Question

Please need help understanding how this code works

a. Analyze the following ode and find out what exactly it is doing. Explaining what each MIPS instruction does literally is not enough, you need to explain how it contributes to the overall logic. (4 Marks) adddu $t1, St2, $t3 nor $t4, St2, Szero sltu St5, St4, St3 bne St5, Szero, exception1_processing What is the specific exception that is being processed at the label exception1_processing? b. Indicate with a diagram how a very simplified implementation of the same logic is possible in hardware. (1 Mark)

Answer 1

Please find the answer below.
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code execution is as below

2 addu $t1,$t2, $t3 #add both numbers 3 nor $t4, $t2, $zero#get once compliment of ti 4. sltu $t5,$t4, $t3 #now check if once compliment of ti is less than t3 5 bne $t5, $zero, exception1_processing #if yes through exception 6 nộp 7 exception1_processing: 8

addu $t1,$t2,$t3 #add both numbers
nor $t4,$t2,$zero#get once compliment of t1
sltu $t5,$t4,$t3 #now check if once compliment of t1 is less than t3
bne $t5,$zero,exception1_processing #if yes through exception
nop
exception1_processing:

Hence the given code is checking if adding t2 and t3 causes over flow or not. If it will cause , Than it will throw exception1_processing.

And exception would be . . Overflow occured.