[Vm-dev] Is bytecodePrimMultiply correct?

[Stefan Marr]

Hi Levente:

On 04 Oct 2011, at 22:25, Levente Uzonyi wrote:

> On Tue, 4 Oct 2011, Stefan Marr wrote:
> 
> snip
> 
>> 0x00032f04  <+0068>  sar    %ebx               // oop_int_t ri = rcvr.integerValue();
>> 0x00032f06  <+0070>  sar    %edx               // oop_int_t ai = arg.integerValue();
>> 0x00032f08  <+0072>  imul   %edx,%ebx          // oop_int_t r  = ri * ai
>> // The following should be: if (ai == 0  ||  (((r / ai)  ==  ri)    &&    Oop::isIntegerValue(r))) {
>> 0x00032f0b  <+0075>  lea    (%ebx,%ebx,1),%eax
> 
> This is the first part of the isIntegerValue check (assign the value of r1 + (r1 * 1) to the eax register).
> 
>> 0x00032f0e  <+0078>  test   %edx,%edx
>> 0x00032f10  <+0080>  je     0x32f1a <_ZN18Squeak_Interpreter20bytecodePrimMultiplyEv+90>
> 
> This is the ai == 0 check.
Thanks for the disassembling :)


>> 0x00032f12  <+0082>  xor    %eax,%ebx
>> 0x00032f14  <+0084>  js     0x3301f <_ZN18Squeak_Interpreter20bytecodePrimMultiplyEv+351>
> 
> This is the second part of the #isIntegerValue check (xor 2 * r1 with r1)
> So the (r / ai) == ri was assumed to be true by the compiler, therefore not checked.

Ok, now the question is probably whether that is a legal transformation, and if so, what the best alternative is.

Using the `long long` to be able to check for overflow might not be the most efficient one.

Thanks
Stefan



-- 
Stefan Marr
Software Languages Lab
Vrije Universiteit Brussel
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