Java 这种转变还在继续
我正在做一个简单的项目,只是为了让自己保持新鲜。决定做一台自动售货机。好吧,我坚持到底,并测试它。它在案例1和案例2中运行得很好,但当我进入案例3时,它要么在我选择零食选项时完成整个程序,要么在案例3中将其更改为if/else,当我选择零食时,它将停止运行。为什么要这样做?我能做些什么来修复它Java 这种转变还在继续,java,if-statement,switch-statement,Java,If Statement,Switch Statement,我正在做一个简单的项目,只是为了让自己保持新鲜。决定做一台自动售货机。好吧,我坚持到底,并测试它。它在案例1和案例2中运行得很好,但当我进入案例3时,它要么在我选择零食选项时完成整个程序,要么在案例3中将其更改为if/else,当我选择零食时,它将停止运行。为什么要这样做?我能做些什么来修复它 import java.util.Scanner; public class Vending { public static void main(String[] args) { i
import java.util.Scanner;
public class Vending
{
public static void main(String[] args)
{
int choice;
double deposit = 0;
double total = 0;
Scanner keyboard = new Scanner(System.in);
boolean test = true;
do
{
System.out.println("\nVending Machine Menu");
System.out.println("\n \n1. View Items");
System.out.println("2. Put Money into Machine");
System.out.println("3. Select an Item");
System.out.println("4.Recieve Change");
System.out.println("5.Exit");
System.out.println("\nPlease Make a Selection: ");
choice = keyboard.nextInt();
switch(choice)
{
case 1 :
System.out.println("\n 1. Fizzy soda : $0.50 \n 2. Gummi Possums : $0.25"
+ "\n 3. Doggy Bones : $0.25 \n 4. Fruity Punch : $0.50");
System.out.println("Press ENTER to continue");
try{System.in.read();}
catch(Exception e){}
break;
case 2:
System.out.println("Enter amount to deposit: ");
deposit = keyboard.nextDouble();
total = total + deposit;
break;
case 3:
Scanner keypad = new Scanner(System.in);
System.out.println("\n 1. Fizzy soda : $0.50 \n 2. Gummi Possums : $0.25"
+ "\n 3. Doggy Bones : $0.25 \n 4. Fruity Punch : $0.50");
System.out.println("\nSelect an item: ");
int snack = keypad.nextInt();
if (total > 0)
{
if(snack == 1)
{
if (total >= .5)
{
System.out.println("The Vending Machine dispenses a FIZZY SODA");
total = total - .5;
}
else
{
System.out.println("Deposit more money");
}
}
else if(snack == 2)
{
if (total >= .25)
{
System.out.println("The Vending Machine dispenses a package of GUMMI POSSUMS");
total = total - .25;
}
else
{
System.out.println("Deposit more money");
}
}
else if(snack == 3)
{
if(total >= .25)
{
System.out.println("The Vending Machine Dispenses a package of DOGGY BONES");
total = total - .25;
}
else
{
System.out.println("Deposit more money");
}
}
else if(snack == 4)
{
if(total >= .5)
{
System.out.println("The Vending Machine Dispenses a can of FRUITY PUNCH");
total = total - .5;
}
else
{
System.out.println("Deposit more money");
}
}
}
else
{
System.out.println("Please deposit some money");
}
keypad.close();
case 4:
System.out.println("You recieve " + total + " in change from the Vending Machine");
total = 0;
case 5:
System.exit(0);
}
}
while (test == true);
keyboard.close();
}
}
您忘记在案例末尾添加
break
关键字,break将阻止执行下一个案例
开关盒应类似于此模板:
case 1:
// do something
break;
case 2:
// do something
break;
default:
// do something
break;
因此,在每个案例末尾添加break
,例如:
case 4:
System.out.println("You recieve " + total + " in change from the Vending Machine");
total = 0;
break;
阅读有关java的更多信息。如果您忘记在案例末尾添加
break
关键字,break将阻止执行下一个案例
开关盒应类似于此模板:
case 1:
// do something
break;
case 2:
// do something
break;
default:
// do something
break;
因此,在每个案例末尾添加break
,例如:
case 4:
System.out.println("You recieve " + total + " in change from the Vending Machine");
total = 0;
break;
阅读有关java的更多信息。乍一看:您的
案例
s3和4结尾没有break
语句。乍一看:您的案例
s3和4结尾没有break
语句。如果不想,您需要在案例
语句结尾使用break
执行下一个case
语句:
case 3:
// do something
break; // break to avoid executing case 4
case 4:
// ...
如果不想执行下一个
case
语句,则需要在case
语句末尾使用break
:
case 3:
// do something
break; // break to avoid executing case 4
case 4:
// ...
如果你是依靠休息;若要脱离案例,则案例3、4或5中没有中断语句。如果您依赖中断;为了摆脱这种情况,在情况3、4或5中没有break语句。不需要将test与true进行比较,比如
while(test==true)
可以直接做while(test)代码>谢谢。我要澄清的是,这里不需要比较test和true,比如while(test==true)
可以直接做while(test)代码>谢谢。我会把它清理干净谢谢你。我甚至没看到我忘了休息。谢谢。我甚至没看到我忘记休息了。