Assume that k corresponds to register $s0, n corresponds to register $s2 and the base of the array v is in $s1. What is the MIPS assembly code corresponding to this C segment

hello your question lacks the C segment so here is the C segment

while ( k<n )

{v[k] = v[k+1];

     k = k+1; }

Answer : while:

   bge $s0, $s2, end   # while (k < n)

   addi $t0, $s0, 1    # $t0 = k+1

   sll $t0, $t0, 2     # making k+1 indexable

   add $t0, $t0, $s1   # $t0 = &v[k+1]

   lw $t0, 0($t0)      # $t0 = v[k+1]

   sll $t1, $s0, 2     # making k indexable

   add $t1, $t1, $s1   # $t1 = &v[k]

   sw $t0, 0($t1)      # v[k] = v[k+1]

   addi $s0, $s0, 1

   j while

end:

Explanation:

The MIPS assembly code corresponding to the C segment is

while:

   bge $s0, $s2, end   # while (k < n)

   addi $t0, $s0, 1    # $t0 = k+1

   sll $t0, $t0, 2     # making k+1 indexable

   add $t0, $t0, $s1   # $t0 = &v[k+1]

   lw $t0, 0($t0)      # $t0 = v[k+1]

   sll $t1, $s0, 2     # making k indexable

   add $t1, $t1, $s1   # $t1 = &v[k]

   sw $t0, 0($t1)      # v[k] = v[k+1]

   addi $s0, $s0, 1

   j while

end:

that without the operating systems you would have to control and synchronize computer hardware.

solution:

this situation is called an ip conflict.

data from a server is sent in small bits called packets. in an ip conflict, some packets will end up in one computer while some other packets will end up in another computer.

so…overall none of the computers get the data as the original message or picture being sent by the server cannot be retrieved with only half the packets. so it will show up as an error.

thus the final answer is d.