Python中的比较运算符未按预期工作答案

【问题标题】：Comparison Operator in Python Not Working as ExpectedPython中的比较运算符未按预期工作
【发布时间】：2020-05-20 16:17:55
【问题描述】：

我正在努力熟悉 python。为此，我正在开发一个石头、纸、剪刀游戏。我有一个类来处理值的比较以确定获胜者。问题是每当我使用重载的大于运算符进行比较时，只有在涉及 rock 时结果才会出错。我做了一个比较方法来测试我的逻辑是否正确。 python中重载运算符的行为是否会导致这种情况？

我的代码：

import random

rock = "rock"
paper = "paper"
scissors = "scissors"
options = (rock, paper, scissors)

class RcpValue:
    def __init__(self, value = rock):
        self.__value = value

    @property
    def value(self):
        return self.__value

    def __eq__(self, other):
        if self.__value == other.__value:
            return True

        return False

    def compare(self, other):
        if (self.__value == rock) and (other.__value == paper):
            return False

        if (self.__value == rock) and (other.__value == scissors):
            return True

        if (self.__value == paper) and (other.__value == scissors):
            return False

        if (self.__value == paper) and (other.__value == rock):
            return True

        if (self.__value == scissors) and (other.__value == rock):
            return False

        if (self.__value == scissors) and (other.__value == paper):
            return True

    def __gt__(self, other):
        if (self.__value == rock) and (other.__value == paper):
            return False

        if (self.__value == rock) and (other.__value == scissors):
            return True

        if (self.__value == paper) and (other.__value == scissors):
            return False

        if (self.__value == paper) and (other.__value == rock):
            return True

        if (self.__value == scissors) and (other.__value == rock):
            return False

        if (self.__value == scissors) and (other.__value == paper):
            return True

def determine_winner(p1: RcpValue, p2: RcpValue):
    if p1.value == p2.value:
        return "Tied!"

    if p1.value > p2.value:
    #if p1.compare(p2):
        return "Player 1 wins"

    return "Player 2 wins"

def test():
    for i in options:
        for j in options:
            player1 = RcpValue(i)
            player2 = RcpValue(j)
            print(f"Player 1 chose: {player1.value}")
            print(f"Player 2 chose: {player2.value}")
            print(determine_winner(player1, player2))
            print()

test()

使用__gt__输出：

Player 1 chose: rock
Player 2 chose: rock
Tied!

Player 1 chose: rock
Player 2 chose: paper
Player 1 wins

Player 1 chose: rock
Player 2 chose: scissors
Player 2 wins

Player 1 chose: paper
Player 2 chose: rock
Player 2 wins

Player 1 chose: paper
Player 2 chose: paper
Tied!

Player 1 chose: paper
Player 2 chose: scissors
Player 2 wins

Player 1 chose: scissors
Player 2 chose: rock
Player 1 wins

Player 1 chose: scissors
Player 2 chose: paper
Player 1 wins

Player 1 chose: scissors
Player 2 chose: scissors
Tied!

使用compare() 输出：

Player 1 chose: rock
Player 2 chose: rock
Tied!

Player 1 chose: rock
Player 2 chose: paper
Player 2 wins

Player 1 chose: rock
Player 2 chose: scissors
Player 1 wins

Player 1 chose: paper
Player 2 chose: rock
Player 1 wins

Player 1 chose: paper
Player 2 chose: paper
Tied!

Player 1 chose: paper
Player 2 chose: scissors
Player 2 wins

Player 1 chose: scissors
Player 2 chose: rock
Player 2 wins

Player 1 chose: scissors
Player 2 chose: paper
Player 1 wins

Player 1 chose: scissors
Player 2 chose: scissors
Tied!

【问题讨论】：

标签： python class operator-overloading

【解决方案1】：

这是一个简单的问题：比较if p1.value > p2.value 比较每个RcpValue 实例中分配的字符串"rock" 等，因此永远不会调用__gt__() 方法。

您需要做的就是将该行替换为if p1 > p2。由于p1 和p2 是RcpValue 对象，解释器使用RcpValue.__gt__() 来比较它们。你也可以用if p1 == p2 替换if p1.value == p2.value，因为你写了RcpValue.__eq__()。

【讨论】：