将分屏多人游戏添加到c++游戏中
Adding split-screen multiplayer to c++ game
我正在用libnds为c++中的NDS编码,但这个问题不是NDS特有的。我现在有一个基于文本的游戏,上面的屏幕只显示一个标志,你在下面的屏幕上玩。
所以我想添加一种类型的单DS多人游戏,其中一个玩家在顶部屏幕上玩,另一个在底部屏幕上玩。我对设置两个屏幕的文本引擎没有问题,我只需要找到一种在多人游戏中高效编码的方法。下面我写了一个摘要或它的简化版本。
注意:consoleClear()清除屏幕,游戏停止的唯一位置是暂停功能。
//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gotos很令人困惑,可以被认为是一个坏习惯,但我想不出更好的方法了。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糟糕的做法之外,在实际的源代码中,我有数百个goto需要添加,这太耗时了。
感谢您的帮助。如果有人有任何问题或答案,请询问/回答。
编辑:虽然我不喜欢这样做,但如果有人有办法的话,我愿意从头开始重写游戏。
为每个情况使用if-else条件语句是一个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在将根据每种条件调用的其他函数中。这对你的案子来说可能已经足够了。
一个更灵活的解决方案(但要复杂得多)是调度表。这个想法是将每个所需的功能都有单独的函数,并将它们的指针放在一个数组中。然后,您可以通过使用这些函数指针对表进行索引来调用它们。如果您有一系列的执行(函数调用)要完成,并且您希望轻松地设置它,或者您希望根据您的输入获得不同的结果,而不更改程序,那么这将非常有用。
下面是一个例子。如果将std::cout替换为printf,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;
}
这将在输出端打印:
case1
case1
case2
case3
case4
case3
case2
case1
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