Python入门练习 - Debugging

  Problem 1

Consider the following function definition:

def integerDivision(x, a):
    """
    x: a non-negative integer argument
    a: a positive integer argument

    returns: integer, the integer division of x divided by a.
    """
    while x >= a:
        count += 1
        x = x - a
    return count

When we call

print integerDivision(5, 3)

we get the following error message:

Traceback (most recent call last):
  File "L7_Problem_5.py", line 15, in <module>
    print integerDivision(5, 3)
  File "L7_Problem_5.py", line 9, in integerDivision
    count += 1
UnboundLocalError: local variable 'count' referenced before assignment

Your task is to modify the code for integerDivision so that this error does not occur.

  Analyze

常量count未定义，应在while之前定义。

  Answer

def integerDivision(x, a):
    """
    x: a non-negative integer argument
    a: a positive integer argument

    returns: integer, the integer division of x divided by a.
    """
    count = 0
    while x >= a:
        x = x - a
        count += 1
    return count

  Problem 2

Consider the following function definition:

def rem(x, a):
    """
    x: a non-negative integer argument
    a: a positive integer argument

    returns: integer, the remainder when x is divided by a.
    """
    if x == a:
        return 0
    elif x < a:
        return x
    else:
        rem(x-a, a)

When we call

rem(2, 5)

==> IDLE returns 2.

When we call

rem(5, 5)

==> IDLE returns 0. 

But when we call

rem(7, 5)

==> IDLE does not return anything!

  Analyze

在递归时，要形成递归关系应当return到函数本身而并非调用。

  Answer

def rem(x, a):
    """
    x: a non-negative integer argument
    a: a positive integer argument

    returns: integer, the remainder when x is divided by a.
    """
    if x == a:
        return 0
    elif x < a:
        return x
    else:
        return rem(x-a, a)

  Problem 3

Consider the following function definition:

def f(n):
   """
   n: integer, n >= 0.
   """
   if n == 0:
      return n
   else:
      return n * f(n-1)

When we call f(3) we expect the result 6, but we get 0.
When we call f(1) we expect the result 1, but we get 0.
When we call f(0) we expect the result 1, but we get 0.
So, what's the problem?

  Analyze

先分析这个函数想要做什么。
此处的函数期望结果为n!，但是在终止时的返回值是错误的，应改为return 1。

  Answer

def f(n):
   """
   n: integer, n >= 0.
   """
   if n == 0:
      return 1
   else:
      return n * f(n-1)