Comp 303 Computer Architecture A Pipelined Datapath Control. Lecture 13

Transcription

1 Comp 33 Compter Architectre A Pipelined path Lectre 3

2 Pipelined path with Signals PCSrc IF/ ID ID/ EX EX / E E / Add PC 4 Address Instrction emory RegWr ra rb rw Registers bsw [5-] [2-6] [5-] bsa bsb Sign etent Shift left 2 Src 6 Add RegDst Eqal op Branch Addr. ot emory in emrd emwr emtoreg

3 Recall: Bits Instrction opcode op Instrction operation Fnction field Desired action control LW load word XXXXXX add SW store word XXXXXX add beq branch eqal XXXXXX sbtract R-type add add R-type sbtract sbtract R-type AND and R-type OR or R-type set on less than set on less than opcode 6 ain fnc 6 op 2 (Local) ctr 3

4 The Vales of Lines for The Last Three Pipeline Stages Eection / Address Calclation stage control lines emory Access stage control lines Write Back stage control lines Instrctions Reg Dst Op Op Src Branch em Rd em Wr Reg Wr em to Reg R-type lw sw X X Beq X X Instrction EX IF/ ID ID/ EX EX / E E /

5 The Pipelined path with Signals IF/ ID ID/ EX EX / E E / EX Add PC 4 Address Instrction emory RegWr ra rb rw Registers bsw [5-] [2-6] [5-] bsa bsb Sign etent Shift left 2 Src Add Eqal op RegDst Branch Addr. ot emory in emrd emwr emtoreg

6 An Eample to Clarify Pipelined Let s look at what s happing in the pipeline for the following program. lw $, 2($) sb $, $2, $3 and $2, $4, $5 or $3, $6, $7 add $4, $8, $9 Code does not have any data, control, or strctral hazard.

7 IF: lw $,2($) IF/ ID ID/ EX EX / E E / EX Add PC 4 Address Instrction emory Clock RegWr ra rb rw Registers bsw [5-] [2-6] [5-] bsa bsb Sign etent Shift left 2 Src Add Eqal op RegDst Branch Addr. ot emory in emrd emwr emtoreg

8 IF: sb $,$2,$3 ID: lw $,2($) IF/ ID ID/ EX EX / E E / lw EX Add PC 4 Address Instrction emory Clock 2 X 2 X RegWr ra rb rw Registers bsw [5-] [2-6] [5-] $ bsa $X bsb Sign etent 2 X Shift left 2 Src Add Eqal op RegDst Branch Addr. ot emory in emrd emwr emtoreg

9 IF: and $2,$4,$5 ID: sb $,$2,$3 EX: lw $,2($) IF/ ID ID/ EX EX / E E / sb EX Add PC 4 Address Instrction emory Clock X X RegWr ra rb rw Registers bsw [5-] [2-6] [5-] $2 bsa bsb Sign etent $3 X X $ 2 Shift left 2 Src Add Eqal op RegDst Branch Addr. ot emory in emrd emwr emtoreg

10 IF: or $3,$6,$7 ID: and $2,$4,$5 EX: sb $,$2,$3 E: lw $,2($) IF/ ID ID/ EX EX / E E / and EX Add PC 4 Address Instrction emory Clock X X 2 RegWr ra rb rw Registers bsw [5-] [2-6] [5-] $4 bsa $5 bsb Sign etent X X 2 $2 $3 Shift left 2 Src Add Eqal op RegDst Branch Addr. ot emory in emrd emwr emtoreg

11 IF: add $4,$8,$9 ID: or $3,$6,$7 EX: and $2,$4,$5 E: sb $,$2,$3 : lw $ IF/ ID ID/ EX EX / E E / or EX Add PC 4 Address Instrction emory Clock X X 3 RegWr ra rb rw Registers bsw [5-] [2-6] [5-] $6 bsa $7 bsb Sign etent X X 3 $4 $5 2 Shift left 2 Src Add Eqal op RegDst Branch Addr. ot emory in emrd emwr emtoreg

12 Hazards Previos eample shows s how independent instrctions that do not se the reslts calclated by prior instrctions are eected. This is not the case with real programs. Let s look at the following code seqence. sb $2, $, $3 and $2, $2, $5 or $3, $6, $2 add $4, $2, $2 sw $5, ($2) The last for instrctions are all dependent on the register $2 of the first instrction. Assme that register $2 had the vale of before the sbtract instrction and -2 afterwards.

13 Hazards (Cont ) CC CC2 CC3 CC4 CC5 CC6 CC7 CC8 CC9 The vale of $2: / sb $2, $, $3 and $2, $2, $5 or $3, $6, $2 add $4, $2, $2 sw $5, ($2)

14 Hazards (Cont ) Simple soltion: Compiler inserts nop instrctions between the sb-and instrctions. nop (no operation) instrction neither modifies data nor writes a reslt. sb $2, $, $3 nop nop and $2, $2, $5 or $3, $6, $2 add $4, $2, $2 sw $5, ($2) Reslt: It works bt 2 clock cycles will be wasted. Performance decrease

15 Hazard Detection & Forwarding It is possible to detect data hazard and then forward the proper vale to resolve the hazard. When an instrction tries to read a register in its EX stage that an earlier instrction intends to write in its stage. This is the case between sb-and instrction below: sb $2, $, $3 and $2, $2, $5 This hazard can be detected by simply checking: EX/E.RegisterRd = ID/EX.RegisterRs = $2

16 Hazard Detection & Forwarding (Cont ) Another hazard is between sb-or instrctions: sb $2, $, $3 and $2, $2, $5 or $3, $6, $2 This hazard can be detected by simply checking: E/.RegisterRd = ID/EX.RegisterRt = $2 There is no data hazard between sb-add and sb-sw instrctions.

