For the MIPS assembly instructions below, what is the corresponding C statement? 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. sll $t0, $s0, 2 # $t0 = f * 4add $t0, $s6, $t0 # $t0 = &A[f]sll $t1, $s1, 2 # $t1 = g * 4add $t1, $s7, $t1 # $t1 = &B[g]lw $s0, 0($t0) # f = A[f]addi $t2, $t0, 4lw $t0, 0($t2)add $t0, $t0, $s0sw $t0, 0($t1)

Answer / Explanation:

Before we answer this question, let us try to understand the meaning of some basic terms as this we help us in better understand the answer provided:

MIPS: This can be refereed to as the assembly language assigned to MIPS processor.

The abbreviated word MIPS means Microprocessor without Interlocked Pipeline Stages and it is a reduced-instruction set architecture.

Now, going back to the question, the answer goes thus:

sll $t0, $s0, 2          // $t0 = f * 4;

add $t0, $s6, $t0   // $t0 = &A[f];

sll $t1, $s1, 2           // $t1 = g * 4;

add $t1, $s7, $t1     // $t1 = &B[g];

lw $s0, 0($t0)         // f = A[f];

addi $t2, $t0, 4       // $t2 = &A[f + 1];

We therefore note that:

                                          // Note here f is still of the original value.

w $t0, 0($t2)                     // $t0 = A[f + 1];

add $t0, $t0, $s0             // $t0 = A[f + 1] + A[f]; // Here assumes a nop.

sw $t0, 0($t1)                    // B[g] = A[f + 1] + A[f];

Now, considering for the overall register:

Overall:

B[g] = A[f + 1] + A[f];

f = A[f];

It can now therefore be concluded that the two statements in the narrative in the question cannot swap order.

brought all tribes under one rule, created first mongol written language, legal code, strong conquest