Higher Order Functions

The Name Game (su16-mt1-2)

Draw the the environment diagram that results from executing the code below until the entire program is finished or an error occurs.

def person(name, age):
  def brain(ask):
    if ask == 'name':
      return name
    elif ask == 'age':
      return age
  return brain
def name(guy):
    return guy('name')
def age(gal):
    return gal('age')
my_name, my_age = 'Marv', 21
marv = person(my_name, my_age)
dan = person('Dan', age(marv) + 7)

The Name Game (fa14-mt1-2)

Draw the the environment diagram that results from executing the code below until the entire program is finished or an error occurs.

def peace(today):
    harmony = love + 2
    return harmony + today(love + 1)
def joy(peace):
    peace, love = peace + 2, peace + 1
    return love // harmony
love, harmony = 3, 2
peace(joy)

Draw the the environment diagram that results from executing the code below until the entire program is finished or an error occurs.
```
def k(g, b):
  def n(s, a):
    return g - p
  return b(n(b, p))
g, p = 3, 7
k(p + 1, lambda s: g + 3)
```

Environmental Policy (sp16-mt1-2)

Draw the the environment diagram that results from executing the code below until the entire program is finished or an error occurs.

y = 3
def out(h, m):
  y = 5 * m
  def inner():
    return y
  if m == 0:
    return h
  else:
    return out(inner, m - 1)
v = out(None, 1)()

Vulcans (sp15-mt2-2)

Draw the environment diagram that results from executing the code after the entire program is finished. No errors occur during the execution of this example.
```
def scramble(egg):
  return [egg, over(egg)]

def over(easy):
  easy[1] = [[easy], 2]
  return list(easy[1])

egg = scramble([12, 24])
```

Pointers (sp16-mt2-1)

For each of the following code fragments, draw the final state of the program in boxes and arrows.

Code:

L = Link(1, Link(2))
P = L
Q = Link(L, Link(P))
P.rest.rest = Q

Code:

L = Link.empty
for i in range(3):
  L = Link(i, L)

Code:

P = Link(0, Link(1, Link(2)))
def crack1(L):
  if L is Link.empty:
    return (Link.empty, Link.empty)
  L1, L2 = crack1(L.rest)
  return (Link(L.first, L2), L1)
Q, R = crack1(P)

Code:

P = Link(0, Link(1, Link(2)))
def crack2(L):
  if L is Link.empty:
    return (Link.empty, Link.empty)
  L1, L2 = crack2(L.rest)
  L.rest = L2
  return (L, L1)
Q, R = crack2(P)