17 Smmary of Hazard Conditions a. EX/E.RegisterRd = ID/EX.RegisterRs b. EX/E.RegisterRd = ID/EX.RegisterRt 2a. E/.RegisterRd = ID/EX.RegisterRs 2b. E/.RegisterRd = ID/EX.RegisterRt This actally refers to destination field of an instrction. It is rd field in R-type instrctions and rt field in I-type instrctions. in the EX stage chooses the correct one, therefore, EX/E and E/ pipeline registers store this information as a rd field (EX/E.RegisterRd and E/.RegisterRd). Since some of the instrctions (i.e. sw, beq) do not write to register file, the above policy is inaccrate. Consider the following code seqence: sw $, ($5) add $3, $, $2 EX/E.RegisterRd = ID/EX.RegisterRs $ = +$5 This problem can be solved simply by checking RegWr signal.

18 Smmary of Hazard Conditions (Cont ) Another problem: What happens if $ is sed as a destination register? A non-zero vale wold not be forwarded. Therefore, hazard detection shold be the following: EX hazard: if (EX/E.RegWr and (EX/E.RegisterRd = ) and (EX/E.RegisterRd = ID/EX.RegisterRs)) ForwardA= if (EX/E.RegWr and (EX/E.RegisterRd = ) and (EX/E.RegisterRd = ID/EX.RegisterRt)) ForwardB= E hazard: if (E/.RegWr and (E/.RegisterRd = ) and (E/.RegisterRd = ID/EX.RegisterRs)) ForwardA= if (E/.RegWr and (E/.RegisterRd = ) and (E/.RegisterRd = ID/EX.RegisterRt)) ForwardB=

19 Smmary of Hazard Conditions (Cont ) Consider the following seqence: add $, $, $2 add $, $, $3 add $, $, $4... In this case, the reslt is forwarded from E stage becase the reslt in the E stage is the more recent reslt than the reslt in stage. Ths, the control for the E hazard: E hazard: if (E/.RegWr and (E/.RegisterRd = ) and (EX/E.RegisterRd = ID/EX.RegisterRs) and (E/.RegisterRd = ID/EX.RegisterRs)) ForwardA= if (E/.RegWr and (E/.RegisterRd = ) and (EX/E.RegisterRd = ID/EX.RegisterRt) and (E/.RegisterRd = ID/EX.RegisterRt)) ForwardB=

20 Smmary of Hazard Conditions (Cont )

21 The Pipelined path with Forwarding ID/ EX EX / E E / IF/ ID EX PC emory Registers IF/ID.RegisterRs Rs emory IF/ID.RegisterRt Rt IF/ID.RegisterRt IF/ID.RegisterRd Rt Rd EX/E.RegisterRd Forwarding nit E/.RegisterRd

22 Signed mediate Enhancement

23 Hazards and Stalls Program eection order (in instrctions) Time (in clock cycles) CC CC2 CC3 CC4 CC5 CC6 CC7 CC8 CC9 lw $2, 2($) Forwarding does not solve the problem. We need a hazard detection nit. and $4, $2, $5 and instrction needs to be stalled one cycle. or $8, $2, $6 add $9, $4, $2 slt $, $6, $7

24 Hazard Detection The control for the hazard detection nit is: if (ID/EX.emRd and ((ID/EX.RegisterRt = IF/ID.RegisterRs) or (ID/EX.RegisterRt = IF/ID.RegisterRt))) stall the pipeline Checks if the instrction is a load

25 Hazards and Stalls (Cont ) Program eection order (in instrctions) Time (in clock cycles) CC CC2 CC3 CC4 CC5 CC6 CC7 CC8 CC9 CC lw $2, 2($) and $4, $2, $5 Reg Reg Dm Reg or $8, $2, $6 add $9, $4, $2 Bbble slt $, $6, $7 and and or instrctions repeat in CC4 what they did in CC3

26 Hazards and Stalls (Cont )

27 The Pipelined path with Forwarding and Hazard Detection Unit IF/IDWrite Hazard detection nit ID/EX.emRead ID/ EX EX / E E / PCWrite IF/ ID EX PC emory Registers IF/ID.RegisterRs Rs emory IF/ID.RegisterRt Rt IF/ID.RegisterRt IF/ID.RegisterRd Rt Rd EX/E.RegisterRd IF/ID.RegisterRt Forwarding nit E/.RegisterRd

28 Branch Hazards Program eection order (in instrctions) Addr. 4 beq $, $3, 7 Time (in clock cycles) CC CC2 CC3 CC4 CC5 CC6 CC7 CC8 CC9 44 and $2, $2, $5 48 or $3, $6, $2 52 add $4, $2, $2 72 lw $4, (5)$7 Actally, the nmber of instrctions needs to be flshed can be redced from 3 to instrction (shown in the following slide) when the direction of branch is mispredicted.

29 path for Branch (inclding HW to flsh the pipeline) 4 + IF/ ID Hazard detection nit Sign_e & shift left 2 + ID/ EX EX EX / E E / PC emory Registers = IF/ID.RegisterRs Rs emory IF/ID.RegisterRt Rt IF.Flsh IF/ID.RegisterRt IF/ID.RegisterRd Rt Rd EX/E.RegisterRd IF/ID.RegisterRt Forwarding nit E/.RegisterRd

30 Reading Assignment Read Chapter 4 Find the error in one of the Hazard Detection Fnctions in the chapter.