How to calculate an expression from left to right

Question

I am currently creating a maths quiz game which outputs random arithmetic questions to the user. I have a function in which it creates the questions and stores them in a string array. The questions can vary in the amount of integers present in the question; e.g. a question can be 3+4-9/3= or 4*2/9= but the minimum amount of integers that must be present is 2. e.g. 3+4= etc.

What I am trying to do is evaluate the question from reading it left to right ignoring order of operations. E.g. 4+2*5= should equal 30 instead of 14.

My thinking to this problem is this:

Get a question, loop through the string and extract individual characters and store them in an arraylist. Convert those characters into an int and first evaluate the first three elemtns in the arraylist, store the answer in a temporary int value. Get the next 2 elements and check if it does not contain an equal sign, if it doesn't then evaluate the the temporary int value with these two elements. Continue to do this whilst it doesn't find an equals sign. If it find the equals sign then stop the iteration and produce the result.

What my problem is that I cant seem to convert the operations +,-,*,/ to int values. I thought I could as every char could be converted into an int.

public void initAnswers(){
   for(int i =0; i < questions.length; i++){
       ArrayList<Integer> indivialIntegers = new ArrayList<>();
       String questionToGetIndividual = questions[0];
       //extracting individual chars
       for(int j = 0; j < questionToGetIndividual.length(); j++){
           indivialIntegers.add(Character.getNumericValue(questionToGetIndividual.charAt(j)));
       }
   }

EDIT = This is my updated code. I am getting a number format exception error. Caused by: java.lang.NumberFormatException: Invalid int

Can anybody see why? By the way, I wrote a method to generate the questions (the amount of integes in the questions are randomly generated). They are stored in an array. Im just simply iterating through each question in that array and evaluating the answer: My starting function is initAnswers().

public int computeAnswerThreeTerms(int numberOne, int numberTwo, String operation){
   if(operation.equals("+")){
       Log.d("onc","in addition");
       return numberOne + numberTwo;
   }else if(operation.equals("-")){
       Log.d("onc","in subtraction");
       return numberOne - numberTwo;
   }else if(operation.equals("/")){
       Log.d("onc","in division");
       return numberOne / numberTwo;
   }else if(operation.equals("*")){
       Log.d("onc","in multiplication");
       return numberOne * numberTwo;
   }else{
       return 0;
   }
}

public int computeAnswer(int numberOne, String operation, int runningTotal){

    if(operation.equals("+")){
        Log.d("onc","in addition");
        return numberOne + runningTotal;
    }else if(operation.equals("-")){
        Log.d("onc","in subtraction");
        return numberOne - runningTotal;
    }else if(operation.equals("/")){
        Log.d("onc","in division");
        return numberOne / runningTotal;
    }else if(operation.equals("*")){
        Log.d("onc","in multiplication");
        return numberOne * runningTotal;
    }else{
        return 0;
    }
}

public void getAnswer(ArrayList<String> toBeExamined){
    //get numbers and put them in an array
    //get operations and put them in an array
    for(int i =0; i < toBeExamined.size(); i++){
        String extractingIntegers = toBeExamined.get(i);
        // /d matches digits /D matches non digits
        String[] numbersInStringformat = extractingIntegers.split("/d");
        ArrayList<Integer> numbersInNumberformat = new ArrayList<>();
        for(int counter = 0; counter < numbersInStringformat.length; counter++){
            numbersInNumberformat.add(Integer.parseInt(numbersInStringformat[counter]));
        }
        String[] operationsInStringformat = extractingIntegers.split("/D");

        int tempAnswer;
        int tempIntCounter =0;
        int stringCounter =0;
        tempAnswer = computeAnswerThreeTerms(numbersInNumberformat.get(tempIntCounter), numbersInNumberformat.get(tempIntCounter+1), operationsInStringformat[stringCounter]);
        tempIntCounter = 1;
        while(!operationsInStringformat[1].equals("=")){
            if(operationsInStringformat[stringCounter].equals("=")){
                break;
            }
            tempAnswer = computeAnswer(numbersInNumberformat.get(tempIntCounter++),operationsInStringformat[stringCounter],tempAnswer);
        }
        answers[i] = tempAnswer;

    }
}

public void initAnswers(){
    ArrayList<String> individualCharacters = new ArrayList<>();

    for(int i =0; i < questions.length; i++){
        //change i to 0 for testing purposes
       String questionToGetIndividual = questions[i];
       //extracting individual chars
       individualCharacters.add(questionToGetIndividual);
    }
    getAnswer(individualCharacters);
}

if youre ignoring order of operations, why not just read the string from left to right char-by-char until you reach a non-digit char and thats your first number, etc etc — Ariel Pinchover, Feb 06 '16 at 15:09
@Infested but how will I know if I have to add,multiply, subtract that number etc? — Tarikh Chouhan, Feb 06 '16 at 15:11
you can read and proof every char and if is not a number is one of the 4 mat operations — ΦXocę 웃 Пepeúpa ツ, Feb 06 '16 at 15:12
you can always save that operation as a string and when you get your second number use a switch case — Ariel Pinchover, Feb 06 '16 at 15:12
Why would you want to do it incorrectly? Why not use a recursive descent expression parser, or the Disjkstra Shunting-yard algorithm, like everybody else? — user207421, Feb 06 '16 at 16:04

score 0 · Answer 1 · edited May 23 '17 at 12:31

0

What you basically need to do is to:

Parse int[] of operands from your question (let's say n items);
Parse Operator[] of operators (+-/*). It'd be n-1 items. Operator is a simple enum with convertor from char to value & function to apply Operator to 2 operands;
Go in cycle through the array of operators applying them to current value & operand;

Aslo, check this question: What's a good library for parsing mathematical expressions in java?

edited May 23 '17 at 12:31

Community

1
1

answered Feb 06 '16 at 15:12

Konstantin Loginov

15,802
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what if the in int is a double digit like 33? That would go fill two indexes rather than one index – Tarikh Chouhan Feb 06 '16 at 15:28
@TarikhChouhan - just write a proper parser for it. For example, you can split string by "+-/*" chars and convert all substrings to integers – Konstantin Loginov Feb 06 '16 at 16:00

score 0 · Answer 2 · answered Feb 06 '16 at 15:44

Here you go

String question = "11+2*3=";
        int first = Integer.MAX_VALUE;
        int second = 0;
        int result = 0;
        boolean firstNumber = true;
        String operator = "";
        String number = "";

    for(char c : question.toCharArray()) {
        if(c != '+' && c != '-' && c != '*' && c != '/' && c != '=') {
            number += c;
        } else {
            if(firstNumber) {
                first = Integer.parseInt(number);
                firstNumber = false;
                operator += c;
                number = "";
                result = first;
            } else {
                second = Integer.parseInt(number);
                //firstNumber = true;
                switch(operator) {
                case "+":
                    result = result + second;
                    break;
                case "-":
                    result = result - second;
                    break;
                case "*":
                    result = result * second;
                    break;
                case "/":
                    result = result / second;
                    break;
                }
                number = "";
                operator = "";
                operator += c;
                if(operator.equals("=")) {
                    System.out.println(result);
                    return;
                }
            }
        }
    }

How to calculate an expression from left to right

2 Answers2