Consider the following code segment. CLR MOV.W MOV.W 01 02 03 04 05 1nxt: 06 07...

   

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Consider the following code segment. CLR MOV.W MOV.W 01 02 03 04 05 1nxt: 06 07 MOV.B BIT.W JNZ INC 08 1skip: RRA.W 09 DEC.B JNZ MOV. B PPO OHNM 10 11 R12 mylw, R6 #1, R5 #16, R7 R5, R6 lskip R12 R6 R7 Inxt R12, P2OUT 12 13 mylw: .long 0x8C9E2A7D " 30 30 30 30 30 (15 points) What does this code segment do? Explain your answer. 3.D. (5 points) Calculate the total execution time in seconds for the code sequence from above (line 01 - line 11). We know the following: the average CPI is 2 clocks per instruction. Assume the clock frequency is 1 MHz. What is MIPS rate for this code? Consider the following code segment. CLR MOV.W MOV.W 01 02 03 04 05 1nxt: 06 07 MOV.B BIT.W JNZ INC 08 1skip: RRA.W 09 DEC.B JNZ MOV. B PPO OHNM 10 11 R12 mylw, R6 #1, R5 #16, R7 R5, R6 lskip R12 R6 R7 Inxt R12, P2OUT 12 13 mylw: .long 0x8C9E2A7D " 30 30 30 30 30 (15 points) What does this code segment do? Explain your answer. 3.D. (5 points) Calculate the total execution time in seconds for the code sequence from above (line 01 - line 11). We know the following: the average CPI is 2 clocks per instruction. Assume the clock frequency is 1 MHz. What is MIPS rate for this code?

Answer rating: 100% (QA)

The given code appears to be written for an assembly language which is a lowlevel programming language typically used to write programs that are very close to the computer hardware Lets first analyze ... View the full answer