我正在开发Processing(在Java上运行)的双人游戏。一个用户将使用WASD键控制他的角色,另一个用户将使用箭头键控制移动。我遇到的问题是当按下箭头时使用keyPressed
否定WASD,反之亦然。我一直在搞乱它很长一段时间。有没有人知道一个工作或注意到我做错了什么?
//global variables
int wide = 600; //canvas width
int tall = 600; //canvas height
int s = 50; //player size
float speed = 2.5; //player movement speed
//colors
int redColor = #CB4646; //player 1 color
int blueColor = #4652CB; //player 2 color
int backgroundColor = #DBE3B3; //background color
float player1X = 600/3-s; //HOW COME width/3 DOESN'T WORK??????????
float player2X = 600*2/3;
float playerY = 600/2-(s/2);
//players
Player player1 = new Player(player1X, playerY, s, speed, "wasd", redColor); //player 1
Player player2 = new Player(player2X, playerY, s, speed, "arrows", blueColor); //player 2
//setup
void setup(){
background(backgroundColor);
size(wide, tall);
smooth();
println(player2.controls);
}
//draw
void draw(){
background(backgroundColor);
player1.usePlayer();
player2.usePlayer();
}
class Player{
//class variables
float x; // x position
float y; // y position
int s; //size
float speed; //speed
String controls; //controls
int colors; //player color
char keyControls [] = new char [4];
//construct
Player(float tempX, float tempY, int tempS , float tempSpeed, String tempControls, int tempColors){
x = tempX;
y = tempY;
s = tempS;
speed = tempSpeed;
controls = tempControls;
colors = tempColors;
}
void usePlayer(){
// draw player
fill(colors);
rect(x, y, s, s);
//move player
keyPressed();
//wraparound
boundaries();
}
void keyPressed(){
//sets controls for wasd
if(controls == "wasd"){
if(key == 'w' || key == 'W'){
y -= speed; //move forwards
}
if(key == 's' || key == 'S'){
y += speed; //move backwards
}
if(key == 'd' || key == 'D'){
x += speed; //move right
}
if(key == 'a' || key == 'A'){
x -= speed; //move left
}
}
//sets controls for arrows
if(controls == "arrows"){
if(key == CODED){
if(keyCode == UP){
y -= speed; //move forwards
}
if(keyCode == DOWN){
y += speed; //move backwards
}
if(keyCode == RIGHT){
x += speed; //move right
}
if(keyCode == LEFT){
x -= speed; //move left
}
}
}
}
//pacman style wraparound
void boundaries(){
if(x == width) x = 2;
if(y == height) y = 2;
if(x == 0) x = width-s;
if(y == 0) y = height-s;
}
}
答案 0 :(得分:3)
独立跟踪您的密钥,不要依赖事件全局。
boolean[] keys = new int[255];
void keyPressed() {
keys[keyCode] = true;
}
void keyReleased() {
keys[keyCode] = false;
}
void draw() {
updatePlayers();
drawStuff();
}
void updatePlayers() {
if(keys[LEFT]) { p1.move(-1,0); }
if(keys[RIGHT]) { p1.move(1,0); }
if(keys[UP]) { p1.move(0,-1); }
if(keys[DOWN]) { p1.move(0,1); }
if(keys['a']) { p2.move(-1,0); }
if(keys['d']) { p2.move(1,0); }
if(keys['w']) { p2.move(0,-1); }
if(keys['s']) { p2.move(0,1); }
}
请注意,这必须是一系列if语句,因为您要处理所有按下的键。如果有人左右握住,p1将向左和向右移动。
另请注意,此示例代码不会过滤特殊键的高于255的代码,因此您可能希望在事件处理程序的开头添加“if(keyCode> 255)return”
答案 1 :(得分:0)
key
是全局变量吗?我没有看到它传递给播放器。如果它是全局的,那么它一次只能容纳一个键,这就不能同时控制两个玩家。
答案 2 :(得分:0)
这是我用来处理同时按键(对角移动)的Arduino / Processing代码。它修复了boolean[] cannot be cast to int[]
错误的问题,如@Brannon所示,并使用keyCodes而不是key。
import processing.serial.*;
boolean[] keys = new boolean[255];
Serial port;
void setup() {
port = new Serial(this, Serial.list()[1], 9600);
}
void draw() {
// loop through boolean array and see which ones (index = keyCode)
// are true, then write to them.
for(int i = 0; i < 255; i++) {
if(keys[i]) {
if (i == 87) { port.write('w'); }
if (i == 65) { port.write('a'); }
if (i == 83) { port.write('s'); }
if (i == 68) { port.write('d'); }
}
}
}
void keyPressed() {
keys[keyCode] = true;
}
void keyReleased() {
keys[keyCode] = false;
}