Lecture VII: The debugger gdb and Branching. Xuan Guo. CSC 3210 Computer Organization and Programming Georgia State University.

Transcription

1 Lecture VII: The debugger gdb and Branching CSC 3210 Computer Organization and Programming Georgia State University February 3, 2015

2 This lecture Plan for the lecture: Recap: Pipelining Filling delay slot Example The Debugger gdb Branching Examples

3 Instruction Cycle von Neumann cycle broken into 4 machine cycles: Instruction fetch Execute Memory access Store results F E M W Fetch & decode instruction Execute arithmetic instruction Access memory for a load or store instruction Write instruction results back to register file obtain any operands compute branch target address fetch instruction at target of branch instruction compute memory address

4 Pipelining.global main main: 1. mov 2, %l0 2. mov 3, %l1 3. add %l0, %l1, %l2 4. mov 1, %g1 5. ta 0 load the instruction indicated by main Pipelining (Stages) F E M W Sequential: each h/w stage idle 75% of the time. time ex = 4 * i Parallel: each h/w stage working after filling the pipeline. time ex = 3 + i Two types of delay slots

5 Data Dependencies Load Delay Problem load [%o0], %o1 add %o1, %o2, %o

6 Branch Delay Problem call.mul add %o0, 2, %l0 Insert branch delay slot instruction (Ex: nop) does nothing to change the state of the machine

7 pc & npc

8 Without nop (no operation) 1. mov 2, %o0 2. mov 3, %o1 3. call.mul 4. mov %o0, %l0 PC A: PC = NPC 1 1 B: 1 2 C: PC->execute 1 2 A 2 2 B 2 3 C 2 3 A 3 3 B 3 4 C 3 4 A 4 4 NPC B 4 mul C 4 mul

9 Without nop 1. mov 2, %o0 2. mov 3, %o1 3. call.mul 4. mov %o0, %l0 PC A: PC = NPC 1 1 B: 1 2 C: PC->execute 1 2 A 2 2 B 2 3 C 2 3 A 3 3 B 3 4 C 3 4 A 4 4 NPC B 4 mul C 4 mul Incorrect the instruction 4 has been execute before get the results from multiplication

10 With nop 1. mov 2, %o0 2. mov 3, %o1 3. call.mul 4. nop 5. mov %o0, %l0 PC A: PC = NPC 1 1 B: 1 2 C: PC->execute 1 2 A 2 2 B 2 3 C 2 3 A 3 3 B 3 4 C 3 4 A 4 4 NPC B 4 mul C 4 mul

11 With nop 1. mov 2, %o0 2. mov 3, %o1 3. call.mul 4. nop 5. mov %o0, %l0 PC A: PC = NPC 1 1 B: 1 2 C: PC->execute 1 2 A 2 2 B 2 3 C 2 3 A 3 3 B 3 4 C 3 4 A 4 4 NPC B 4 mul C 4 mul correct nop just do nothing

12 Filling Delay Slots delayed instruction: instruction in the delay slot caused by delayed control transfer instruction, ex. call waste one cycle of instruction by filling in nop move one instruction to fill delay slot Tip: for.mul and.div, move the instruction setting argument for multiplication or division to fill in delay slot fill all possible delay slots when writing assembly language programs

13 main: y = (x-1)*(x-7)/(x-10) if x = 9.global main save %sp, -96, %sp mov 9, %l0 sub %l0, 1, %o0 sub %l0, 7, %o1 call.mul nop sub %l0, 7, %o1 call.div nop mov %o0, %l1 mov 1, %g1 ta 0

14 The debugger gdb program name simple gdb simple debugging program simple gdb r run the program gdb b *address create a breakpoint, ex. gdb main gdb ni run next step gdb p $l0 show the contents of register %l0 gdb q quit gdb gdb x/i $pc gdb x/i $npc gdb command n gdb command 1 gdb command 2

