Question

除了askCalcChoice1之外，这个计算器一切正常。由于askCalcChoice1是一个字符串，我称之为错（显然）。错误说它不能将字符串转换为int，以及将int转换为boolean。但是，当我将inputOperation作为字符串时，它会破坏askCalcChoice1下面的其他2个调用。（它会破坏displayRedults和askTwoValues，因为它们不是字符串）。我不知道如何格式化askCalcChoice以便调用这个在另一个类中编写的方法而不会破坏任何东西。 askCalcChoice被写成一个字符串，我粘贴在oopCalculator代码下面。有没有办法，有人可以告诉我如何在oopCalculator中编写该部分代码吗？

    int inputOperation; // user to choose the function
    askCalcChoice1 myAskCalcChoice1 = new askCalcChoice1(); 

    //menu becomes a complete string below
    String menu = "Welcome to Hilda Wu's Calculator\t\t"
            + "\n1. Addition\n"
            + "2. Subtraction\n"
            + "3. Multiplication\n"
            + "4. Division\n"
            + "5. Exit\n\n";
    calculatorCommands.pickNewSymbol(menu); //complete menu will be picked up as a string and display
    calculatorCommands.putDownSymbol();

    while (inputOperation = myAskCalcChoice1.calcChoice()) { //this will call for myAskCalcChoice1 class
        calculatorCommands.pickNewSymbol("\n"); //pick up the class 
        calculatorCommands.putDownSymbol(); //display the class

        askTwoValues myAskTwoValues = new askTwoValues();           
        float[] myFloats = myAskTwoValues.inputFloats(inputOperation);


        displayResults myDisplayResults = new displayResults();
        float result = myDisplayResults.showResults(inputOperation, myFloats);

        String strFormat = "The answer is:  " + result + "\n\n"; //print out The answer is as a string
        calculatorCommands.pickNewSymbol(strFormat); //pick up string from above
        calculatorCommands.putDownSymbol(); //display string
        calculatorCommands.pickNewSymbol(menu); // pick up menu from the beginning of code, loop to calculator menu
        calculatorCommands.putDownSymbol(); //display menu as loop
    }
    calculatorCommands.pickNewSymbol("\nThank you for using Hilda Wu's Calculator\n"); //when user choose to exit calculator
    calculatorCommands.putDownSymbol();
}






String calcChoice() {
    String input;
    do { //do loop will continue to run until user enters correct response
        System.out.print("Please enter a number between 1 and 5, A for Addition, S for Subtraction, M for Multiplication, or D for Division, or X for Exit: ");
        try { 
            input = readInput.nextLine(); //user will enter a response
            if (input.equals("A") || input.equals("S") || input.equals("M") || input.equals("D") || input.equals("X")) {
                System.out.println("Thank you");
                break; //user entered a character of A, S, M, or D
            } else if (Integer.parseInt(input) >= 1 && Integer.parseInt(input) <= 5) { 
                System.out.println("Thank you");
                break; //user entered a number between 1 and 5
            } else {
                System.out.println("Sorry, you have entered an invalid choice, please try again.");
            }
            continue;
        }
        catch (final NumberFormatException e) {
            System.out.println("You have entered an invalid choice. Try again.");
            continue; // loop will continue until correct answer is found
        }
    } while (true);
    return input;
}

}

Answer 1

首先，您使用两个参数调用showResults： int选择和 float [] f

从不使用选择。您在交换机中使用输入变量，但默认情况下返回显示选项的错误。

最好将选择作为参数传递给函数，并确保它是char而不是其他类型。

这也不是一个好的陈述问题的形式。我不会评价它，但请重新制作它，以便正确显示整个代码。我无法轻易理解它。我可能已经误解了它。请不要在它们之间添加注释，请确保您有正确的缩进，并且您已获得所有代码。如果您需要发表评论，请在之后进行评论。它不是很复杂，只需告诉我们代码，然后再问一下是什么问题;）

如果选择是为了传递开关...那么就这样做，但不是作为int而是作为char。

控制台计算器错误java

1 个答案: