When I try your examples using a VM compiled in 64-bit mode, the VM crashes.

  (1 << 63) negated / -1 ==> crash

So there is a problem in the primitive.

Dave


On Tue, Aug 28, 2012 at 11:36:43PM +0200, Nicolas Cellier wrote:
> Sorry to post here, this bounced at vm-dev
> But I think it's important
> 
> Nicolas
> 
> 
> ---------- Forwarded message ----------
> From: Nicolas Cellier <nicolas.cellier.aka.nice at gmail.com>
> Date: 2012/8/28
> Subject: doesn't Primitive 33 primitiveQuoLargeIntegers relies on C
> Undefined Behaviour ? uncover a BUG in Primitive 32
> primitiveDivLargeIntegers and maybe some others
> To: vm-dev-request at lists.squeakfoundation.org
> 
> 
> (1 << 63) negated is converted to signed long long
> 0x8000000000000000LL by signed64BitValueOf:
> 
> So far so good...
> Then in primitiveQuoLargeIntegers, we will try to take absolute value
> of this beast...
> 
> integerRcvr > 0 ifTrue: [
>                 integerArg > 0
>                         ifTrue: [result := integerRcvr // integerArg]
>                         ifFalse: [result := 0 - (integerRcvr // (0 -
> integerArg))].
>         ] ifFalse: [
>                 integerArg > 0
>                         ifTrue: [result := 0 - ((0 - integerRcvr) //
> integerArg)]
>                         ifFalse: [result := (0 - integerRcvr) // (0 -
> integerArg)].
>         ].
> 
> That is translated straightfully in C...
> 
>     sqLong integerArg;
>     sqLong integerRcvr;
>     sqLong result;
> 
> snip...
> 
>         if (integerRcvr > 0) {
>                 if (integerArg > 0) {
>                         result = integerRcvr / integerArg;
>                 }
>                 else {
>                         result = 0 - (integerRcvr / (0 - integerArg));
>                 }
>         }
>         else {
>                 if (integerArg > 0) {
>                         result = 0 - ((0 - integerRcvr) / integerArg);
>                 }
>                 else {
>                         result = (0 - integerRcvr) / (0 - integerArg);
>                 }
>         }
> 
> Isn't this relying on UB? like
> http://stackoverflow.com/questions/2539178/why-is-abs0x80000000-0x80000000
> At worse, I would expect 0 -  0x8000000000000000LL to be converted to
> itself  0x8000000000000000LL...
> So can someone explain why this works?
> 
>     ((1 << 63) negated quo: 2) = (1 << 62) negated
> 
> I guess it must be some compiler optimization, like:
> 
>     0 - (0-a)/b -> 0 - (0/b-a/b) -> a/b
> 
> But then, why doing the sign checks at all?
> Isn't it just result = integerRcvr / integerArg; whatever the signs?
> 
> Maybe the compiler is able to just do that, but it sounds like obfuscation...
> 
> Anyway, that won't work in primitiveDivLargeIntegers, i would expect:
>     ((1 << 63) negated // -2) =  (1 << 62)
> But I get
>     ((1 << 63) negated // -2) =  (1 << 62) negated
> 
> Ah Ah, it looks like a bug this time, no Compiler optimization did save us
> 
> We have several solutions...
> - let signed64BitValueOf: (1 << 63) negated fail
> - protect each and every negation of those
> - perform most arithmetic with unsigned, handle sign separately
> 
> Nicolas