15 main: (x-1)*(x-7)/(x-10) if x = 9.global main save %sp, -96, %sp mov 9, %l0 sub %l0, 1, %o0 call.mul sub %l0, 7, %o1 call.div sub %l0, 10, %o1 mov %o0, %l1 mov 1, %g1 ta 0

16 Branching Assembler Mnemonic ba bn Unconditional Branches Branch always, goto Branch never Assembler Mnemonic Unconditional Branches Complement bl Branch on less than zero bge ble Branch on less or equal to zero bg be Branch on equal to zero bne bne Branch on not equal to zero be bge Branch on greater or equal to zero bl bg Branch on greater than zero ble

17 Branching Testing store state of execution: conditional codes Z: whether the result was zero N: whether the result was negative V: whether execution resulted in a number too large to store in the register C: whether execution resulted in a number that generated a carry out of register addcc subcc cmp reg, reg_or_imm, reg reg, reg_or_imm, reg reg, reg_or_imm subcc reg, reg_or_imm, %g0

18 Branching also delayed control transfer instructions delayed instruction will be executed before branch takes place test the condition codes b icc label b icc stands for one of the branches testing the integer condition codes

19 Example y = (x 1)/(x 10) for 0 <= x <= 9 C program: int x; int y; for(x = 0; x<=9; x++){ y = (x-1)/(x-10); } x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 y = 0, 0, 0, 0, 0, 0, -1, -2, -3, -8

20 .file "ex3.c".global.div.section ".text".align 4.global main.type main, #function.proc 04 main:!#prologue# 0 save %sp, -120, %sp!#prologue# 1 mov 2, %g1 st %g1, [%fp-20].ll2: ld [%fp-20], %g1 cmp %g1, 9 bg.ll3 nop ld [%fp-20], %g1 add %g1, -1, %o5 ld [%fp-20], %g1 add %g1, -10, %g1 mov %o5, %o0 mov %g1, %o1 call.div, 0 nop mov %o0, %g1 st %g1, [%fp-24] ld [%fp-20], %g1 add %g1, 1, %g1 st %g1, [%fp-20] b.ll2 nop.ll3: mov %g1, %i0 ret restore.size main,.-main.ident "GCC: (GNU) 3.4.6"

21 Example y = (x 1)/(x 10) for 0 <= x <= 9.global main main: mov 0, %x_r ba test loop: sub %x_r, 1, %o0 sub %x_r, 10, %o1 call.div nop mov %o0, %y_r add %x_r, 1, %x_r test: cmp %x_r, 10 bl loop nop C program: for(x = 0; x<=9; x++){ } y = (x-1)/(x-10); x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 y = 0, 0, 0, 0, 0, 0, -1, -2, -3, -8

22 Example y = (x 1)/(x 10) for 0 <= x <= 9.global main main: mov 0, %x_r ba test loop: sub %x_r, 1, %o0 call.div sub %x_r, 10, %o1 mov %o0, %y_r add %x_r, 1, %x_r test: cmp %x_r, 10 bl loop nop y = (x 1)/(x 10) for 0 <= x <= 9.global main main: mov 0, %x_r ba test loop: sub %x_r, 1, %o0 sub %x_r, 10, %o1 call.div nop mov %o0, %y_r add %x_r, 1, %x_r test: cmp %x_r, 10 bl loop nop

23 Example y = (x 1)/(x 10) for 0 <= x <= 9.global main main: mov 0, %x_r ba test loop: call.div sub %x_r, 10, %o1 mov %o0, %y_r add %x_r, 1, %x_r test: cmp %x_r, 10 bl loop sub %x_r, 1, %o0 y = (x 1)/(x 10) for 0 <= x <= 9.global main main: mov 0, %x_r ba test loop: sub %x_r, 1, %o0 call.div sub %x_r, 10, %o1 mov %o0, %y_r add %x_r, 1, %x_r test: cmp %x_r, 10 bl loop nop