干燥剪刀石头布 [英] DRYing up Rock Paper Scissors

问题描述

我是一个红宝石新手，尽管此代码有效，但我想知道如何改进它.我对lambda和proc之类的知识非常有限，但是任何建议都是不错的选择.有没有办法简化每种情况下的if else语句?另外，是否有任何替代方法可以跳过case语句，而不是将几乎整个代码都变成大的if else语句?

I'm a novice ruby programmer and although this code works, I'm wondering how I can improve it. I have very limited knowledge on lambdas and procs and the like, but any advice would be great. Is there any way to simplify the if else statements in each case? Also, is there any alternative way to the case statement skipped instead of making almost the entirety of the code one big if else statement?

def rps(roll)
    roll_ops = ["rock", "paper", "scissors"]
    pick = roll_ops.sample
    result = nil
    if roll == pick
        result = "tie"
    else
    case roll
    when "scissors" then
        if pick == "paper"
            result = "win"
        else
            result = "lose"
        end
    when "rock" then
        if pick == "scissors"
            result = "win"
        else
            result = "lose"
        end
    when "paper" then
        if pick == "rock"
            result = "win"
        else
            result = "lose"
        end
    else
        puts "Please input rock paper or scissors"
    end
    end

    puts "#{pick}, #{result}"
end

rps("scissors")

推荐答案

您可以构建一个散列，其中包含选择权以及针对该选择权丢失的选项:

You can build a hash that contains the pick and which option lose against that pick:

hash = {'scissors' => 'paper', 'rock' => 'scissors', 'paper' => 'rock'}

然后，您检查机器选择是否与您的选择相同:

Then you check if the machine pick is the same like you did:

roll_ops = ["rock", "paper", "scissors"]
pick = roll_ops.sample
if roll == pick

获胜/失败的条件变成这样:

And the win/lose condition becomes something like this:

if hash[roll] == pick
  "win"
else
  "lose"
end

仅需2种条件就可以清洁干净.

Nice and clean with just 2 conditions.

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