> UB problems happen only when compilers detect it, understand it is an opportunity for optimization, because they are allowed to do anything, then do that optimization.
Perhaps I'm misunderstanding the comment, but I took "detect it, understand it is an opportunity for optimization, because they are allowed to do anything, ..." to mean "detect that undefined behavior occurs and then ..."
Instead, it should be "detect that an optimization can be applied and that the optimized code will do the right thing in all cases that undefined behavior does not occur, and apply the optimization."
I've met programmers who seem to believe that compilers intentionally apply weird transformations when they detect undefined behavior. But really, the compilers are applying sensible transformations for well-defined code and ignoring whether those transformations are sensible for code that isn't well-defined.
The compiler sees a variable x that's allocated on the stack. It looks at all the uses of the address of x, and sees that they are all loads and stores. It therefore decides to promote x into a register variable.
Where's the UB, you ask. Well, since the address of x was never leaked, it is UB to compute another pointer to that address (say, via an integer-to-pointer conversion). The compiler never checked for that possibility to affirm that it was UB; it knew that any code that did that would be UB and simply ignored the possibility.
This makes arithmetic overflow a very poor vehicle for UB because it's unusual in that you can't really take advantage of the UB without pointing to the specific operation that caused the UB to occur. This is why I believe that arithmetic overflow UB is gratuitous, and people objecting to UB because of what happens specifically with arithmetic overflow go on to make suggestions that are completely untenable because they don't have familiarity with how most UB works.
The problem causing this is that C loops are overspecified. If there was a “for…in” loop that didn’t make you declare and increment a variable, then manual increments could trap on overflow without causing so many problems.
If you can turn the loop from upward to downward then you do have a bound for it, and then you can tell if it overflows or not.
Also why is signed overflow a problem and not unsigned?
Surely you want unsigned loop optimized also?
Typically not a problem for i=0...n loops but eg `for (i=0;i<n;i+=2)` could overflow.
No, the OP wrote specifically that "the compiler detects [UB]" and then does optimizations exploiting that. Not "detects potential UB". The former is a common misconception, the latter is basically a truism.
Tell me how these use UB: integer constant folding, unused load removal, optimal use of GP registers to minimize stack spilling
>sane assumptions
"sane" here is doing a lot of work. Assuming overflow won't happen is not a sane assumption, assuming some other thread won't randomly write into your stack memory is.
int a = 10;
int var1;
int b = 20;
foo(&var1);
int c = a + b;
You'd like to constant fold c? Better assume no UB: void foo(int* x) { *(x-1) = 0; }
> unused load removalSame idea as above.
> Assuming overflow won't happen is not a sane assumption
If the alternative is loops being much slower when you request optimizations, then maybe it is. Consult your compiler manuals if this default irks you.
My original point was that compilers don't specifically abuse UB to sleight you. You're trying hard to draw a line in the sand between different degrees of UB so that you can label one side as "not sane", with (what appears to be) the sole intent of concluding malice on the compiler author side. Please accept that the related tradeoffs have been discussed, and that a default was chosen (if you are curious for a recent instance of this, see [0]).
If you are ok with leaving performance on the table by disabling overflow being UB, you are literally free to do that. You're also free to lobby the standard bodies to alter their carefully considered choices. But you're not going to get anywhere by labeling any of this "insane".
[0]: http://www.open-std.org/jtc1/sc22/wg21/docs/papers/2018/p090...
Oh come on, this is like the most UB-centric thing in the entire compiler. The compiler uses UB to know that it has enumerated all uses of a memory location, as any pointer that didn't get its value from the data dependency tree that compiler used had to use UB to compute the address. (Be it from buffer overflow, use after end-of-lifetime, random integer converted to pointer, etc.)
Everybody agrees that compilers should only do sane optimizations. Nobody agrees on which subset is sane.
It isn't; that sentence is even a type error. UB is not a property of compiler output. It's a property of compiler input.
The use of registers to store variables between uses (when those variables never have their address taken) relies on the lack of any defined behavior which would allow the program to determine this is being done. The fact you can't find this in the spec is precisely because it is not defined.