Question

为什么我必须将x的歌词传递为4而不是两个字符串？我终于理解了参数以及如何满足它们，但我不明白为什么我必须将x的歌词传递给除了内存地址之外的其他内容，这与bulls_on_parade和happy_bday不同。

class Song():

    def __init__(self, lyrics, x):
        self.lyrics = lyrics
        #self.x = x

    def sing_me_a_song(self):
        for line in self.lyrics:
            print line

    def print_x(self):
        print x.lyrics

happy_bday = Song(["Happy birthday to you,",
               "I don't want to get sued",
               "So I'll stop right there"], 'x-value')

bulls_on_parade = Song(["They'll rally around the family",
                    "With pockets full of shells"], 'x-value')

happy_bday.sing_me_a_song()

bulls_on_parade.sing_me_a_song()

def lyrics(args):
    pass

x = Song(lyrics= 4, x = lyrics)

x.print_x()

Answer 1

在此代码段中，您将全局名称lyrics定义为函数。将lyrics作为x = lyrics传递给x的对象构造函数会将函数的地址传递给它（这不完全正确，但现在它会这样做）作为x参数。

lyrics = 4是不同的，因为Python区分函数调用参数名和变量名。

def lyrics(args):
    pass

x = Song(lyrics= 4, x = lyrics)

事实上，如果在您的代码中执行此操作：

x = Song(lyrics=lyrics)
print x.lyrics

你会看到这样的事情：

<function lyrics at 0x107b9d1b8>

使用Python中的OOP更加困惑

1 个答案: