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我正在使用 libnds 为 C++ 中的 NDS 编码，但这个问题不是 NDS 特定的。我目前有一个基于文本的游戏，其中顶部屏幕只显示一个徽标，而您在底部屏幕上玩。

所以我想添加一种单DS多人游戏，其中一个玩家在顶部屏幕上播放，另一个在底部屏幕上。我在设置两个屏幕的文本引擎时没有问题，我只需要找到一种在多人游戏中有效编码的方法。下面我写了一个摘要或简化版本。

注意：consoleClear() 清除屏幕，游戏停止的唯一地方是暂停函数。

//Headers

void display(int x,int y,const char* output))
{
    printf("\x1b[%d;%dH%s", y, x,output);
}

void pause(KEYPAD_BITS key) //KEYPAD_BITS is an ENUM for a key on the NDS
{
    scanKeys();
    while (keysHeld() & key)
    {
        scanKeys();
        swiWaitForVBlank();
    }
    while (!(keysHeld() & key))
    {
        scanKeys();
        swiWaitForVBlank();
    }
    return;
}

void pause() //Only used to simplify coding
{
    pause(KEY_A);
    return;
}

int main(void)
{
    //Initializations/Setup
    while (1)
    {
        if (rand()%2==1) //Say Hello
        {
            if (rand()%3!=1) //To Friend (greater chance of friend than enemy)
            {
                display(6,7,"Hello Friend!");
                display(6,8,"Good greetings to you.");
                pause();
                consoleClear(); //Clears text
                display(6,7,"Would you like to come in?");
                pause();
                //Normally more complex complex code (such as interactions with inventories) would go here
            }
            else //To enemy
            {
                display(6,7,"Hello enemy!");
                display(6,8,"I hate you!");
                pause();
                consoleClear();
                display(6,7,"Leave my house right now!!!");
                pause();
            }
        }
        else //Say goodbye
        {
            if (rand()%4==1) //To Friend (lesser chance of friend than enemy)
            {
                display(6,7,"Goodbye Friend!");
                display(6,8,"Good wishes to you.");
                pause();
                consoleClear();
                display(6,7,"I'll see you tomorrow.");
                pause();
                consoleClear();
                display(6,7,"Wait, I forgot to give you this present.");
                pause();
            }
            else //To enemy
            {
                display(6,7,"Goodbye enemy!");
                display(6,8,"I hate you!");
                pause();
                consoleClear();
                display(6,7,"Never come back!!");
                pause();
                consoleClear();
                display(6,7,"Good riddance!"); //I think I spelt that wrong...
                pause();
            }
        }
    }
}

我知道 goto 令人困惑并且可以被认为是一个坏习惯，但我想不出更好的方法。我的集成多人游戏版本：

//Headers and same functions

int game(int location)
{
    switch (location)
    {
    case 1: goto one; break;
    case 2: goto two; break;
    case 3: goto three; break;
    case 4: goto four; break;
    case 5: goto five; break;
    case 6: goto six; break;
    case 7: goto seven; break;
    case 8: goto eight; break;
    case 9: goto nine; break;
    case 10: goto ten; break;
    default: break;
    }

    if (rand()%2==1) //Say Hello
    {
        if (rand()%3!=1) //To Friend (greater chance of friend than enemy)
        {
            display(6,7,"Hello Friend!");
            display(6,8,"Good greetings to you.");
            return 1;
one:;
            consoleClear(); //Clears text
            display(6,7,"Would you like to come in?");
            return 2;
two:;
            //Normally more complex complex code (such as interactions with inventories) would go here
        }
        else //To enemy
        {
            display(6,7,"Hello enemy!");
            display(6,8,"I hate you!");
            return 3;
three:;
            consoleClear();
            display(6,7,"Leave my house right now!!!");
            return 4;
four:;
        }
    }
    else //Say goodbye
    {
        if (rand()%4==1) //To Friend (lesser chance of friend than enemy)
        {
            display(6,7,"Goodbye Friend!");
            display(6,8,"Good wishes to you.");
            return 5;
five:;
            consoleClear();
            display(6,7,"I'll see you tomorrow.");
            return 6;
six:;
            consoleClear();
            display(6,7,"Wait, I forgot to give you this present.");
            return 7;
seven:;
        }
        else //To enemy
        {
            display(6,7,"Goodbye enemy!");
            display(6,8,"I hate you!");
            return 8;
eight:;
            consoleClear();
            display(6,7,"Never come back!!");
            return 9;
nine:;
            consoleClear();
            display(6,7,"Good riddance!"); //I think I spelt that wrong...
            return 10;
ten:;
        }
        return -1;
    }
}
int main(void)
{
    //Initializations/Setup
    int location1 = -1, location2 = -1;
    location1 = game(location1);
    location2 = game(location2);
    while (1)
    {
        scanKeys(); //Whenever checking key state this must be called
        if (keysDown() & KEY_A) //A key is used to continue for player1
            location1 = game(location1);
        if (keysDown() & KEY_DOWN) //Down key is used to continue for player2
            location2 = game(location2);
    }
}

除了这种方法是一种不好的做法之外，在实际的源代码中，我需要添加数百个 goto，这太耗时了。

任何帮助表示赞赏。如果有人有任何问题或答案，请询问/回复。

编辑：虽然不喜欢这样做，但如果有人有办法，我愿意从头开始重写游戏。

4

1 回答 1

1

对每种情况使用 if-else 条件语句是首先想到的简单解决方案。

例如：

int game(int i){
  if(i == 1){
    //first case code here.
  }
  else if(i == 2){
    //second case code here.
  }
  //....
  return 0;
}

每种情况下的代码甚至可以放入将根据每个条件调用的其他函数中。对于您的情况，这可能就足够了。

更灵活的解决方案（但更复杂）是调度表。这个想法是为每个所需的功能提供单独的函数，并将它们的指针放在一个数组中。然后，您可以使用这些函数指针通过索引表来调用它们。如果您有一系列执行（函数调用）要完成并且您希望轻松完成，或者您希望根据您的输入获得不同的结果，而无需更改程序，这将非常有用。

下面有一个例子。如果您将 std::cout 替换为 printf 并将iostream替换为stdio库，则此代码也可以在 C 中使用。

#include <iostream>

using namespace std;

// Arrays start from 0.
// This is used for code
// readability reasons.
#define CASE(X) X-1 

typedef void (*chooseCase)();

// Functions to execute each case.
// Here, I am just printing
// different strings.
void case1(){
    cout<< "case1" << endl;
}

void case2(){
    cout<< "case2" << endl;
}

void case3(){
    cout<< "case3" << endl;
}

void case4(){
    cout<< "case4" << endl;
}

//Put all the cases in an array.
chooseCase cases[] = {
    case1, case2, case3, case4
};

int main()
{
    //You can call each scenario
    //by hand easily this way:
    cases[CASE(1)]();
    cout << endl;

    //Idea: You can even set in another
    // array a sequence of function executions desired.
    int casesSequence[] = {
        CASE(1), CASE(2), CASE(3), CASE(4),CASE(3),CASE(2),CASE(1)
    };
    //Execute the functions in the sequence set.
    for(int i = 0; i < (sizeof(casesSequence)/sizeof(int)); ++i){
        cases[casesSequence[i]]();
    }

    return 0;
}

这将在输出中打印：

case1

case1
case2
case3
case4
case3
case2
case1
于 2013-08-12T00:57:50.960 回答