为什么这个功能不起作用？

Question

只是想玩一些LPTHW函数。我在这里编的：

def character_class(intel, str, agil):
    print "A barbarian's main attribute starts at: %d." % str
    print "A wizard's main attribute is intel, and it starts at: %d." % intel
    print "An archer's main attribute is agility, and has the default agility speed of: %d.\n" % agil


character_class(20, 40, 60)
print character_class

character_class(20 + 40, 40 + 50, 100 + 100)
print character_class

input1 = raw_input("Barbarian str:")
input2 = raw_input("Wizard intel:")
input3 = raw_input("Archer agil:")
character_class % (input1, input2, input3)
print character_class

下面是我在Powershell中得到的结果：

A barbarian's main attribute starts at: 40.
A wizard's main attribute is intel, and it starts at: 20.
An archer's main attribute is agility, and has the default agility speed of: 60.

<function character_class at 0x025078B0>
A barbarian's main attribute starts at: 90.
A wizard's main attribute is intel, and it starts at: 60.
An archer's main attribute is agility, and has the default agility speed of: 200.

<function character_class at 0x025078B0>
Barbarian str:200
Wizard intel:300
Archer agil:400
Traceback (most recent call last):
  File "test19.py", line 16, in <module>
    character_class % (input1, input2, input3)
TypeError: unsupported operand type(s) for %: 'function' and 'tuple'

首先，在每次调用函数character_class之后出现的<function character_class at 0x025078B0>是什么？这在LPTHW练习19中没有出现。你知道吗

另外，我还尝试从用户那里获取原始输入，以插入到函数中。这是不可能的，还是我做错了？你知道吗

修订：将最后一行代码更改为：character_class(input1, input2, input3)

这就是我现在得到的错误：

A barbarian's main attribute starts at: 40.
    A wizard's main attribute is intel, and it starts at: 20.
    An archer's main attribute is agility, and has the default agility speed of: 60.

    A barbarian's main attribute starts at: 90.
    A wizard's main attribute is intel, and it starts at: 60.
    An archer's main attribute is agility, and has the default agility speed of: 200.

    Barbarian str:1000
    Wizard intel:2000
    Archer agil:3000
    Traceback (most recent call last):
      File "test19.py", line 16, in <module>
        character_class(input1, input2, input3)
      File "test19.py", line 2, in character_class
        print "A barbarian's main attribute starts at: %d." % str
    TypeError: %d format: a number is required, not str

最终修订：

def character_class(intel, str, agil):
    print "A barbarian's main attribute starts at: %d." % str
    print "A wizard's main attribute is intel, and it starts at: %d." % intel
    print "An archer's main attribute is agility, and has the default agility speed of: %d.\n" % agil


character_class(20, 40, 60)


character_class(20 + 40, 40 + 50, 100 + 100)


input1 = raw_input("Barbarian str:")
input2 = raw_input("Wizard intel:")
input3 = raw_input("Archer agil:")
inputa = int(input1)
inputb = int(input2)
inputc = int(input3)

character_class(inputa, inputb, inputc)

Answer 1

首先，让我们看看您的输出：

A barbarian's main attribute starts at: 40.
A wizard's main attribute is intel, and it starts at: 20.
An archer's main attribute is agility, and has the default agility speed of: 60.

<function character_class at 0x025078B0>.

然后生成这个输出的代码：

character_class(20, 40, 60)
print character_class

您已经在上面定义了character\u类函数来获取三个参数并分别调用print函数三次。当您调用character_class(20, 40, 60)时，您将传入20、40和60作为参数。然后，您的函数将调用print三次，在最后一次调用中使用换行符，结果显示输出的前四行。你知道吗

调用print character_class时，将函数定义作为参数传递给print函数。输出<function character_class at 0x025078B0>，是所定义函数的参考位置。关键是，您不需要在character_class调用之前调用print，因为您已经定义了调用print三次的函数。当python执行character_class(20,40,60)时，它将进入函数并执行您定义的每一行代码，并插入使用字符串替换传递的参数。附加的print character_class不是必需的，因为函数代码为您执行print-ing。你知道吗

最后，您正确地获得了原始输入。raw_input函数将从控制台获取您的输入并返回结果，您已将结果赋给变量。您的调用（character_class % (input1, input2, input3)）几乎是正确的，但是我认为您将调用函数与字符串变量替换混为一谈。你知道吗

您只需通过传递从input获得的三个参数来调用函数：character_class(input1, input2, input3)，您的函数将为您进行打印，正如您在上面定义的那样！你知道吗

Answer 2

我对unicode知识不太了解。我相信它是键盘上特定字符的参考代码。另外，你的问题也不太清楚。不过，我写了这篇文章，希望它能重新格式化并回答您的几个问题：

class character_class(object):
def __init__(self, intel, str, agil):
    self.intel = intel
    self.str = str
    self.agil = agil

def input(self):
    str = input1
    intel = input2
    agil = input3
    print "A barbarian's main attribute starts at: %r." % str
    print "A wizard's main attribute is intel, and it starts at: %r." % intel
    print "An archer's main attribute is agility, and has the default agility speed of: %r.\n" % agil
    cont = raw_input("")

input1 = raw_input("Barbarian str:")
input2 = raw_input("Wizard intel:")
input3 = raw_input("Archer agil:")
character_class = character_class(input1, input2, input3)
character_class.input()

我对编程还是比较陌生的，所以我的代码可能有点“脱节”或者马虎之类的。无论如何，我将intel、str和agil分配给一个类属性，然后通过原始输入定义它们。你知道吗

当您运行此脚本时，它将提示您输入input 1-3下的属性，然后通过character显示它们_类输入(). 这些输入值被分配给str、intel和agil。这回答了你的问题吗？如果不告诉我。你知道吗

Answer 3

您成功地调用了执行打印的函数，然后每次还告诉Python打印函数本身，当您执行print character_class时。别那么做。你知道吗

另外，我也不知道为什么在上一篇文章中使用了%，在这里您要传递来自原始输入的数据。同样，不要这样做：

character_class(input1, input2, input3)