高复杂性防止功能

Question

首先，因为我使用的代码很大，所以 here 是代码的 link。

我有这个功能运行如果给出特定输入，其他功能：

allowed_commands = ['help', 'joblist', 'job', 'jobresign', 'work', 'bal', 'balance', 'dep', 'deposit', 'with', 'withdraw', 'fish', 'hunt', 'shop', 'buy', 'sell', 'hunt', 'fish', 'multiply']

def gamePlay():
    while True:
        command = input(f"{bright_green}Command (type [help] to see list of all commands):\n>> ")
        while command in allowed_commands:
            # <-- Display Rules -->
            if command == 'help':
                rules()
                break

            # <-- Display Jobs -->
            elif command == 'joblist':
                joblist_function()
                break

            # <-- Get Jobs -->
            elif command == 'job' and working == False:
                job_funtion()
                break

            elif command == 'job' and working == True:
                print(f"\n{red}You are already doing a job. You can't work on two jobs,that is dumb...\n")
                break

            # <-- Resign Job -->
            elif command == 'jobresign':
                job_resign()
                break

            # <-- Work -->
            elif command == 'work' and working == True:
                work()
                break
            
            elif command == "work" and working == False:
                print(f"{red}\nLOL, you don't have a job, how you gonna work?\n")
                break

            # <-- Deposit -->
            elif command == 'dep' or command == 'deposit' and deposit_allowed != deposited:
                dep_money()
                break

            elif command == 'dep' or command == 'deposit' and deposit_allowed == deposited:
                print("You have a full bank kiddo...")
                break

            # <-- Balance -->
            elif command == 'bal' or command == 'balance':
                display_balance()
                break

            # <-- Withdraw -->
            elif command == 'with' or command == 'withdraw' and deposited != 0:
                withdraw_money()
                break
            
            elif command == 'with' or command == 'withdraw' and deposited == 0:
                print(f"{red}\nNo money deposited. What are you even trying to wothdraw LOL?\n")
                break

            elif command == 'shop':
                shop()
                break

            elif command == 'beg':
                beg()
                break
def beg():
    global money
    random_number2 = random.choice([0, 1, 2])
    random_money = random.choice(range(100, 500))

    if random_number2 == 1:
        print("Ewwww beggar. No stonks for u")
    
    if random_number2 == 2:
        print(f"Mr.beggar, you can have ⏣ {random_money}.")
        money += random_money

但是函数下方绿线上的工具提示显示“圈复杂度太高：17（阈值 15）”。

通常情况下，即使复杂度高达 30，这也适用于我的代码。但是最后一个 elif 之后的代码不起作用。即使我输入 'beg'，函数也不会运行:

Command (type [help] to see list of all commands):
>> beg
Command (type [help] to see list of all commands):
>> beg
Command (type [help] to see list of all commands):
>>

为什么会这样，我该如何解决？

Answer 1

你可以这样使用字典：

{
    'help': rules,
    'joblist': joblist_function,
}[command]()

但您也可以像这样进行一些重构：

from typing import Callable, Dict


class Game:
    def __init__(self) -> None:
        self.deposit_allowed = 1000
        self.deposited = 0

    def play(self) -> None:
        commands = self._get_commands()

        while True:
            command = input(f"{bright_green}Command (type [help] to see list of all commands):\n>> ")

            if command not in commands:
                print('Wrong command!')
                continue

            commands[command]()

    def _get_commands(self) -> Dict[str, Callable[[], None]]:
        return {
            'help': self._rules,
            'joblist': self._joblist_function,
            'deposit': self._deposit_money,
            'dep': self._dep_money,
            # ...
        }

    def _rules(self) -> None:
        pass

    def _joblist_function(self) -> None:
        pass

    def _deposit_money(self) -> None:
        if self.deposit_allowed == self.deposited:
            print("You have a full bank kiddo...")
            return

        self._dep_money()

    def _dep_money(self) -> None:
        pass


Game().play()

Answer 2

根据 sonarQube 文档，

Cyclomatic Complexity calculated based on the number of paths through the code. Whenever the control flow of a function splits, the complexity counter gets incremented by one. Each function has a minimum complexity of 1. This calculation varies slightly by language because keywords and functionalities do.

尽管这个圈复杂度可能因语言而异，但通常情况下，循环和条件一起形成一个 30 的圈复杂度。通过总结以下关键字，我们可以估计圈复杂度为 30（请注意，实际算法可能不同）。

    while 
        while  
            if ... 
            elif ... 
            elif ... and ... 
            elif ... and ...  
            elif ... 
            elif ... and ... 
            elif ... and ... 
            elif ... or  ... and ... 
            elif ... or  ... and ... 
            elif ... or  ... 
            elif ... or  ... and ... 
            elif ... or  ... and ... 
            elif ... 
            elif ...

虽然较高的圈复杂度在代码中不是一个好的做法，但这只是一个警告，您的代码将照常编译。

在您的情况下，不执行您的 beg 命令的问题是，您的 allowed_commands.

中不存在 beg

高复杂性防止功能

High complexity preventing function from

python

function

if-statement

cyclomatic-complexity