Question

Question: Translate the following C code to MIPS. Assume that the variables f, g, h, i , and j are assigned to registers \(s0,\)s1,\(s2,\)s3, and \(s4, respectively. Assume that the base address of the arrays A and B are in registers \)s6 and $s7 respectively. Assume that the elements of the arrays A and B are 4 byte words:

B[8]=A[i]+A[j];

Short answer

MIPS code:

sll $t0,$s1,2

add $t0,$t0,$s7

lw $t0, 0($t0)

addi $t0,$t0,1

sll $t0, $t0,2

lw $s0,0($t0)

Step-by-step solution

Define MIPS assembly language.

The MIPS assembly language is the instruction-based programming language. MIPS assembly language uses multiple instructions to perform an operation.

MIPS has operands for 32 registers and in MIPS data must be in registers to perform operations on them.

Some example instructions are as follows:

add $s1,$s2,$s3

The above instruction will add the values in the registers s2 and s3. The result will be stored in the s1 register.

Likewise, MIPS performs all the operations through the registers and memory locations.

Determine the MIPS code for the given C statement.

The given C statement:

The registers, $s0, $s1, $s2, $s3, and $s4 are assigned with f, g, h, i, and j respectively.

The base address of the arrays A and B are stored in the registers $s6 and $s7. The elements of the arrays are 4 byte words.

With above information, the MIPS code for the given C statement is as follows:

sll $t0,$s1,2

add $t0,$t0,$s7

lw $t0, 0($t0)

addi $t0,$t0,1

sll $t0, $t0,2

lw $s0,0($t0)

Instruction	Explanation
sll $t0,$s1,2	Shift left logical instruction will shift 4 spaces in $s1 register that is g. $t0🡨-4*g
add $t0,$t0,$s7	The values of $t0 and $s7 will be added and the result will be stored in $t0.The $t0 will have the address of array B with value g. $t0🡨Address of B[g]
lw $t0,0($t0)	Now, the value of B[g] will be loaded to $t0. $t0🡨B[g]
addi $t0,$t0,1	The value of B[g] will be added with the immediate 1. $t0🡨B[g]+1
sll $t0,$t0,2	Now the logical shift of 4 will be pointed to B[g]+1 and the address of B[g]+1 will be assigned.
lw $s0,0($t0)	The f will be assigned with the value of A[B[g]+1]