重构重复后卫声明

问题描述

我定义，计算获胜的几率，并返回所许给玩家数量的方法。我原本写了这个完成此任务的最明显的方式，但我真的想知道这个重构，以尽量减少计算机在执行的工作量的最佳方式。这里是我的功能：

  INT getPay（INT winningOdds，诠释派息）
{
    winningOdds =兰特（）％MAXIMUM_ODDS + MINIMUM_ODDS;    如果（winningOdds == 1）
    {
        支出= 4000;
        的printf（！大奖 - ％d个\\ N，派息）;
    }
    否则如果（winningOdds＆下; = 2）
    {
        支出= 1000;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则如果（winningOdds＆下; = 3）
    {
        支出= 500;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则，如果（winningOdds＆LT; = 40）
    {
        支出= 100;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则，如果（winningOdds＆LT; = 50）
    {
        支出= 50;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则，如果（winningOdds＆LT; = 75）
    {
        支出= 25;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则如果（winningOdds＆下; = 1000）
    {
        支出= 10;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则，如果（winningOdds＆LT; = 3000）
    {
        支出= 5;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则，如果（winningOdds＆LT; = 5000）
    {
        支出= 3;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }
    否则，如果（winningOdds＆LT; = 10000）
    {
        支出= 1;
        的printf（赢得的硬币内容 - ％d \\ n，派息）;
    }    返回支付;
 }
 

解决方案

尝试这种风格，你需要填写的阵列中的奖金

 结构赔率{
    INT winningOdds，派息;
};INT getPay（INT winningOdds，诠释派息）
{
    赔付支出[] = {{1,4000}，{} 2,1000 ....};
    INT numElem = sizeof的（支出）/ sizeof的（派息[0]）;    winningOdds =兰特（）％MAXIMUM_ODDS + MINIMUM_ODDS;
    的for（int i = 0; I＆LT; numElem;我++）{
        如果（winningOdds＆LT; =派息[I] .winningOdds）{
            支出=支出[I] .payout;
            打破;
        }
    }
    的printf（赢得的硬币内容 - ％d \\ n，派息）;}
 

I've defined a method that calculates the odds of winning and returns the amount payed to the player. I originally had written this the most obvious way of accomplishing this task, but I really would like to know the best way of refactoring this to minimize the amount of work the computer is performing. Here's my function:

int getPay(int winningOdds, int payOut)
{
    winningOdds = rand() % MAXIMUM_ODDS + MINIMUM_ODDS;

    if(winningOdds == 1)
    {
        payOut = 4000;
        printf("Jackpot!  - %d\n", payOut);
    }
    else if(winningOdds <= 2)
    {
        payOut = 1000;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 3)
    {
        payOut = 500;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 40)
    {
        payOut = 100;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 50)
    {
        payOut = 50;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 75)
    {
        payOut = 25;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 1000)
    {
        payOut = 10;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 3000)
    {
        payOut = 5;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 5000)
    {
        payOut = 3;
        printf("Coins won - %d\n", payOut);
    }
    else if(winningOdds <= 10000)
    {
        payOut = 1;
        printf("Coins won - %d\n", payOut);
    }

    return payOut;
 }

解决方案

Try this style, you need to fill in the array of Payout

struct Payout{
    int winningOdds, payOut;
};

int getPay(int winningOdds, int payOut)
{
    Payout payout[] = {{1,4000},{2,1000}....};
    int numElem = sizeof(payout)/sizeof(payout[0]);

    winningOdds = rand() % MAXIMUM_ODDS + MINIMUM_ODDS;
    for(int i=0;i<numElem ;i++){
        if(winningOdds <=payout[i].winningOdds){
            payOut = payout[i].payout;
            break;
        }
    }
    printf("Coins won - %d\n", payOut);

}

这篇关于重构重复后卫声明的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！