abrash-black-book/12-04.md
2013-12-30 20:26:41 +11:00

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You don't need to understand every corner of the 486 universe unless
you're a diehard ASMhead who does this stuff for fun. Just learn enough
to be able to speed up the key portions of your programs, and spend the
rest of your time on a fast design and overall implementation.
#### More Fun with Byte Registers {#Heading10}
Rule \#4: Don't load *any* byte register exactly 2 cycles before using
*any* register to address memory.
This, the last of this chapter's rules, is the strangest of the lot. If
any byte register is loaded, and then two cycles later any register is
used to point to memory, one cycle is lost. So, for example, this code
mov al,bl
mov cx,dx
mov si,[di]
takes four rather than the expected three cycles to execute. Note that
it is *not* required that the byte register be part of the register used
to address memory; any byte register will do the trick.
Worse still, loading byte registers both one and two cycles before a
register is used to address memory costs two cycles, as in
mov bl,al
mov cl,3
mov bx,[si]
which takes five rather than three cycles to run. However, there is *no*
penalty if a byte register is loaded one cycle but not two cycles before
a register is used to address memory. Therefore,
mov cx,3
mov dl,al
mov si,[bx]
runs in the expected three cycles.
In truth, I do not know why this happens. Clearly, it has something to
do with interrupting the start of the addressing pipeline, and I have my
theories about how this works, but at this point they're pure
speculation. Whatever the reason for this rule, ignorance of it—and of
its interaction with the other rules—could lead to considerable
performance loss in seemingly air-tight code. For instance, a casual
observer would expect the following code to run in 3 cycles:
mov bx,offset MemVar
mov cl,al
mov ax,[bx]
A more sophisticated programmer would expect to lose one cycle, because
BX is loaded two cycles before being used to address memory. In fact,
though, this code takes 5 cycles—2 cycles, or 67 percent, longer than
normal. Why? Well, under normal conditions, loading a byte register—CL
in this case—one cycle before using a register to address memory
produces no penalty; loading 2 cycles ahead is the only case that
normally incurs a penalty. However, think of Rule \#4 as meaning that
loading a byte register disrupts the memory addressing pipeline as it
starts up. Viewed that way, we can see that **MOV BX,OFFSET MemVar**
interrupts the addressing pipeline, forcing it to start again, and then,
presumably, **MOV CL,AL** interrupts the pipeline again because the
pipeline is now on its first cycle: the one that loading a byte register
can affect.
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![](images/i.jpg) *I know—it seems awfully complicated. It isn't, really. Generally, try not to use byte destinations exactly two cycles before using a register to address memory, and try not to load a register either one or two cycles before using it to address memory, and you'll be fine.*
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#### Timing Your Own 486 Code {#Heading11}
In case you want to do some 486 performance analysis of your own, let me
show you how I arrived at one of the above conclusions; at the same
time, I can warn you of the timing hazards of the cache. Listings 12.1
and 12.2 show the code I ran through the Zen timer in order to establish
the effects of loading a byte register before using a register to
address memory. Listing 12.1 ran in 120 µs on a 33 MHz 486, or 4 cycles
per repetition (120 µs/1000 repetitions = 120 ns per repetition; 120 ns
per repetition/30 ns per cycle = 4 cycles per repetition); Listing 12.2
ran in 90 µs, or 3 cycles, establishing that loading a byte register
costs a cycle only when it's performed exactly 2 cycles before
addressing memory.
**LISTING 12.1 LST12-1.ASM**
; Measures the effect of loading a byte register 2 cycles before
; using a register to address memory.
mov bp,2 ;run the test code twice to make sure
; it's cached
sub bx,bx
CacheFillLoop:
call ZTimerOn ;start timing
rept 1000
mov dl,cl
nop
mov ax,[bx]
endm
call ZTimerOff ;stop timing
dec bp
jz Done
jmp CacheFillLoop
Done:
**LISTING 12.2 LST12-2.ASM**
; Measures the effect of loading a byte register 1 cycle before
; using a register to address memory.
mov bp,2 ;run the test code twice to make sure
; it's cached
sub bx,bx
CacheFillLoop:
call ZTimerOn ;start timing
rept 1000
nop
mov dl,cl
mov ax,[bx]
endm
call ZTimerOff ;stop timing
dec bp
jz Done
jmp CacheFillLoop
Done:
Note that Listings 12.1 and 12.2 each repeat the timing of the code
under test a second time, to make sure that the instructions are in the
cache on the second pass, the one for which results are displayed. Also
note that the code is less than 8K in size, so that it can all fit in
the 486's 8K internal cache. If I double the **REPT** value in Listing
12.2 to 2,000, making the test code larger than 8K, the execution time
more than doubles to 224 µs, or 3.7 cycles per repetition; the extra
seven-tenths of a cycle comes from fetching non-cached instruction
bytes.
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![](images/i.jpg) *Whenever you see non-integral timing results of this sort, it's a good bet that the test code or data isn't cached.*
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### The Story Continues {#Heading12}
There's certainly plenty more 486 lore to explore, including the 486's
unique prefetch queue, more optimization rules, branching optimizations,
performance implications of the cache, the cost of cache misses for
reads, and the implications of cache write-through for writes.
Nonetheless, we've covered quite a bit of ground in this chapter, and I
trust you've gotten a feel for the considerable extent to which 486
optimization differs from what you're used to. Odd as 486 optimization
is, though, it's well worth mastering, for the 486 is, at its best, so
staggeringly fast that carefully crafted 486 code can do more than twice
as much per cycle as the best 386 code—which makes it perhaps 50 times
as fast as optimized code for the original PC.
Sometimes it *is* hard to believe we're still in Kansas!