CodeMonkey过关学习笔记系列：特技关卡 14-1 ～ 14-15 关

特技关卡 14-1
yummyBanana = (y) ->
    if not y.green() and not y.rotten()
        goto y
for b in bananas
    yummyBanana b

特技关卡 14-2
#修复这个函数:
yummyBanana = (y) ->
    if y.green() or y.rotten()
        return no
    return yes

#一旦你修复yummyBanana这个函数，这个编码是好的。
x = 0
3.times ->
    2.times ->
        if yummyBanana bananas[x]
            goto bananas[x]
        x = x + 1
        goto turtle
    turtle.step 12

特技关卡 14-3
#nearestZone函数传回健康地带
#那更接近猴子
nearestZone = () ->
    d0 = distanceTo healthZones[0]
    d1 = distanceTo healthZones[1]
    if d0 < d1
        #修复函数来传回正确的对象
        return healthZones[0]
    else
        return healthZones[1]

#一旦你修复nearestZone函数，这个编码是有效的
for b in bananas
    goto b
    if health() < 50
        zone = nearestZone()
        goto zone
        until health() == 100
            wait()

特技关卡 14-4
#nearestBridge 函数传回最近的桥
nearestBridge = () ->
    d0 = distanceTo bridges[0]
    d1 = distanceTo bridges[1]
    #完成这个函数来传回最近的桥
    if d0 <= d1
        return bridges[0]
    else
        return bridges[1]

#当健康值很低时，猴子后巷最近的桥
#然后去健康地带
#这个函数是正确的！
getHealthy = () ->
    goto nearestBridge()
    goto healthZone
    until health() == 100
        wait()
    goto nearestBridge()

#这里不需要更改
for b in bananas
    goto b
    if health() < 70
        getHealthy()

特技关卡 14-5
#safeFrom函数传回yes如果老虎
#睡觉或是玩耍
safeFrom = (a) ->
    return a.sleeping() or a.playing()

for t in tigers
    until safeFrom t
        wait()
    step 8
    #现在轮到谁了？
    until safeFrom t
        wait()
    goat.step 8

特技关卡 14-6
safeFrom = (a) ->
    #这个函数应该传回什么呢？
    return a.sleeping() or a.playing()

#一旦你修复了 safeFrom这个函数，这个编码就是正确的。
x = 0
for stepper in [monkey, goat]
    #首先猴子先过去，然后是山羊。
    until safeFrom(tigers[x])
        wait()
    stepper.step 12
    #等待熊
    until safeFrom(bears[x])
        wait()
    stepper.step 12
    x = x + 1

特技关卡 14-7
# nearestGoat函数会获得一个对象
#并比较每只山羊到对象的距离
#函数传回哪只山羊距离对象最近
nearestGoat = (y) ->
    d0 = goats[0].distanceTo y
    d1 = goats[1].distanceTo y
    if d0 < d1
        return goats[0]
    else
        return goats[1]

for b in bananas
    #这里有一个小修复
    if b.green()
        g = nearestGoat b
        say g
        g.goto b
    else
        goto b

特技关卡 14-8
whichAnimal = (y) ->
    if y.green()
        return goat
    return monkey
for b in bananas
    mover = whichAnimal b
    mover.goto b

特技关卡 14-9
whichAnimal = (y) ->
    if y.green()
        return goat
    else
        return monkey


for b in bananas
    mover = whichAnimal b
    mover.goto b

特技关卡 14-10
#修复函数useGoat传回yes
#当香蕉是绿色的时候
useGoat = (y) ->
    return y.green()

nearestGoat = (y) ->
    d0 = goats[0].distanceTo y
    d1 = goats[1].distanceTo y
    if d0 < d1
        return goats[0]
    else
        return goats[1]

for b in bananas
    mover = monkey
    if useGoat b
        mover = nearestGoat b
    say mover
    mover.goto b

特技关卡 14-11
#这个函数传回哪只乌龟距离香蕉更近
#这个函数的编码是正确的！
nearestTurtle = (y) -> 
    d0 = turtles[0].distanceTo y
    d1 = turtles[1].distanceTo y
    if d0 < d1
        return turtles[0]
    else
        return turtles[1]

for b in bananas
    t = nearestTurtle(b)
    t.goto b

特技关卡 14-12
#这个函数传回多少次
#奶牛吃：
leftToEat = (c) ->
    return 35 - c.weight()


say leftToEat(cow)

#修复这个条件，什么时候循环应该停止？
until leftToEat(cow) == 0
    cow.eat()

goto banana

特技关卡 14-13
#这个函数检测奶牛的重量是否
#和门上的条件相同
shouldNotEat = (c) ->
    return c.weight() == 100

say shouldNotEat(cow)

until shouldNotEat(cow)
    cow.eat()
for b in bananas
    goto b

特技关卡 14-14
#修复这个函数传回如果奶牛的重量是
#和门上的条件相同
shouldNotEat = (c) ->
    return c.weight()==60

#使用shouldNotEat函数打开大门
for i in [0,1,2]
    until shouldNotEat(cows[i]) 
        cows[i].eat()
    goto bananas[i]

特技关卡 14-15
startWithBananaZero = () ->
    d0 = distanceTo bananas[0]
    d2 = distanceTo bananas[2]
    if d0 < d2
        return yes
    else
        return no

if startWithBananaZero()
    x = 0
    3.times ->
        goto bananas[x]
        x = x + 1
else
    x = 2
    3.times ->
        goto bananas[x]
        x = x - 1