[理工] [計組] control hazard
Given the following MIPS instruction code segment, please answer each question
below.
L1: addi $t0 ,$t0,4
lw2 $s1 ,0($t0)
sw $s1 ,32($t0)
lw1 $t1 ,64($t0)
slt $s0 ,$t1,$zero
bne $s0 ,$zero,L1
Assume a forwarding unit is available to only forward data from ME and/or WB
to EX. Please reorder/rewrite the code to maximize its performance.Note that
you should consider maximizing the performance based on the assumption that
the loop might be iterated a few times.You may insert no-ops in the code
segment to resolve inevitable hazards if any.
解答：
L1: addi $t0,$t0,4
lw1 $t1,64($t0)
lw2 $s1,0($t0)
slt $s0,$t1,$zero
nop
nop
bne $s0,$zero,L1
sw $s1,32($t0)
疑問：想請問大家在解類似這種reorder code時，尤其有control hazard時，
是都如何思考，下手去做的。雖然課本上有三種from before，from target，
from fall through方法，可是碰到這種題目真的不是很清楚該如何去作答。
不曉得有沒有能說明一下您在碰到這種題目的做法，讓我參考看看...
我只會先解control hazard，然後在用課本三種方法隨便亂試...或是交換指令
然後找到data hazard，就畫成pipeline stage了...看要不要插入nop
問題：請問為什麼不是插一個nop，而是兩個。我只會畫圖，以下是我畫的圖
，不曉得又是哪裡出錯了...(看完上篇d大解說，現在懂了。可是都打了就...
請問大家都是怎麼做這類題型吧。)
addi IF ID EX ME-| WB
lw2 IF ID EX | ME-| WB
lw1 IF ID |- EX | ME WB
slt IF ID |-EX ME WB-|
nop X X X X | X
nop X X X | X X
bne IF ID-| EX ME WB
sw IF ID EX ME WB
(addi和lw用有hazard：forwarding解，slt和bne有hazard)
※ 編輯: numin 來自: 123.193.221.223 (09/20 22:19)
推
09/20 23:02, , 1F
09/20 23:02, 1F
→
09/20 23:03, , 2F
09/20 23:03, 2F
→
09/20 23:18, , 3F
09/20 23:18, 3F
→
09/20 23:19, , 4F
09/20 23:19, 4F
→
09/20 23:19, , 5F
09/20 23:19, 5F
推
09/20 23:57, , 6F
09/20 23:57, 6F
→
09/20 23:57, , 7F
09/20 23:57, 7F
→
09/20 23:58, , 8F
09/20 23:58, 8F
→
09/20 23:58, , 9F
09/20 23:58, 9F
→
09/20 23:59, , 10F
09/20 23:59, 10F
→
09/20 23:59, , 11F
09/20 23:59, 11F
→
09/21 00:00, , 12F
09/21 00:00, 12F
→
09/21 00:00, , 13F
09/21 00:00, 13F
→
09/21 00:00, , 14F
09/21 00:00, 14F
→
09/21 00:32, , 15F
09/21 00:32, 15F
推
09/21 00:39, , 16F
09/21 00:39, 16F
→
09/21 00:40, , 17F
09/21 00:40, 17F
→
09/23 20:45, , 18F
09/23 20:45, 18F