使用循环的Python程序

Question

Given this Python program:

  num = input("Enter a number: ")
  result = 1024
  for i in range(num):
     result = result / 2
  print result

If the number you enter is 4, why is the output of this program 64?

Answer 1

Trace through the program to see what happens. range(num) here is range(4) , which gives the values 0, 1, 2, and 3.

When i = 0, we divide 1024 by 2 to get 512.

When i = 1, we divide 512 by 2 to get 256.

When i = 2, we divide 256 by 2 to get 128.

When i = 3, we divide 128 by 2 to get 64.

And voila! There's your 64.

More generally, each iteration of the loop will divide result by 2, so after num iterations of the loop, result will be 1024 / 2 ^num . Since 1024 = 2 ¹⁰ , this means that the result is 2 ¹⁰ / 2 ^num = 2 ^{10 - num} . That said, if num > 10 , because result is an integer, Python will round down to zero. In other words:

• If the number entered is negative, range(num) is the empty range and the program prints 1024.
• If the number entered is in the range of 0 to 10, the result is 2 ^{10 - num} .
• If the number entered is greater than 10, the result is 0.

Hope this helps!

Answer 2

You can see what's happening by simply adding some debug statements to your code:

num = input("Enter a number: ")
result = 1024
print "Starting result %d"%(result)
print range(num)
for i in range(num):
    result = result / 2
    print "Looping result %d"%(result)
print result

If you run that and enter 4, you'll see:

Enter a number: 4
Starting result 1024
[0, 1, 2, 3]
Looping result 512
Looping result 256
Looping result 128
Looping result 64
64

The reason is that range(4) gives you the list [0,1,2,3] with four elements in it, so that's how many times the body of the loop is being executed.

On each execution of the body, you simply halve the current value of the result:

iteration 1, 1024 -> 512
iteration 2,  512 -> 256
iteration 3,  256 -> 128
iteration 4,  128 ->  64