递归解析计算器java

nqwrtyyt 于 2021-06-30 发布在 Java

关注(0)|答案(2)|浏览(398)

我正在尝试用java创建一个递归解析计算器，用于加法、乘法和阶乘，但是在第一部分，我正在努力阅读用户输入，以便将输入拆分为数字和运算符。在调试时，我试着看它哪里出错了，我发现当“+”遍历if else语句时，它只是跳过了它。我真的不确定是什么问题，我最初尝试使用令牌，并分裂成子字符串，但它也不顺利，然后。任何帮助都将不胜感激。谢谢您

package com.company;
import java.util.Scanner;

class Main {

    public static void main(String[] param) {
        String input = input("Please enter an expression");
        int n = input.length()-1;
        String[] splitter = input.split("(?<=\\G.)");
        split(input, n);
        //int result = calculate(input);
        //String[] splitter = input.split("(?<=\\G.)");
    }
    public static String split(String input, int n) {
        String[] splitter = input.split("(?<=\\G.)");
        System.out.println(splitter[n]);
        String symbol = splitter[n];
        if (symbol.equals("+")) {
            evalADD(n, splitter);
        }
        if (symbol.equals("*")) {
            evalMULT(n, splitter);
        }
        if (symbol.equals("!")) {
            evalFACT(n, splitter);
        }
        else if (Integer.parseInt(splitter[n]) >= 0 && Integer.parseInt(splitter[n]) <=9)
        {
            if (n != 0) {
                n = n - 1;
                split(input, n);
            }
        }

        if (n != 0)
            n = n - 1;
        split(input, n);
        return input;
    }
    public static int evalADD(int n, String [] splitter){
        int arg1;
        int arg2;
        int result;
        arg1 = Integer.parseInt(splitter[n+1]);
        arg2 = Integer.parseInt(splitter[n+2]);
        result = arg1 + arg2;
        return result;
    }
    public static int evalMULT(int n, String [] splitter){
        int arg1;
        int arg2;
        int result;
        arg1 = Integer.parseInt(splitter[n+1]);
        arg2 = Integer.parseInt(splitter[n+2]);
        result = arg1 * arg2;
        return result;
    }
    public static int evalFACT(int n, String [] splitter){
        int arg1;
        int arg2;
        int result;
        arg1 = Integer.parseInt(splitter[n+1]);
        arg2 = Integer.parseInt(splitter[n+2]);
        result = arg1 - arg2;
        return result;
    }
    public static String input(String message) {
        Scanner scanner = new Scanner(System.in);
        System.out.println(message);
        return (scanner.nextLine());
    }
}

Java parsing recursion procedural-programming calculator

来源：https://stackoverflow.com/questions/53662873/recursive-parsing-calculator-java

2条答案

按热度按时间

um6iljoc1#

我注意到你在用 java.util.Scanner . 我写了一个脚本，它应该按照您的所有标准为您完成任务：

import java.util.Scanner;

class recursiveParsingCalculator {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        // Ask user to input the expression
        System.out.println("Please input the expression");
        String userInput = scanner.nextLine();
        System.out.println(
                "And the final result is: " + recursiveCalculation(userInput, userInput.length() - 1, 0, 0, 0));
        scanner.close();
        System.exit(0);
    }

    // Identify the type of character at a specific position
    public static char charOfString(String userInput, int i) {
        return userInput.charAt(i);
    }

    /*
     * Position must be userInput.length() - 1 initially. currentResults, operand1
     * and operand2 are also meant to be initilized with 0.
     */
    public static int recursiveCalculation(String userInput, int position, int operand1, int operand2,
            int currentResults) {
        // If position is zero, just output the operand.
        if (position == 0) {
            if (Character.isDigit(charOfString(userInput, position))) {
                return charOfString(userInput, position) - '0';
            } else {
                System.out.println("Invalid input.");
            }
        }
        if (position > -1) {
            // Check if it is a number or an operator
            if (Character.isDigit(charOfString(userInput, position))) {
                operand1 = charOfString(userInput, position) - '0'; // First operand

                // Check if 2nd char is a number or an operator.
                if (Character.isDigit(charOfString(userInput, position - 1))) {
                    operand2 = charOfString(userInput, position - 1) - '0';
                    position = position - 1;
                }
            } else {
                // If it is an operator, then proceed to compute the results so far
                char operator = charOfString(userInput, position);

                // If it is a binary situation
                if (operator == '+' || operator == '*') {
                    currentResults = binaryOperator(operator, operand1, operand2);
                    operand2 = currentResults;
                }
                // If it is an unary situation
                else if (operator == '!') {
                    if (currentResults == 0) {
                        currentResults = operand1;
                    }
                    currentResults = unaryOperator(currentResults);
                    operand2 = currentResults;
                } else {
                    System.out.println("Invalid operator");
                    return 0; // Return zero by default
                }
            }
            position = position - 1;
        }
        if (position > -1) {
            return recursiveCalculation(userInput, position, operand1, operand2, currentResults);
        } else {
            return currentResults;
        }
    }

    public static int binaryOperator(char operator, int operand1, int operand2) {
        switch (operator) {
        case '+':
            return operand1 + operand2;
        case '*':
            return operand1 * operand2;
        default:
            System.out.println("Invalid binary Operator");
            return 0; // Return zero by default
        }
    }

    // Calculate the factorial
    public static int unaryOperator(int operand) {
        if (operand <= 1)
            return 1;
        else
            return operand * unaryOperator(operand - 1);
    }
}

用法示例：对于二进制运算符，输入+21，它将为您添加它们。一元输入！它将产生阶乘。现在，您可以使用一元和二元运算符尝试任何组合链和数字排列，它将递归地为您计算值。
例如，考虑输入*3+12：它将1和2相加，然后乘以3，最后计算出整个表达式的阶乘，从而得到预期的362880。

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7vux5j2d2#

为什么不将输入计算字符串赋给一个字符数组，然后遍历数组并匹配字符“+”、“-”、“*”？

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