Compiling MIPS Code Give the shortest sequence of MIPS assembler instructions to
ID: 3600900 • Letter: C
Question
Compiling MIPS Code
Give the shortest sequence of MIPS assembler instructions to implement the C code
for (i=0; i<=10; i=i+1)
for (j=0; j<=10; j=j+1)
a[i][j] = b[j][i] + c;
Assume that a and b are 11 x 11 arrays of word values (integers), and that the base address of a is in $a0 and the base address of b is in $a1. Register $t0 holds the variable i (that has been initialized to zero prior to entering your code). Register $t1 holds the variable j (that has been initialized to zero) and register $s0 holds the byte variable c. The constant 11 has been loaded into register $s1. (The only pseudoinstruction you can use is mul dest src1 src2).
Outer: slt $t2, $t0, $s1
beq $t2, $zero, exit # 0<=i<=10
inner: slt $t2, $t1, $s1
beq $t2, $zero, end_inner # 0 <= j <= 10
add $t2, $t0, $t0 # 2*i
add $t2, $t2, $t2 # 4*i
mul $t3, $t1, $s1 # 11*j
add $t3, $t3, $t3 # 22*j
add $t3, $t3, $t3 # 44*j
add $t3, $t3, $a1 # b[j][i]
lw $t4, 0($t3) # load b[j][i]
add $t4, $t4, $s0 # b[j][i]+c
add $t2, $t1, $t1 # 2*j
add $t2, $t2, $t2 # 4*j
mul $t3, $t0, $s1 # 11*i
add $t3, $t3, $t3 # 22*i
add $t3, $t3, $t3 # 44*i
add $t3, $t3, $a0 # a[i][j]
sw $t4, 0($t3) # a[i][j] = b[j][i] + c
addi $t1, $t1, 1 # j ++
j inner
end_inner: addi $t0, $t0, 1 # i ++
j outer
How can this code be changed to just have a one dimensional array. ( a[i] = b[i] + c )
Explanation / Answer
slt $t1,$t0,$s1
beq $t1,$0,exit
inner:
slt $t1,$t0,$s1
beq $t1,$zero,end-inner
add $t1,$t0,$t0
add St1,$t1,$t1
add $t2,$t2,$a1
lw $t3,0($t2)
add $t3,$t3,$s0
mul $t2,$t0,$s1
add $t2,$t2,$t2
add $t2,$t2,$t2
add $t2,$t2,$a0
sw $t3,0($t2)
addi $t0,$t0,1
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.