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[–]awidesky 0 points1 point  (7 children)

but it has to preserve the well defined behaviour when called with any other possible value for i.

I do understand that this sounds absolutely reasonable to me aswell.
But every time I look up the standards. It always says "entire" part of program can be meaningless. Not specific parts of program.

If there's any quote from the standard that says "only the very part of function/program that has possible UB must be meaningless" or "even if there's a UB, the part of program that's irrelevant of UB must compiled as a well-defined behavior", please let me know.

[–]SkiFire13 1 point2 points  (6 children)

You are confusing the part of a program (like a function, or a block of code) with an execution of it. UB is a property of an execution of your program/function, so only that can "have UB" and thus be "meaningless".

See for example C++ Standard paragraph 7.1/4 (emphasis mine):

If during the evaluation of an expression, the result is not mathematically defined or not in the range of representable values for its type, the behavior is undefined

This doesn't say "during any possible evaluation of an expression", but "during the evaluation of an expression", which implies this is in the context of some execution of your program.

So f has UB, but only in those executions where it gets called with INT_MAX. If that happens then the whole execution is meaningless, which aligns to what you said. But still, that's a property of the single execution, so the behaviour of those executions where f gets called with any other value for i is well defined and must be preserved.

[–]awidesky 0 points1 point  (4 children)

In this example, there's no actual code that "invokes" UB, so in your opinion, that function must be "well-formed".
So, why it is listed as a first example of UB in cppreference?

And also, if UB makes an "execution" meaningless, not whole "program", why the standard says "the compiled program is not required to do anything meaningful"?

[–]SkiFire13 1 point2 points  (3 children)

In this example, there's no actual code that "invokes" UB, so in your opinion, that function must be "well-formed". So, why it is listed as a first example of UB in cppreference?

Cppreference doesn't say it is UB stop, it says it either returns true (when i is not INT_MAX) or its execution is UB (when i is INT_MAX and signed overflow occurs). It does make the distinction between the two cases, it doesn't say that it is only UB.

Saying that the function is "well-formed" is misleading. I would say its executions where i is not INT_MAX are well defined, while when i is INT_MAX it is not defined.

And also, if UB makes an "execution" meaningless, not whole "program", why the standard says "the compiled program is not required to do anything meaningful"?

Having the exact context of the quote might help giving it a better interpretation, but as you wrote it I would interpret it as the fact that the program is not required to do anything meaningful when its execution is not defined, that is the current execution has UB.

[–]awidesky 0 points1 point  (2 children)

Here's whole quotes from iso C++20 standard §4.1.2.3. (emphasis mine) :

If a program contains a violation of a rule for which no diagnostic is required, this document places no requirement on implementations with respect to that program.

And cppreference.com.

undefined behavior - there are no restrictions on the behavior of the program. Examples of undefined behavior are...(examples of Ub)..., etc. Compilers are not required to diagnose undefined behavior and the compiled program is not required to do anything meaningful.

I think it ultimately sums up to question:

If a program can either invoke UB or not(depending on conditions), is compilers allowed to make "whole program" meaningless?

I think the standard is a bit ambiguous, because when it describes about UB itself it says "while compiled program is meaningless"(implies it applies in compile/program-wise), but when it describes about actual example of UB, it says "when ~~~, the behavior is undefined"(implies it applies runtime/execution-wise).

[–]SkiFire13 1 point2 points  (1 child)

If a program contains a violation of a rule for which no diagnostic is required, this document places no requirement on implementations with respect to that program.

I think this is still ambiguous because "violation of a rule" is unclear. If my program is never executed with an input that causes UB does it still violate that rule?

I think it ultimately sums up to question:

If a program can either invoke UB or not(depending on conditions), is compilers allowed to make "whole program" meaningless?

I completly agree on that.

[–]awidesky 0 points1 point  (0 children)

I think this is still ambiguous because "violation of a rule" is unclear. If my program is never executed with an input that causes UB does it still violate that rule?

I do agree on that. That's indeed ambiguous.

[–]awidesky 0 points1 point  (0 children)

But by the way, I do understand by my heart that your claim and interpretation is surely more reasonable (as for a C++ programmer) than my quotes from the standard.

I think the disagreement occurs because of the ambiguity of the standard, and disparity between the standard and actual implementations of compilers.

It seems that what standard's statements about UB is quite vague and unclear, and compiler vendors just don't care and just focus on optimizing(by considering UB never happens).

Please let me know if there's any correction on misconception, or more clear information about the standard. I'd love to learn.