Consider the Memory layout: int a, b, c, d[10] in a program's data segment.

Convert the high level language statement below to MIPS assembly:

d[3] = d[2] + a;

Let $s1 contain the base address of the memory segment.

Let $s4 = starting memory address of the d array.

	A.	lw $t0, 0($s4) lw $t1, 0($s1) add $t0, $t0, $t1 sw $t0, 4($s4)
	B.	lw $t0, 8($s4) lw $t1, 0($s1) add $t0, $t0, $t1 sw $t0, 12($s4)
	C.	lw $t0, 2($s4) lw $t1, 0($s1) add $t0, $t0, $t1 sw $t0, 3($s4)
	D.	lw $t0, 4($s4) lw $t1, 0($s1) add $t0, $t0, $t1 sw $t0, 8($s4)

Consider the Memory layout: int a, b, c, d[10] in a program's data segment. Convert the high level language statement below to MIPS assembly: d[3] = d[2] + a; Let $s1 contain the base address of the memory segment. Let $s4-starting memory address of the d array. lw $t1, 0($s1) add $t0, $t0, $t1 sw $to, 4($s4) lw $t1, 0($s1) add $t0, $to, $t1 sw $t0, 12($s4) Iw $t1, 0($s1) add $t0, $t0, $t1 sw $to, 3($s4) lw $t1, 0($s1) add $t0, $t0, $t1 sw $to, 8($s4)