Show all your work, I WILL RATE!!
Consider the following code:
li $t0, 99
li $t1, 44
loop: addi $t0, $t0, -1
lw $t1, 0($t1)
bne $t0, $zero, loop
add $v0, $t0, $zero
What is the static instruction count of this code? What is the dynamic instruction count? (Assume any pseudoinstructions are counted as 1 instruction).
How many times is the instruction memory accessed? How many times is the data memory (i.e. lw/sw) accessed?
Answer:-------------
a). Static Instruction Count: 6 (Since there
are only 6 instructions in the code.)
Dynamic Instruction Count: (99 * 3) - 1 + 3 = 299 (The two li
instructions and the add instruction only executed once.
The loop itself executes 99 times but bne instructions
executes only 98 times since at the 99th iteration, $t0 = 0 and bne
instruction won't executed.)
b). Instruction Memory Accessed: 299 (Number of
times program counter updated)
Data Memory Accessed: 99 (Number of times lw or
load instruction was executed)
Show all your work, I WILL RATE!! Consider the following code: li $t0, 99 li $t1,...
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Question 11 The classic five-stage pipeline MIPS architecture is used to execute the code fragments in this problem. Assume the followings: • The architecture fully supports forwarding, • Register write is done in the first half of the clock cycle; register read is performed in the second half of the clock cycle, • Branches are resolved in the third stage of the pipeline and the architecture does not utilize any branch prediction mechanism, • Register R4 is initially 200....
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The classic five-stage pipeline MIPS architecture is used to
execute the code fragments in this problem. Assume the
followings:
The architecture fully supports forwarding,
Register write is done in the first half of the clock cycle;
register read is performed in the second half of the clock
cycle,
Branches are resolved in the third stage of the pipeline and
the architecture does not utilize any branch prediction
mechanism,
Register R4 is initially 100.
L1: lw R1,
0(R4)
add R3, R1, R2
sw ...
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