程序代写代做代考 python interpreter CS 61A Structure and Interpretation of Computer Programs – cscodehelp代写

CS 61A Structure and Interpretation of Computer Programs
Spring 2019
INSTRUCTIONS
• You have 55 minutes to complete the exam.
Midterm 1
• The exam is closed book, closed notes, closed computer, closed calculator, except one hand-written 8.5″ × 11″ crib sheet of your own creation and the official CS 61A midterm 1 study guide.
• Mark your answers on the exam itself. We will not grade answers written on scratch paper.
Last name
First name
Student ID number
CalCentral email
TA
Name of the person to your left
Name of the person to your right
All the work on this exam is my own.
(please sign)
POLICIES & CLARIFICATIONS
• If you need to use the restroom, bring your phone and exam to the front of the room.
• You may use built-in Python functions that do not require import, such as min, max, pow, len, and abs.
• You may not use example functions defined on your study guides unless a problem clearly states you can.
• For fill-in-the blank coding problems, we will only grade work written in the provided blanks. You may only write one Python statement per blank line, and it must be indented to the level that the blank is indented.
• Unless otherwise specified, you are allowed to reference functions defined in previous parts of the same question.

2
1. (10 points, 15 minutes) What Would Python Display
For each of the expressions in the table below, write the output displayed by the interactive Python interpreter when the expression is evaluated. The output may have multiple lines. If an error occurs, write “Error”, but include all output displayed before the error. If evaluation would run forever, write “Forever”. To display a function value, write “Function”. The first two rows have been provided as examples.
The interactive interpreter displays the value of a successfully evaluated expression, unless it is None. Assume that you have first started python3 and executed the statements on the left.
n=1
f = lambda: n g=f
n=2
x, y, loops = 0, 21, 0
while y:
loops += 1
if x <= y: if x % 2 == 1: x, y = y, x elif x == 0: x += (2 ** 2) else: x=x+3 y = y // 2 else: y=0 def foo(bar, z): def bar(w): z=w+2 return z print(bar(z)) print(z) return bar Expression pow(10, 2) Interactive Output 100 print(4, 5) + 1 45 Error print(1, print(print(2), 3 or 4 // 0)) g() x loops foo(lambda m: m + 1, 4)(7) Name: 3 2. (10 points, 15 minutes) Environment Diagram Fill in the environment diagram that results from executing the code on the right until the entire program is finished, an error occurs, or all frames are filled. You may not need to use all of the spaces or frames. A complete answer will: • Add all missing names and parent annotations to all local frames. • Add all missing values created or referenced during execution. • Show the return value for each local frame. 4 3. (10 points, 25 minutes) Zip It We would like to create a function make_zipper that takes two functions f1(x) and f2(x) and a “zipper sequence”, which is a number that contains a series of 1s and 2s. It returns a function that is the equivalent of f1(f2(f2(...f1(x)...))) in which the exact sequence of f1s and f2s is given by the digits of the sequence. As an example, if the sequence were 1211, that would mean return a function of x that is the equivalent to f1(f2(f1(f1(x)))). Neither recursion nor containers (lists, dictionaries, sets, etc) are allowed in your solution. def make_zipper(f1, f2, sequence): """ Return a function of f1 and f2 composed based on sequence. >>> def increment(x):
return x + 1
>>> def square(x):
return x * x
>>> do_nothing = make_zipper(increment, square, 0)
>>> do_nothing(2) # Don’t call either f1 or f2, just return your input untouched 2
>>> incincsq = make_zipper(increment, square, 112)
>>> incincsq(2) # increment(increment(square(2))), so 2 → 4 → 5 → 6
6
>>> sqincsqinc = make_zipper(increment, square, 2121)
>>> sqincsqinc(2)
100
“””
zipper =
helper =
while
if
zipper
else: zipper
sequence =
return zipper
# square(increment(square(increment(2)))), so 2 → 3 → 9 → 10 → 100
:
= helper(f1,
= helper(f2,
)
)
== 1: