用字符串解析数学表达式

Question

I have been trying to find a solution for this for days now and I cannot figure this out. I am writing a program that (at one point) takes in mathematical expressions and displays the answer. I am using the beanshell parser to do this, but the evaluation isn't the problem. When the user presses the "undo" button, this method is supposed to undo the last input (either an operation (+ - * /) or a number. It gives me errors at the strangest time and I can't figure out why. Can anybody help? I would like to thank anybody who helps in advance!

public void undo(View v) throws EvalError{
    Interpreter interpreter1 = new Interpreter(); // interpreter to evaluate user solution
    userExpressionList.remove(userExpressionList.size()-1); // remove last element of userExpressionList
    String tempExp3 = "";
    if (userExpressionList.size() != 0){
        for (String element:userExpressionList) {
            tempExp3 = tempExp3 + element;
            if (tempExp3.substring(tempExp3.length() - 1).equals("+") || tempExp3.substring(tempExp3.length() - 1).equals("-") ||
                tempExp3.substring(tempExp3.length() - 1).equals("*") || tempExp3.substring(tempExp3.length() - 1).equals("/")) {
                    displaySolution = (Double)interpreter1.eval(tempExp3.substring(0, tempExp3.length() - 2));
                    userSolution = tempExp3.substring(tempExp3.length() - 1);
                }
            else {
                displaySolution = (Double) interpreter1.eval(tempExp3.substring(0, tempExp3.length() - 1));
                userSolution = "";
                }
            Log.i("tempExp3", tempExp3);
            Log.i("displaySolution", displaySolution.toString());
        }
        textViewUserSolution.setText(displaySolution.toString());
    }
    else {
        clear(findViewById(R.id.clearButton));
        textViewUserSolution.setText("");
    }
    Log.i("isExpectingNumber before invert", String.valueOf(isExpectingNumber));
    isExpectingNumber = !isExpectingNumber;
    Log.i("isExpectingNumber after invert", String.valueOf(isExpectingNumber));
    textViewUserExpression.setText(tempExp3);
}

If you need any more information, please ask. I really appreciate any help you guys can offer.

Answer 1

The problem is very likely based on the assumption that "going back" character by character will always result in something that can be evaluated without a problem.

Consider:

1+23 - you can undo '3' and eval "1+2"
1+2  - you can undo '2' but you can't eval "1+", just "1"
1+2+ - you can undo '+' and eval "1+2"

Therefore,

switch(exp3.charAt(exp3.length()-1)){
case '+': case '-': case '*': case '/':
    // 1+2+
    sol = int.eval( exp3.substring( 0, exp3.length()-1 ) ); // not -2!       
    usrSol = exp3.substring(exp3.length() - 1);
    break;
default:
    if( exp3.length() > 2 ){
        char d = exp3.charAt(exp3.length()-2);
        if( '0' <= d && d <= '9' ){
             // 1+23
             sol = int.eval(exp3.substring(0, exp3.length() - 1));
        } else {
             // 1+2
             sol = int.eval(exp3.substring(0, exp3.length() - 2));
        }
    } else {
        // "12"
        sol = int.eval(exp3.substring(0, exp3.length() - 1));
    }
    usrSol = "";
    break;
}