blocks|key|860911|text|我建议你看看model-view-controller模型，它将帮助你解决这个问题，以及如果你像这样继续你的程序将会遇到的所有其他问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|860912|编辑:快捷方式|860913|要不断地移动你的吃豆人，你需要一个单独的线程来控制它并让它移动。在pthread中可以看到这一点。|860914|如果只在程序的主运行循环中保留用户输入，则必须按几次键的问题也会消失(这里的问题是，当您按键时，处理器必须在getch()行上，否则它将不会被检测到。|860915|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Model%25E2%2580%2593view%25E2%2580%2593controller^0|6|L|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@$A|S|B|T|1|U]]|C|$]]|$1|D|3|E|5|6|7|V|8|@]|9|@]|C|$]]|$1|F|3|G|5|6|7|W|8|@]|9|@]|C|$]]|$1|H|3|I|5|6|7|X|8|@]|9|@]|C|$]]|$1|J|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|K|$L|$5|M|N|O|C|$P|Q]]]]

I suggest you take a look at the <a href="http://en.wikipedia.org/wiki/Model%E2%80%93view%E2%80%93controller" rel="nofollow">model-view-controller</a> model, it will help you with this problems and all the other problems that you will have if you continue your program like this.

Edit: shortcut

To move your pacman continuously you will need a separate thread to control it and make it move. Take a look at pthreads for that.

If you keep only the user input in the main run loop of your program, the problem that you have to press the keys a few times will go away too (the problem here is that the processor has to be on the getch() line when you press the key, otherwise it will not be detected.

blocks|key|860932|text|对此的标准解决方案是finite-state+machine。角色有六种可能的状态:站着，向上移动，向右移动，向下移动，向左移动，死亡。按键应该改变角色的状态，而不是直接按键移动角色。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|860933|在这样一个小型应用程序中，您可以使用一个非常简单的实现，而不是实现一个非常灵活的有限状态机：|860934|enum+PlayerState+{
++++Standing,
++++MovingUp,
++++MovingRight,
++++MovingDown,
++++MovingLeft,
++++Dead
};|code-block|syntax|javascript|860935|然后，在您的游戏循环中，您可以添加一个非常简单的状态检查，然后对帧采取适当的操作。|860936|switch(state)+{
case+Standing:
++++break;
case+MovingUp:
++++player.PositionY+%2B=+1;
++++break;
//+...
}|860937|最后一步是挂钩输入，这取决于您检索输入的方法。使用回调的示例如下：|860938|void+KeyDown(Key+k)+{
++++switch(k)+{
++++case+UpArrow:
++++++++if(state+!=+Dead)
++++++++++++state+=+MovingUp;
++++++++break;
++++case+RightArrow:
++++++++if(state+!=+Dead)
++++++++++++state+=+MovingRight;
++++//+...
++++}
}|860939|您可以看到为什么在更大的项目中实现更灵活的有限状态机是很重要的，但希望这个例子能给您一个易于理解的想法。|860940|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Finite-state_machine^0|A|K|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|12|8|@]|9|@$A|13|B|14|1|15]]|C|$]]|$1|D|3|E|5|6|7|16|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|17|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|18|8|@]|9|@]|C|$]]|$1|M|3|N|5|H|7|19|8|@]|9|@]|C|$I|J]]|$1|O|3|P|5|6|7|1A|8|@]|9|@]|C|$]]|$1|Q|3|R|5|H|7|1B|8|@]|9|@]|C|$I|J]]|$1|S|3|T|5|6|7|1C|8|@]|9|@]|C|$]]|$1|U|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|V|$W|$5|X|Y|Z|C|$10|11]]]]

The standard solution to this is the <a href="http://en.wikipedia.org/wiki/Finite-state_machine" rel="noreferrer">finite-state machine</a>. The character has six or so possible states: standing, moving up, moving right, moving down, moving left, dead. Rather than a keypress directly moving the character, the keypress should change the character's state.

In such a small application, rather than implementing an incredibly flexible finite-state machine, you may use a very simple implementation as such:

<pre><code>enum PlayerState {
 Standing,
 MovingUp,
 MovingRight,
 MovingDown,
 MovingLeft,
 Dead
};
</code></pre>

Then, inside your game loop you can add a very simple state check which then takes the appropriate action for the frame.

<pre><code>switch(state) {
case Standing:
 break;
case MovingUp:
 player.PositionY += 1;
 break;
// ...
}
</code></pre>

The last step is to hook input, which depends on your method of retrieving input. Using callbacks, an example is:

<pre><code>void KeyDown(Key k) {
 switch(k) {
 case UpArrow:
 if(state != Dead)
 state = MovingUp;
 break;
 case RightArrow:
 if(state != Dead)
 state = MovingRight;
 // ...
 }
}
</code></pre>

You can see why in a larger project it would be important to implement a more flexible finite-state machine, but hopefully this example has given you an easy-to-understand idea.

blocks|key|1032218|text|在按下某个键之前，函数getch将被阻塞。如果你不想被阻塞，那么在getch之前调用_kbhit，确保在输入缓冲区中有一些东西。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1032219|编辑:看看ncurses函数nodelay和cbreak。它们支持异步输入。|1032220|entityMap^0|B|5|X|5|16|6|0|E|7|M|6|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|I|$]]

Function <code>getch</code> is blocked until some key is pressed. If you don't want to be blocked then call <code>_kbhit</code> before <code>getch</code> too make sure that there is something in input buffer.

EDIT: Take a look at ncurses functions <code>nodelay</code> and <code>cbreak</code>. They enable asynchronous input.

blocks|key|861007|text|你需要一个kbhit的等价物，即。一个非阻塞的getch。这实际上提供了解决方案，在输入上设置O_NONBLOCK。请参阅示例here。一旦你有了这个，只需连续循环，只需检查按键，而不是等待实际的按键。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|861008|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cprogramming.com/fod/kbhit.html|1|http://linux.die.net/man/2/fcntl|2|http://cboard.cprogramming.com/c-programming/63166-kbhit-linux.html^0|5|5|N|5|1B|A|5|5|0|1B|A|1|1R|4|2|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]|$9|U|A|V|B|C]|$9|W|A|X|B|C]]|D|@$9|Y|A|Z|1|10]|$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]|N|$5|I|J|K|E|$L|O]]|P|$5|I|J|K|E|$L|Q]]]]

You need an equivalent of <a href="http://www.cprogramming.com/fod/kbhit.html" rel="nofollow"><code>kbhit</code></a>, ie. a non-blocking <code>getch</code>. Which really gives the solution, set <a href="http://linux.die.net/man/2/fcntl" rel="nofollow"><code>O_NONBLOCK</code></a> on the input. See an example <a href="http://cboard.cprogramming.com/c-programming/63166-kbhit-linux.html" rel="nofollow">here</a>. Once you have this, simply loop contiguously and just check for the key hit, w/o waiting on actual key press.

blocks|key|4644741|text|这很简单。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4644742|对于每个方向的make+4函数，在函数内部放入其他3个方向的函数，这些函数由kbhit激活。|4644743|然后放入一个else语句，如果你没有按下按钮，它就会继续向前移动，即(!kbhit())；然后把所有这些放在一个循环中。|4644744|4644745|如果你对所有的方向函数都这样做，你应该能够得到想要的结果。|4644746|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|M|8|@]|9|@]|A|$]]|$1|F|3|-4|5|6|7|N|8|@]|9|@]|A|$]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|J|$]]

It is pretty easy 
for each direction make 4 function 
and inside the function,put in the other 3 direction function which get activated by kbhit.

and put a else statement in which it keeps moving forward if you do not hit a button i.e (!kbhit());
and then put all this in a loop.

<hr>

If you do this for all the direction functions you should be able to get the desirable outcome.

I am making Pacman in C++ with the Ncurses library. I am able to move Pacman with my code, but I want to move it so that pacman keeps moving even when I am not pressing any key and when I press another direction key, it changes direction. Right now, the pacman only takes one step when I press a key. Also I have to press a key 2 or 3 times before pacman moves in that direction.

<pre><code>if (ch==KEY_LEFT)
{
 int b,row,column;
 getyx(stdscr,row,column);
 int h; 
 do // do-whileloop to move the pacman left until it hits the wall
 {
 column-=1;
 mvprintw(row,column,"&gt;"); //print the "&gt;" symbol
 refresh();
 waitf(0.1); //this pauses the game for 0.1sec
 attron(COLOR_PAIR(1)); 
 mvprintw(row,column,"&gt;");
 attroff(COLOR_PAIR(1));
 refresh();
 waitf(0.1);
 mvprintw(row,(b),"O"); //showing the open mouth of pacman
 refresh();
 waitf(0.1);
 attron(COLOR_PAIR(1));a
 mvprintw(row,column,"O");
 attroff(COLOR_PAIR(1));
 h = getch();
 }
 while(h == KEY_LEFT);
}

right = getch();
</code></pre>

loop to move right in an if condition

<pre><code>up = getch();
</code></pre>

loop to move up in an if condition

<pre><code>down = getch();
</code></pre>

oop for moving down in an if condition.

How do I make something happen on the screen even without a key being pressed?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

我正在用Ncurses库在C++中制作吃豆人。我可以用我的代码移动吃豆人，但我想移动它，这样即使我没有按任何键，当我按下另一个方向键时，吃豆人也会继续移动，它会改变方向。现在，当我按下一个键时，吃豆人只会走一步。另外，在吃豆人朝那个方向移动之前，我必须按一个键2到3次。if (ch==KEY_LEFT){    int...

问我怎样才能在没有按键的情况下在屏幕上做一些事情？
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问我怎样才能在没有按键的情况下在屏幕上做一些事情？EN