Answer:
Check the explanation
Explanation:
Below is the approx assembly code for above `for loop` :-
1). mov ecx, 0
2). loop_start :
3). cmp ecx, ARRAY_LENGTH
4). jge loop_end
5). mv temp_a, array[ecx]
6). cmp temp_a, 0
7). branch on nge
8). mv array[ecx], temp_a*2
9). add ecx, 1
10). jmp loop_start
11). loop_end :
Assumptions :-
*ARRAY_LENGTH is register with value 1000000
*temp_a is a register
Frequency of statements :-
1) will be executed one time
3) will be executed 1000000 times
4) will be executed 1000000 times
5) will be executed 1000000 times
6) will be executed 1000000 times
7) `nge` will be executed 1000000 times, branch will be executed 500000 times
8) will be executed 500000 times
9) will be executed 1000000 times
10) will be executed 1000000 times
Cost of statements :-
1) 10 ns
3) 10ns + 10ns + 10ns [for two register accesses and one cmp]
4) 10ns [for jge ]
5) 10ns + 100ns + 10ns [10ns for register access `ecx`, 100ns for memory access `array[ecx]`, 10ns for mv]
6) 10ns + 10ns [10ns for register_access `temp_a`, 10ns for mv]
7) 10ns for nge, 10ns for branch
8) 30ns + 110ns + 10ns
10ns + 10ns + 10ns for temp_a*2 [10ns for moving 2 into a register, 10ns for multiplication],
110ns for array[ecx],
10ns for mv
9) 10ns for add, 10ns for `ecx` register access
10) 10ns for jmp
Total time taken = sum of (frequency x cost) of all the statements
1) 10*1
3) 30 * 1000000
4) 10 * 1000000
5) 120 * 1000000
6) 20 * 1000000
7) (10 * 500000) + (10 * 1000000)
8) 150 * 500000
9) 20 * 1000000
10) 10 * 1000000
Sum up all the above costs, you will get the answer.
It will equate to 0.175 seconds
"