2012-04-19 57 views
1

我觉得我错过了非常非常基本的东西,所以为了避免撕掉更多的头发,我来这里问为什么p1,p1t和p1t2正在用“hand = newgame”进行修改。查看p1“。我甚至不知道为什么P1会被修改,所以我在这里完全困惑。任何帮助将非常感激。变量的基本问题

# encoding: utf-8 

class Cards 

def view hand 
    x = 0 
    hand.each do |card| 
     if card[0] == 's' 
      card[0] = '♠' 
     elsif card[0] == 'd' 
      card[0] = '♦' 
     elsif card[0] == 'h' 
      card[0] = '♥' 
     elsif card[0] == 'c' 
      card[0] = '♣' 
     else 
      #nil 
     end 
     hand[x] = card 
     x = x + 1 
    end 
    prettyhand = '' 
    hand.each do |card| 
     prettyhand = prettyhand + card[0] + card[1] + ' ' 
    end 
    return prettyhand 
end#view 

end#Cards 
########### 

deck = 0 
up = 0 
p1 = 0 
p2 = 0 
newgame = Cards.new 
p1 = [["c", "4"], ["s", "4"], ["d", "q"], ["c", "8"], ["s", "5"], ["s", "q"], ["h", "3"], ["c", "10"]] 
    p1t = [] 
    p1t2 = [] 
    hand = '' 
    p1.each do |card| 
     p1t.push card 
    end 
    p1t.each do |card| 
     p1t2.push card 
    end 
    p '----------------------------------' 
    p 'fresh p1:' 
    p p1 
    p p1t 
    p p1t2 
    hand = newgame.view p1 
    p 'unfresh p1:'#why is this changing?? 
    p p1 
    p p1t 
    p p1t2 
    p '----------------------------------' 

输出:

"----------------------------------" 
"fresh p1:" 
[["c", "4"], ["s", "4"], ["d", "q"], ["c", "8"], ["s", "5"], ["s", "q"], ["h", "3"], ["c", "10"]] 
[["c", "4"], ["s", "4"], ["d", "q"], ["c", "8"], ["s", "5"], ["s", "q"], ["h", "3"], ["c", "10"]] 
[["c", "4"], ["s", "4"], ["d", "q"], ["c", "8"], ["s", "5"], ["s", "q"], ["h", "3"], ["c", "10"]] 
"unfresh p1:" 
[["♣", "4"], ["♠", "4"], ["♦", "q"], ["♣", "8"], ["♠", "5"], ["♠", "q"], ["♥", "3"], ["♣", "10"]] 
[["♣", "4"], ["♠", "4"], ["♦", "q"], ["♣", "8"], ["♠", "5"], ["♠", "q"], ["♥", "3"], ["♣", "10"]] 
[["♣", "4"], ["♠", "4"], ["♦", "q"], ["♣", "8"], ["♠", "5"], ["♠", "q"], ["♥", "3"], ["♣", "10"]] 
"----------------------------------" 

回答

1

数组通过引用传递,而不是复制。看看这个片段。

a1 = [[1], [2], [3]] 
a2 = [] 

a2 << a1.first # push a reference to the array 
a2 # => [[1]] 


a1[0][0] += 1 # => 2 
a1 # => [[2], [2], [3]] # a1 has changed 
a2 # => [[2]] # a2 changed too! 

# now push a copy of the array 
a2 = [a1.first.dup] # => [[2]] 

a1[0][0] += 1 # => 3 
a1 # => [[3], [2], [3]] # a1 has changed 
a2 # => [[2]] # a2 has not 
+2

特别是,请注意,这也适用于方法调用。因此你的Cards#view方法修改了它的参数。 – 2012-04-19 01:38:09

+0

我想我明白了!那么,这里最好的解决方案是什么?我应该将数组转换为字符串来复制它们吗?谢谢! – catbat 2012-04-19 01:42:22

+0

@catbat:查看我的更新回答 – 2012-04-19 01:46:41

1

您已经定义了一个方法view这需要一个参数hand(顺便说一句,括号都不错,不知道为什么你总是忽略它们在你的代码,即使是在方法签名)。 view方法修改其参数hand。你把它作为

hand = newgame.view p1 

p1由法view修改,这就是为什么它的变化。

我也很好奇你为什么设置p10 ...然后将其设置为一个数组数组。没有理由这样做。此外,尝试使用描述性变量名称,以及其他人。

+0

谢谢 - 当你说'签名'时,这是否意味着返回线? – catbat 2012-04-19 01:41:14

+0

@catbat:对不起。它表示方法的名称和参数(以静态语言表示返回类型)。所以'def view(hand)'是方法签名。 – 2012-04-19 01:46:28

+0

明白了。我通常使用它们,并且将来肯定会继续使用它们。编辑:它看起来很奇怪,因为它只是匆匆被扔在一起。将其设置为0并非出于任何有目的的理由:> – catbat 2012-04-19 02:00:52

1

FWIW,你view方法可以简化很多:

class Cards 
    Suits = { 'c' => '♣', 'd' => '♦', 'h' => '♥', 's' => '♠' } 

def view(hand) 
    hand.map do |suit, rank| [ Suits[suit], rank ] end 
end 
end 

此版本不修改任何东西,所以避免了这个问题。

+0

这非常有趣,我需要在.map上做一些阅读,因为我对它不是很熟悉,但它似乎是一个更简单的方法。格拉西亚斯! – catbat 2012-04-19 02:07:55

+0

map(也被称为'collect')是一个修改数组副本的好方法。您提供了一个在每个元素上运行的代码块;块的返回值将成为副本中的替换元素。例如,要增加列表中的每个数字:'[1,2,3] .map {| x | x + 1}',它返回'[2,3,4]'。 – 2012-04-19 02:11:47

+0

看起来非常有用。再次感谢! – catbat 2012-04-19 02:19:09