G++ could optimize ASM code more

[Daniel Marschall]

Hello,

> Look for the Intel Optimization Manual on intel.com.  The appendixes
> have latency and throughput information for the instruction set on
> various Intel processors.

Uh-oh, that's hard. I tried to find the information, but I did only 
found a part of the informations I was looking for.

First, I used -masm=intel to use the Intel syntax and got.

- for the no-typecast-variant (imull):

imul    ecx, esi   # imull
movsx   rcx, ecx   # movslq

- for the typecast-variant (imulq):

imul    rcx, rsi   # imulq

In the Intel manual I collected following informations from Appendix C, 
Table C-16a:

		Latency		Throughput
		0f_3h	0f_2h	0f_3h	0f_2h
imul r32	10	14	1	3
imul imm32	-	14	1	3
imul		-	15-18	-	5
mov		1	0.5	0.5	0.5
movsb/movsw	1	0.5	0.5	0.5


I have 3 problems:
1. I do not know my DisplayName/DisplayFamily (0f_2h or 0f_3h?).
2. The table does not contain "movsx"
3. Should I compare Latency or Throughput if I want to produce fast 
code? Or doesn't it matter which value I compare?

I assume that movsx has the same latency of movsw (but not sure) and I 
think that "imul" in the table refers to AT&T's "imulq" resp. Intel's 
"imul rcx, rsi" while "imul r32" in the table refers to AT&T's "imull" 
resp. Intel's "imul ecx, esi". Am I right?

Daniel

Am 09.05.2012 20:30, schrieb Ian Lance Taylor:
> Daniel Marschall <daniel-marschall@viathinksoft.de> writes:
>
>> I did understand that the compiler used "signed" multiplication
>> instead of an unsigned one because char*char needs to be extended.
>>
>> Maybe I am wrong, but couldn't the compiler "know" that the result
>> will be at least unsigned because unsigned * unsigned = unsigned ?
>
> Well, but the rules of C say that the unsigned char values are
> zero-extended to int, and then they are multiplied using a signed
> multiplication.  So the result is not unsigned.  The compiler really
> would have to do some sort of type or value based reasoning here to
> determine that an unsigned multiplication would work also.
>
>> Mh... good point. I do not know much about Assembler so I just 
>> thought
>> the shorter the code the better.
>
> Sadly, no.
>
>
>> If imull is faster than imulq, then
>> the question is, if imull+movslq is still faster than a single
>> imulq. Do you know where I can find these informations for my CPU
>> (Intel Xeon X3440)? I was searching for a table which shows how many
>> CPU-ticks the imull, imulq and movslq need, but yet I have not found
>> one.
>>
>> My Linux is 2.6.32-5-amd64 #1 SMP Mon Jan 16 16:22:28 UTC 2012 
>> x86_64
>> GNU/Linux .
>>
>> And the CPU is "Intel(R) Xeon(R) CPU X3440  @ 2.53GHz". (I hope the
>> "amd64" version of Debian is the correct one, or should our admin 
>> have
>> installed the "ia64" variant since it is an Intel CPU?)
>
>
> Ian