如何简化这段代码并将其变成函数？

Question

我刚刚开始编写代码，并且正在尝试重新创建蒙蒂霍尔问题游戏节目。我的程序让用户可以在游戏模式、研究模式和退出游戏之间进行选择。我正在尝试简化我的代码并将游戏模式和研究模式放入函数中，但我所做的所有研究对我来说都没有多大意义。我会很感激一些帮助，谢谢！

#include <stdio.h>
#include <stdlib.h>
#include <time.h>


int winning (int win);
float PERCENTAGE (float win,float playedgames);
int RANDOM ();


int main (void)
{
int choice, choice2, Ngames, Ngames2,chosendoor,chosendoor2;
int gamesplayed,gamesplayed2,winningdoor,winningdoor2,revealeddoor;
int switcheddoor, revealeddoor2,stayeddoor, doorchoice,winningdoor3;
int revealeddoor3, choice3, switcheddoor2;
float percent,percent2;
float win = 0,win2 = 0;
int gamesplayed3 = 0,win3 = 0;


srand(time(NULL)); //randomizes based on time


while (choice!=3)
{
    printf ("Please enter one of the following options\n\n1. Research mode\n2. Game mode\n3. Exit\n\n");
    scanf ("%d", &choice);


    switch(choice)
    {
    case 1:
        printf ("\nYou have chosen research mode.\n");
        printf ("Please indicate whether you will always us the always-    switch strategy or the never-switch strategy.\n");
        printf ("1. Always switch doors\n2. Never switch doors\n\n");
        scanf ("%d", &choice2);


        switch (choice2)
        {
        case 1:
            printf ("\nYou have chosen to always switch doors.\n");
            printf ("How many games would you like to play?\n\n");
            scanf ("%d", &Ngames);


            for (gamesplayed = 0;gamesplayed <= Ngames;gamesplayed++)
            {
                winningdoor = RANDOM();
                chosendoor = 1;


                do
                {
                    revealeddoor = RANDOM();
                } while (revealeddoor == chosendoor || revealeddoor == winningdoor);


                do
                {
                    switcheddoor = RANDOM();
                } while (switcheddoor == chosendoor || switcheddoor == revealeddoor);


                if (switcheddoor == winningdoor)
                {
                    win = winning(win);
                }
            }


            percent = PERCENTAGE(win,gamesplayed);


            printf ("\nYou have chosen the winning door %f percent of the time.\n\n", percent);


            win = 0;
            break;


        case 2:
            printf ("\nYou have chosen to never switch doors.\n");
            printf ("How many games would you like to play?\n\n");
            scanf ("%d", &Ngames2);


            for (gamesplayed2 = 0;gamesplayed2 <= Ngames2; gamesplayed2++)
            {
                winningdoor2 = RANDOM();
                chosendoor2 = 1;


                do
                {
                    revealeddoor2 = RANDOM();
                } while (revealeddoor2 == winningdoor2 || revealeddoor2 == chosendoor2);


                do
                {
                    stayeddoor = RANDOM();
                } while (stayeddoor != chosendoor2);


                if (stayeddoor == winningdoor2)
                {
                    win2 = winning(win2);
                }
            }


            percent2 = PERCENTAGE(win2,gamesplayed2);


            printf ("\nYou chose the winning door %f percent of the time.\n\n", percent2);


            win2 = 0;
            break;


        default:
            printf ("You have entered an incorrect value.");
            break;
        }
        break;


    case 2:
        printf ("\nYou have chosen game mode.\n");


        winningdoor3 = RANDOM();


        printf ("Which door will you choose? (Door 1,2, or 3)\n\n");
        scanf ("%d", &doorchoice);


        if (doorchoice < 1 || doorchoice > 3)
        {
            printf ("That value is not a door.\n\n");
            break;
        }


        do
        {
            revealeddoor3 = RANDOM();
        } while  (revealeddoor3 == doorchoice);


        printf ("\nDoor %d was opened and there was no prize behind it.\n", revealeddoor3);
        printf ("Would you like to:\n1. Switch doors?\n2. Stay with your current door?\n\n");
        scanf ("%d", &choice3);


        switch (choice3)
        {
        case 1:
            printf ("\nYou have chosen to switch doors.\n\n");


            do
            {
                switcheddoor2 = rand()%2 + 1;
            } while (switcheddoor2 == doorchoice);


            if (switcheddoor2 == winningdoor3)
            {
                printf ("Good choice! You have won!\n\n");
                win3 = winning(win3);
            }


            else
            {
                printf ("Unfortunately, you lost.\n\n");
            }


