Question

所以我最近开始编写一个计算器，它将提供1-50的两个随机数和*, +或-符号。但是，我不太确定如何实际检查用户输入的答案是否正确，因为我实际上无法计算答案。任何帮助都会非常感谢。下面的代码（也很抱歉没有注释）

public static void main(String[] args) {
    // TODO Auto-generated method stub

    System.out.println("Hello, what is your name?"); //Asks the childs name
    Scanner demo = new Scanner(System.in); //creates the scanner
    String a = demo.nextLine();
    System.out.println("Hi " + a + " I hope your ready for the quiz");
    System.out.println("Lets begin");


    String [] arr = {"*", "+", "-"};
    Random random = new Random();
    Random no = new Random();
    for(int counter1= 1; counter1 <=1;counter1++){
    int select = random.nextInt(arr.length);
    int firstnumber;
    for(int counter= 1; counter <=1;counter++){
        firstnumber = no.nextInt(50);
        int firstnumber2;
        firstnumber2 = no.nextInt(50);
        System.out.println(firstnumber + " " + arr[select] + " " + firstnumber2);

        int b = demo.nextInt();

Answer 1

您必须使用if（或switch）才能了解如何计算输入。

int result = 0;
if (arr[select].equals("*")
    result = firstnumber * secondnumber; // do not call it firstnumber2
else if (arr[select].equals("+")
    result = firstnumber + secondnumber;
else // if (arr[select].equals("-")) - else if not needed if only three elements are used
    result = firstnumber - secondnumber;

switch：

switch(arr[select]) {
    case "*": result = firstnumber * secondnumber;
              break;
    case "+": result = firstnumber + secondnumber;
              break;
    case "-": result = firstnumber - secondnumber;
              break;
    default: break;
}

Answer 2

有几种方法可以解决这个问题。一个好的风格如下：

@FunctionalInterface
interface CalcFunction{
    int calc(int a , int b);
}

HashMap<String , CalcFunction> operations = new HashMap<>();
operations.put("*" , (a , b) -> a * b);
operations.put("/" , (a , b) -> a / b);
operations.put("+" , (a , b) -> a + b);
operations.put("-" , (a , b) -> a - b);

//select a random operation
String op = generateOperation();

//generate operands
int a = randomNumber();
int b = randomNumber();

//the correct result
int expected = operations.get(op).calc(a , b);

这种方法的优点是它可以通过额外的操作轻松扩展。

编辑：
基本思想是映射每个表达式＆＃34; *＆＃34;，＆＃34; /＆＃34;，＆＃34; +＆＃34;，＆＃34; - ＆＃34;到CalcFunction的实例，它在calc中完全实现了该操作。由于CalcFunction是Functional Interface，因此可以用lambda表示。

(a , b) -> a * b

例如也可以表示为

new CalcFunction(){
    int calc(int a , int b){
        return a * b;
    }
}

Java：如何从string和int计算答案

2 个答案: