我想做什么的总结：我正在做一个非常简单的游戏，它使用do / while功能并切换到检查你所在的房间。如果你在房间里= 1 ，交换机选择room1并运行room1（）函数。在房间内，您可以选择去哪个方向。如果你选择北或东，门1（）或门4（）功能将运行说'＃34;你想打开这扇门吗？＆＃34;你可以说＆＃34; Open＆＃34;并走进隔壁房间，你的房间= 1将更新为新值。
什么是破坏：这一切都运行得很漂亮，功能（虽然它们有点臃肿）似乎正在按照它们应有的方式工作。一个主要的问题是，每当我走进一扇门时，我的变量都会重置，所以我总是在房间1，我的指南针总是等于0，没有一个门或房间等于＆＃34;访问＆＃34; （例如：room1V = 0）。
基本上，我走过一扇门，然后我再次进入房间1，而不是2号或4号房间。
这是代码和演练：
var room1V = 0; //room1 // these variables tell the computer whether I have 'visited' a room before.
var room2V = 0; //room2
var room3V = 0; //room3
var room4V = 0; //room4
var door1V = 0; //room1 - room2 // these variables tell the computer whether I have used a door before.
var door2V = 0; //room2 - room3
var door3V = 0; //room3 - room4
var door4V = 0; //room4 - room1
var compass = 0; // which side of the room am I on?
var room = 1; // what room am I currently in?
var reply = 1; // this is re-declared as a local variable in each function, and it works fine
win = 0; // this will eventually tell the room-check do/while to stop
// This do/while checks what room I am in:
do {
quit = 0;
switch(room) {
case '1':
room1(compass,room1V);
break;
case '2':
room2(compass,room2V);
break;
case '3':
room3(compass,room3V);
break;
case '4':
room4(compass,room4V);
break;
}
} while (win != 1);
由于默认值为room = 1，因此将启动room1（指南针，room1V）功能。
function room1(compass,room1V) {
if (room1V === 1) {
console.log("You are in room 1 again.");
document.write("You are in room 1 again." + "<br>");
var reply = prompt("Where would you like to go? EAST, NORTH?");
switch(reply.toLowerCase()) {
case 'east':
compass = "east"; //because you are trying to open the east door, you will now see room1 from the east. If you make it through, your compass will update to 'west' because you will be on the west side of room2.
door1(compass,door1V);
break;
case 'north':
compass = "north";
door4(compass,door4V);
break;
default:
console.log("Something went wrong.");
document.write("Something went wrong." + "<br>");
}
} else {
console.log("You are in room 1.");
document.write("You are in room 1." + "<br>");
room1V = 1;
reply = prompt("Where would you like to go? EAST, NORTH?");
switch(reply.toLowerCase()) {
case 'east':
compass = "east";
door1(compass,door1V);
break;
case 'north':
compass = "north";
door4(compass,door4V);
break;
default:
console.log("Something went wrong.");
document.write("Something went wrong." + "<br>");
}
}
} // Working
如果我去北方，＆＃39;罗盘会更新，说明我正在查看房间的方向，并且门4（房间，门4V）将会运行。
function door4(room,door4V) {
if (door1V === 1) {
console.log("You approach door 4 again.");
document.write("You approach door 4 again." + "<br>");
var reply = prompt("What would you like to do? OPEN, QUIT?");
switch(reply.toLowerCase()) {
case 'open':
if (room === 4) {
room = 1;
compass = "north";
console.log("You walk through door 4 into room 1...");
document.write("You walk through door 4 into room 1..." + "<br>");
} else {
room = 4;
compass = "south";
console.log("You walk through door 4 into room 4...");
document.write("You walk through door 4 into room 4..." + "<br>");
}
quit = 1;
break;
case 'quit':
quit = 1;
break;
default:
console.log("Something went wrong.");
document.write("Something went wrong." + "<br>");
}
} else {
console.log("You approach door 4.");
document.write("You approach door 4." + "<br>");
var reply = prompt("What would you like to do? OPEN, QUIT?");
switch(reply.toLowerCase()) {
case 'open':
if (room === 4) {
room = 1;
compass = "north";
console.log("You walk through door 4 into room 1...");
document.write("You walk through door 4 into room 1..." + "<br>");
} else {
room = 4;
compass = "south";
console.log("You walk through door 4 into room 4...");
document.write("You walk through door 4 into room 4..." + "<br>");
}
quit = 1;
break;
case 'quit':
quit = 1;
break;
default:
console.log("Something went wrong.");
document.write("Something went wrong." + "<br>");
}
}
} // Working
此时，do / while函数应该说，＆＃34;哦！由于房间= 4，你现在在4号房间。＆＃34;但事情并非如此。 房间确实= 4，但是当do / while重新运行时，我回到了room1，并且所有变量似乎都已重置。
答案 0 :(得分：4)
您的变量“重置”的原因是因为JavaScript中的基元按值传递。在do-while循环中，您将变量room1V .. room4V传递给函数room1 .. room4。变量是JavaScript中的原语，并按值传递。这意味着在被调用函数中对它们所做的任何更改都不会反映在原始参数中。使用door4和compass调用door4v函数时也是如此：这些变量按值传递。要更好地理解这一点，请尝试阅读this chapter。
答案 1 :(得分：1)
您遇到了“变量阴影”（Variable shadowing in JavaScript）
的问题var fancy = 0;
function room1(fancy) {
/* here, there are two variables with the name 'fancy' in scope */
fancy = 1;
}
console.log(fancy); // 0
room1();
console.log(fancy); // 0
您可以通过将room1参数更改为具有不同的名称来解决此问题：
function room1(otherFancy) { fancy = 1; }
现在：
console.log(fancy); // 0
room1();
console.log(fancy); // 1
答案 2 :(得分：0)
您永远不会更改房间变量的值以匹配新房间。它在全局变量中设置并且不会更改。
像你这样在do while循环中更新你的房间：
// This do/while checks what room I am in:
do {
quit = 0;
switch(room) {
case '1':
room1(compass,room1V);
room = 1;
break;
case '2':
room2(compass,room2V);
room = 2;
break;
case '3':
room3(compass,room3V);
room = 3;
break;
case '4':
room4(compass,room4V);
room = 4;
break;
}
} while (win != 1);