lec1-6

记录了lec1-6的零碎内容.

hw与lab运行方式

lab-Test:

$ python ok -q q1 -u # 或者q2, q3, etc.

lab/hw-Problem:

测试全部：

$ python ok --local

测试单个函数：

$ python ok -q <func name> --local

• and/or: 如果第一个为False/True则短路，否则直接返回后者.

• lambda expression:

  

• When is the return expression of a lambda expression executed? When the function returned by the lambda expression is called.

• Which of the following statements describes a difference between a def statement and a lambda expression?
  A lambda expression does not automatically bind the function object that it returns to an intrinsic name.

• 写出这段程序的输出：

  >>> b = lambda x: lambda: x
  >>> c = b(88)
  >>> c
  <function <lambda>.<locals>.<lambda> at 0x7866b5acca40>
  >>> c()
  88
  

  实际上，这是一种嵌套式的调用，改写成def如下：

  >>> def f(x):
  ...     def g(y):
  ...         return x
  

• lambda函数的调用传参：

  >>> higher_order_lambda = lambda f: lambda x: f(x)
  >>> g = lambda x: x * x
  >>> higher_order_lambda(2)(g)
  ? Error
  >>> higher_order_lambda(g)(2)
  ? 4
  

• 参数传递的遮蔽

  >>> a = lambda x: x * 2 + 1
  >>> def b(b, x):
  ...     return b(x + a(x))
  >>> x = 3
  >>> b(a, x)
  21
  

  在函数内部，参数 b 会遮蔽（shadow）外部的函数名 b，因此相当于调用了2次函数a，导致结果上是嵌套函数.

• lab02的最后一个任务：

  def minimal_root(n, f):
      """
      Return the smallest k such that k is a root of f with 0 <= k <= n.
      If no such k exists, return n + 1
  
      >>> minimal_root(6, lambda x: x) # f(0) = 0
      0
      >>> minimal_root(6, lambda x: x * x - 5 * x + 6) # f(2) = 0, f(3) = 0
      2
      >>> minimal_root(6, lambda x: x * x + 1) # no roots
      7
      """
      # return next((k for k in range(n+1) if f(k) == 0), n+1)
      return n+1 if (product_of_trigger(n,f) == 1) else minimal_root(n-1,f)