'int'对象不可调用？

Question

当我尝试运行这个程序（下面的代码），它返回一个TypeError: 'int' object is not callable

代码：

import random 
import math 

def var(): 
    strength = 10 
    skill = 10 
    dice4 = 0 
    dice12 = 0 

    character_name = str(input("Please enter your characters name: ")) 
    skill(strength, skill, dice4, dice12, character_name) 

def skill(strength, skill, dice4, dice12, character_name): 

    print(character_name + "'s attributes are being generated! ... ") 

    dice4, dice12 = random.randrange(1,4), random.randrange(1,12) 

    dice_score = dice12/dice4 
    dice_score = math.floor(dice_score) 
    skill = skill + dicescore 

    strength(strength, skill, dice4, dice12, character_name) 

def strength(strength, skill, dice4, dice12, character_name): 
    dice4, dice12 = random.randrange(1,4), random.randrange(1,12) 

    dice_score = dice12/dice4 
    dice_score = math.floor(dice_score) 
    strength = strength + dicescore 
    file(strength, skill, dice4, dice12, character_name) 

def file(strength, skill, dice4, dice12, character_name): 
    file = open("N:\Controlled Assessment - Ryan Harper\Task Two\attributes.txt", w) 
    file.writelines(character_name + " - Strength = " + str(strength) + ", Skill = " + str(skill)) 

var() 

错误：

Traceback (most recent call last): 
    File "N:\Controlled Assessment - Ryan Harper\Task Two\task 2 v2.py", line 37, in <module> 
    var() 
    File "N:\Controlled Assessment - Ryan Harper\Task Two\task 2 v2.py", line 11, in var 
    skill(strength, skill, dice4, dice12, character_name) 
TypeError: 'int' object is not callable 

Answer 1

的问题是这一行：

技能（力量，技巧，骰子4，骰子12，character_name）

您将技能称为函数，但它是在此行之前几行定义的数字。

Answer 2

您不应该有一个变量和一个函数skill。

对于strength也是一样的。

这真的让翻译和你自己感到困惑。

给他们其他粉丝名字的未读。 :)

Answer 3

在函数var中，skill是绑定到整数的本地名称，该整数会影响全局函数skill()。为其中一个使用不同的名称。