Consider the following fragment of C code: for (i=0; i<=100; i++) { a[i] = b[i]

Question

Consider the following fragment of C code:
for (i=0; i<=100; i++) {
a[i] = b[i] + C;
}
Assume that a and b are arrays of words and the base address of a is in $a0 and the base address of b is in $a1. Register $t0 holds the variable i and register $s0 the constant C. Write the code for this fragment in MIPS. How many total instructions are executed during the running of your code? How many memory data references will be made during execution? Typing your answer is recommended.

Add comments to each line of the following MIPS code. Then, describe what it computes (one or two sentences). Assume that $a0 is used for the input and initially contains n, a positive integer. Assume that $v0 is used for returning the output value.
begin: addi $t0, $zero, 0   
addi $t1, $zero, 1   
loop: slt $t2, $a0, $t1   
bne $t2, $zero, finish   
add $t0, $t0, $t1   
addi $t1, $t1, 2
j loop
finish:    add $v0, $t0, $zero

Explanation / Answer

MIPS CODE FOR LOOP SEGMENT

addi $t6, $zero, 101 # the loop termination value

add $t0, $zero, $zero # i = 0

addi $t2, $a0, 0 # ptr to current A[i]

addi $t3, $a1, 0 # ptr to current B[i]

loop: lw $t4, 0($t3) # load B[i]

add $t4, $t4, $s0 # B[i] + c

sw $t4, 0($t2) # store in A[i]

addi $t0, $t0, 1 # i++

addi $t2, $t2, 4 # ptr to next A[i]

addi $t3, $t3, 4 # ptr to next B[i]

bne $t0, $t6, loop # if i < 101, goto loop

The loop is executed 101 times and the loop contains 7 statements. There are 4 statements outside the loop. Therefore, the total number of instructions executed is 4 + (7*101) = 711.

The total number of memory data reference: 101 * 2 = 202.

Thank You

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