问题描述
大家好，
我这里有这个简单的计算器。目前，它只接受两个整数和一个运算符。我想接受两个以上的整数（无限）。请记住，我不是在寻找操作顺序，只是从左到右。我考虑过递归，但不知道如何去做。
Hi everyone,
I have this simple calculator here. At the moment, it only accepts two integers and an operator. I want to accept more than two integers (infinite). Keep in mind I am not looking for order of operations, just left to right. I thought about recursion, but not sure how to go about it.
import java.util.Scanner;
public class Calc24 {
public static void main(String[] args) {
try (Scanner scanner = new Scanner(System.in)) {
System.out.println("Enter an expression of the form 3 m 5");
int n1 = scanner.nextInt();
String operation = scanner.next();
int n2 = scanner.nextInt();
switch (operation) {
case "a":
System.out.println("Your answer is " + (n1 + n2));
break;
case "s":
System.out.println("Your answer is " + (n1 - n2));
break;
case "d":
System.out.println("Your answer is " + (n1 / n2));
break;
case "m":
System.out.println("Your asnwer is " + (n1 * n2));
break;
default:
System.out.println("Je ne sais pas");
}
}
}
}推荐答案
17 + 3 - 11 * 44 / 6
阅读前三项（ operand1 操作 operand2 ）
进行计算并将答案保存为新的 operand1 。
获取接下来的两项（操作 operand2 ）
进行计算并将答案保存为新 operand1 。
重复直到没有更多项目。
一旦实施并了解到那时你可以考虑使用运算符优先级，括号等。
Read first three items (operand1 operation operand2)
Do the calculation and save the answer as the new operand1.
Get the next two items (operation operand2)
Do that calculation and save the answer as the new operand1.
Repeat until no more items.
Once you have implemented and understood that then you can think about using operator precedence, parentheses etc.
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