Question

please trace this is sort program and base address s0, number of element s1

please also complete the program(..)

explain the program too thanks

sit $t4, $t1, $to beq $t4, $zero, Skip sll $s1, $s1, 2 add $s2, $s0, $s1 addi $t2, $so, 0 Skip: oopl: addi $t3, $t2, 0 oopJ: addi $t3, $t3, 4 beq $t3, $s2, NextI Iw $to, 0($t2) lw NextI: addi $t2, $t2, 4 End: 配列の昇順ソートベースアドレス$50, 空欄を埋めてプログラムを完成させよ。 要素数$51

Answer 1

Complete sorting program along with comment :- (it's actually MIPS implementation of bubble sort)

sll $s1, $s1, 2       #count the offset value of the last element
add $s2, $s0, $s1   #store the address of end of the array
addi $t2, $s0, 0   #store the starting address of the array

LoopI:                        #Outside loop
   addi $t3, $t2, 0       #store the starting address of the array
LoopI:                        #Inside loop
   addi $t3, $t3, 4       #poiniting to next element address
   beq $t3, $s2, NextI   #condition for checking the end of the array
   lw $t0, 0($t2)       #loading the value at address pointed by $t2
   lw $t1, 0($t3)       #loading the value at address pointed by $t3
   slt $t3, $t1, $t0       #comparing two values; if value of $t1 is less than set value in $t4
   beq $t4, $zero,Skip   #checking for value in $t4 is set or not;if not then swap the values in memory
                           #otherwise Skip the swapping part
                           #swapping the values
   sw $t0, 0($t3)
   sw $t1, 0($t2)
Skip:
   j LoopJ                   #jumping back to inner loop
NextI:
   addi $t2, $t2, 4       #pointing to next element
   bne $t2, $s2, LoopI   #if not the end of the array; jump back to LoopI
End: