如何打破无限的 Python 循环？

Question

我正在寻找修复以下代码，以便它可以成功完成列出的所有数字。 我怎样才能打破存在的无限循环？ 此外，如何将 print(is_power_of_two(8)) 返回为 True？

  # Check if the number can be divided by two without a remainder
  while n % 2 == 0:
    n = n / 2
  # If after dividing by two the number is 1, it's a power of two
  if n == 1:
    return True
  return False


print(is_power_of_two(0)) # Should be False
print(is_power_of_two(1)) # Should be True
print(is_power_of_two(8)) # Should be True
print(is_power_of_two(9)) # Should be False

Answer 1

将and n != 0添加到 while 条件，因此当 n 为零时循环将停止。

Answer 2

def is_power_of_two(n): # Check if the number can be divided by two without a remainder while n!=0 and (n % 2 == 0 or 2%n==0): n = n / 2 return True return False

Answer 3

import math    
math.log2(number).is_integer()

用这个。 从检查给定的数字是否是 Python 中的 2 的幂

Answer 4

你的while条件应该是：

# Check if the number can be divided by two without a remainder
  while n % 2 == 0 and n >1:
    n = n / 2

暂时导致您的代码在 n=1 时无限循环导致 1%2 = 1

Answer 5

def is_power_of_two(n):

 #Check Number for Zero
  if n==0:
    return False
  else:
    # Check if the number can be divided by two without a remainder
    while n % 2 == 0:
      n = n / 2 
    # If after dividing by two the number is 1, it's a power of two
    if n == 1:
      return True
    return False

Answer 6

def is_power_of_two(n):
  # Check if the number can be divided by two without a remainder
  while n % 2 == 0 and n!=0:
    n = n / 2
  # If after dividing by two the number is 1, it's a power of two
  if n == 1:
    return True
  return False
  

print(is_power_of_two(0)) # Should be False
print(is_power_of_two(1)) # Should be True
print(is_power_of_two(8)) # Should be True
print(is_power_of_two(9)) # Should be False

Answer 7

While 循环

使 print_prime_factors 函数打印一个数字的所有质因数。 质因数是质数并能整除另一个数而没有余数。