重构重复的保护声明

时间:2013-02-04 07:10:42

标签: c refactoring

我已经定义了一种计算获胜几率的方法,并返回支付给玩家的金额。我最初写这是完成此任务最明显的方法,但我真的想知道重构这个的最佳方法,以最大限度地减少计算机的工作量。这是我的功能:

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;
 }

1 个答案:

答案 0 :(得分:2)

尝试这种风格,您需要填写Payout

数组
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);

}