int opEquals(Object), and other legacy ints (!)

[Bruno Medeiros]

Walter Bright wrote:
> Stewart Gordon wrote:
>> Walter Bright wrote:
>>> Stewart Gordon wrote:
>>>> xs0 wrote:
>>>> <snip>
>>>>> Well, I'm just guessing, but I think something like
>>>>>
>>>>>  > int opEquals(Foo foo)
>>>>>  > {
>>>>>  >     return this.bar == foo.bar;
>>>>>  > }
>>>>>
>>>>> is compiled to something like
>>>>>
>>>>>> return this.bar-foo.bar; // 1 instruction
>>>>>
>>>>> but if the return type is bool, it becomes
>>>>>
>>>>>> return this.bar-foo.bar?1:0; // 3 instructions
>>>>
>>>> If it does this, then there's a serious bug in the compiler.
>>>
>>> What instruction sequence do expect to be generated for it?
>>
>> If anything resembling the above, then
>>
>>     return this.bar-foo.bar?0:1;
> 
> ? Let's look at an example:
> 
> class Foo
> {
>     int foo, bar;
> 
>     int Eq1(Foo foo)
>     {
>         return this.bar-foo.bar?0:1;
>     }
> 
>     int Eq2(Foo foo)
>     {
>         return this.bar-foo.bar;
>     }
> }
> 
> which generates:
> 
>     Eq1:
>                 mov     EDX,4[ESP]
>                 mov     ECX,0Ch[EAX]
>                 sub     ECX,0Ch[EDX]
>                 cmp     ECX,1
>                 sbb     EAX,EAX
>                 neg     EAX
>                 ret     4
>     Eq2:
>                 mov     ECX,4[ESP]
>                 mov     EAX,0Ch[EAX]
>                 sub     EAX,0Ch[ECX]
>                 ret     4
> 
> So we have 4 instructions generated rather than 1. If there's a trick to 
> generate only one instruction for Eq1, I'd like to know about it.
> 
>>> I can. (a == b), where a and b are ints, can be implemented as (a - 
>>> b), and the result is int 0 for equality, int !=0 for inequality.
>>
>> How is this (a == b) rather than (a != b)?
> 
> I don't understand your question.

As per the other posts, Eq2 actually takes 2 instructions:

Eq2:
	...
	sub     EAX,0Ch[ECX]
	not	EAX;


And uuuh.., I've checked gcc's generated code for a C++'s Eq1, and it 
was only 2 instructions too, CMP and SETE ! :

Eq1:
	...
	cmp     EAX,0Ch[ECX]
	sete	EAX;

(http://www.cs.tut.fi/~siponen/upros/intel/instr/sete_setz.html)
It seems to me perfectly valid, is there any problem here?

What does the original Eq1 even do? :

	sub     ECX,0Ch[EDX]
	cmp     ECX,1       // Huh?
	sbb     EAX,EAX
	neg     EAX


-- 
Bruno Medeiros - MSc in CS/E student
http://www.prowiki.org/wiki4d/wiki.cgi?BrunoMedeiros#D