Question

Question:

Consider the following C language instruction.

A[10] = ((f+g) – (h+A[5])) + 100;

Translate the above code into MIPS assembly language code. Assume variables f, g, and h are associated

with registers $S0, $S1, and $S2 respectively. The base address of A is associated with register $S5.

Answer 1

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code snippet

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.text
#f is in $s0
#g is in $s1
#h is in $s2
#Base address of A is in $s5

#get f+g
add $t0,$s0,$s1

#get A[5] using 20, Because int is of 4 byte in size
#hence 5th element will be at 20th position
lw $t1,20($s5)

#get h+A[5]
add $t1,$t1,$s2

#get f+g - (h+A[5])
sub $t0,$t0,$t1

#get f+g - (h+A[5])+100
add $t0,$t0,100

#now store the result at location A[10]
sw $t0,40($s5)