be more difficult for the implementation to prove that input-dependent UB is unconditional: that every possible input results in UB, or that no possible input results in UB. > But the C implementation does not have to obey the sequence point rules or perform observable effects in the correct order for invocations that contain UB Indeed, the standard places no requirements on the observable effects of an execution that eventually results in UB at some point in the future. But if the UB is input-conditional, then a "good" execution and a "bad" execution are indistinguishable until the point that the input is entered. Therefore, the implementation is required to correctly perform all observable effects sequenced prior to the input being entered, since otherwise it would produce incorrect behavior on the "good" input. > E.g. if you write a program to search for a counterexample for something like the Collantz Conjecture, that loops trying successively higher numbers until it finds one and then exits, GCC may compile that into a program that exits immediately (since looping forever is, arguably, undefined behaviour) - there's a real example of a program that does this for Fermat's Last Theorem. That only works because the loop has no observable effects, and the standard says it's UB if it doesn't halt, so the compiler can assume it does nothing but halts. As noted on https://blog.regehr.org/archives/140 https://blog.regehr.org/archives/140, if you try to print the resulting values, then the compiler is actually required to run the loop to determine the results, either at compile time or runtime. (If it correctly proves at compile time that the loop is infinite, only then can it replace the program with one that does whatever.) It's also irrelevant, since my point is about programs with input-conditional UB, but the FLT program has unconditional UB.

saagarjha 4y ago

How this might happen is that one branch of your program may have unconditional undefined behavior, which can be detected at the check itself. This would let a compiler elide the entire branch, even side effects that would typically run.

tpush 4y ago
That example doesn't contradict LegionMammal978's point though, if I understood correctly. He's saying that the 'time-travel' wouldn't extend to before checking the conditional.
LegionMammal978 4y ago
The compiler can elide the unconditional-UB branch and its side effects, and it can elide the check itself. But it cannot elide the input operation that produces the value which is checked, nor can it elide any side effects before that input operation, unless it can statically prove that no input values can possibly result in the non-UB branch.

lmm 4y ago

If it's provably input-conditional then of course it's impossible. But the C implementation does not have to obey the sequence point rules or perform observable effects in the correct order for invocations that contain UB, and it doesn't have to implement "possible" non-UB-containing invocations if you can't find them. E.g. if you write a program to search for a counterexample for something like the Collantz Conjecture, that loops trying successively higher numbers until it finds one and then exits, GCC may compile that into a program that exits immediately (since looping forever is, arguably, undefined behaviour) - there's a real example of a program that does this for Fermat's Last Theorem.

LegionMammal978 4y ago
> If it's provably input-conditional then of course it's impossible. My entire point pertains to programs with input-conditional UB: that is, programs for which there exists an input that makes it result in UB, and there also exists an input that makes it not result in UB. Arguably, it would be more difficult for the implementation to prove that input-dependent UB is unconditional: that every possible input results in UB, or that no possible input results in UB. > But the C implementation does not have to obey the sequence point rules or perform observable effects in the correct order for invocations that contain UB Indeed, the standard places no requirements on the observable effects of an execution that eventually results in UB at some point in the future. But if the UB is input-conditional, then a "good" execution and a "bad" execution are indistinguishable until the point that the input is entered. Therefore, the implementation is required to correctly perform all observable effects sequenced prior to the input being entered, since otherwise it would produce incorrect behavior on the "good" input. > E.g. if you write a program to search for a counterexample for something like the Collantz Conjecture, that loops trying successively higher numbers until it finds one and then exits, GCC may compile that into a program that exits immediately (since looping forever is, arguably, undefined behaviour) - there's a real example of a program that does this for Fermat's Last Theorem. That only works because the loop has no observable effects, and the standard says it's UB if it doesn't halt, so the compiler can assume it does nothing but halts. As noted on https://blog.regehr.org/archives/140 https://blog.regehr.org/archives/140, if you try to print the resulting values, then the compiler is actually required to run the loop to determine the results, either at compile time or runtime. (If it correctly proves at compile time that the loop is infinite, only then can it replace the program with one that does whatever.) It's also irrelevant, since my point is about programs with input-conditional UB, but the FLT program has unconditional UB.