            ++gamesplayed3;


            printf ("You have played a total of %d games and you won a total of %d games.\n", gamesplayed3, win3);
            break;


        case 2:
            printf ("\nYou have decided to stay with your current door.\n");


            if (doorchoice == winningdoor3)
            {
                printf ("Good choice! You have won!\n\n");
                win3 = winning(win3);
            }


            else
            {
                printf ("Unfortunately, you lost.\n\n");
            }


            gamesplayed3 = gamesplayed3 + 1;


            printf ("You have played a total of %d games and you won a total of %d games.\n\n", gamesplayed3, win3);
            break;


        default:
            printf ("You have entered an incorrect value.\n\n");
            break;
        }
        break;


        case 3:
            printf("Thanks for playing!");
            exit(0);


        default:
            printf ("You have entered an incorrect value.\n\n");
            break;
    }
}
return 0;
}


int winning (int win)
{
int newwin;
newwin = win + 1;
return (newwin);
}


float PERCENTAGE (float win,float playedgames)
{
float percentage;
percentage = (win/playedgames)*100;
return (percentage);
}


int RANDOM ()
{
int random;
random = rand()%3;
return (random);
}

Answer 1

这是一个示例，用于转换您的开关案例之一...

这是您的代码

case 1:
    printf ("\nYou have chosen to always switch doors.\n");
    printf ("How many games would you like to play?\n\n");
    scanf ("%d", &Ngames);
    for (gamesplayed = 0; gamesplayed <= Ngames; gamesplayed++)
    {
        winningdoor = RANDOM();
        chosendoor = 1;
        do
        {
            revealeddoor = RANDOM();
        }
        while (revealeddoor == chosendoor || revealeddoor == winningdoor);
        do
        {
            switcheddoor = RANDOM();
        }
        while (switcheddoor == chosendoor || switcheddoor == revealeddoor);
        if (switcheddoor == winningdoor)
        {
            win = winning(win);
        }
    }
    percent = PERCENTAGE(win, gamesplayed);
    printf ("\nYou have chosen the winning door %f percent of the time.\n\n", percent);

    win = 0;
    break;

下面是提取函数后的样子

case 1:
    printf ("\nYou have chosen to always switch doors.\n");
    printf ("How many games would you like to play?\n\n");
    scanf ("%d", &Ngames);
    percent = PERCENTAGE(win, kernel(Ngames));
    printf ("\nYou have chosen the winning door %f percent of the time.\n\n", percent);
    win = 0;
    break;

这是函数的样子

int kernel (int Ngames)
{
    int gamesplayed;
    for (gamesplayed = 0; gamesplayed <= Ngames; gamesplayed++)
    {
        int winningdoor = RANDOM();
        int chosendoor = 1;
        int revealeddoor,switcheddoor;
        do
        {
            revealeddoor = RANDOM();
        }
        while (revealeddoor == chosendoor || revealeddoor == winningdoor);
        do
        {
            switcheddoor = RANDOM();
        }
        while (switcheddoor == chosendoor || switcheddoor == revealeddoor);
        if (switcheddoor == winningdoor)
        {
            win = winning(win);
        }
    }
    return gamesplayed;
}

现在，当您查看特定的开关盒时，您可以更好地理解它的含义。只有当你需要知道内核的逻辑时，你才需要进入。

我想说的是，如果您正在计算具有实际意义的东西，请将其作为函数移出。如果您重用某些代码，请将其设为函数。如果你的函数很长或者有太多的循环（while、for）或条件嵌套（if、switch），那么是时候模块化你的代码了。