3
所以我想说让int Posx;
拥有一个数字,然后当用户浏览一个菜单时,根据他们选择的内容,将添加1,0或-1到int Posx
。我不能让Posx
在添加或减去之后保留数字。我试图通过数组浏览,我想保留2个整数作为标记,让我知道我现在在数组中关于x和y的位置。我怎样才能让计数器在while循环中工作?
tl; dr如何在do/while循环中保留一个计数器?
这里是哪里出了问题:
//this variable will hold if we found or not the string in the puzzle
boolean found = true;
do {
//Print out the array
for (int x = 0; x < maze.length; x++) {
for (int y = 0; y < maze[0].length; y++) {
System.out.print(maze[x][y]);
}
System.out.println();
}
System.out.printf("You may:\n1) Move up\n2) Move down\n3) Move left\n4) Move right\n0) Quit\nYour Choice (0-4):\n");
int usrAns = sc2.nextInt();
if (usrAns == 0){
System.out.println("Bye!\n");
found = false;
break;
}
//arrays to hold the direction.
int[] movx ={ -1, 0, 0, 1};
int[] movy ={ 0, -1, 1, 0};
//Array to hold the position.
int Posx = 0;
int Posy = 0;
if (usrAns == 1){
if(check(Posx, Posy, maze, movx[1], movy[1])){
System.out.println("Cannot move past cave boundary! Try something else.\n");
continue;
}
else{
Posy = Posy - 1;
System.out.printf("This is Posx %d and Posy %d\n", Posx, Posy);
continue;
}
}
if (usrAns == 2){
if(check(Posx, Posy, maze, movx[2], movy[2])){
System.out.println("Cannot move past cave boundary! Try something else.\n");
continue;
}
else{
Posy= Posy + 1;
System.out.printf("This is Posx %d and Posy %d\n", Posx, Posy);
continue;
}
}
if (usrAns == 3){
if(check(Posx, Posy, maze, movx[0], movy[0])){
System.out.println("Cannot move past cave boundary! Try something else.\n");
continue;
}
else{
Posx = Posx - 1;
System.out.printf("This is Posx %d and Posy %d\n", Posx, Posy);
continue;
}
}
if (usrAns == 4){
if(check(Posx, Posy, maze, movx[3], movy[3])){
System.out.println("Cannot move past cave boundary! Try something else.\n");
continue;
}
else{
Posx =Posx + 1;
System.out.printf("This is Posx %d and Posy %d\n", Posx, Posy);
continue;
}
}
while (usrAns >= 5 || usrAns < 0){
System.out.println("Please enter a number between 0 and 4:\n");
usrAns = sc2.nextInt();
if (usrAns == 0){
System.out.println("Bye!\n");
found = false;
break;
}
}
}while(found);
任何意见,建议,或技巧将不胜感激。
不仅如此,我们不需要初始化Posx和Posy,我们不想这样做,因为它每次循环都会重置我们的位置。 – 3yakuya
@Byakuya谢谢,编辑我的帖子;-) –
谢谢!不能相信这件事很简单... – Drieke