4. i. Translate the following loop into C. Assume that the C-level integer i is held...

 

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4. i. Translate the following loop into C. Assume that the C-level integer i is held in register X10, XO holds the C-level integer called result, and X1 holds the base address of the integer MemArray. (10pts) ORR X10, XZR, XZR LOOP: LDUR X11, [X1, #0] ADD XO, XO, X11 ADDI X1, X1, #8 ADDI X10, X10, #1 CMPI X10, 100 B.LT LOOP ii) Rewrite the loop to reduce the number of LEGv8 instructions executed. (10pts) Hint: Notice that variable i is used only for loop control. 4. i. Translate the following loop into C. Assume that the C-level integer i is held in register X10, XO holds the C-level integer called result, and X1 holds the base address of the integer MemArray. (10pts) ORR X10, XZR, XZR LOOP: LDUR X11, [X1, #0] ADD XO, XO, X11 ADDI X1, X1, #8 ADDI X10, X10, #1 CMPI X10, 100 B.LT LOOP ii) Rewrite the loop to reduce the number of LEGv8 instructions executed. (10pts) Hint: Notice that variable i is used only for loop control.