Question

I’m trying to make a program that lets you input any one-step equation, and solve for it. To make it so that you can enter it in any way, I have a for loop running that checks every character to see what it is. But then, I realized that I could not use double digit numbers for this. So I tried adding a little function that checks if there is a character one to the right, to test if it is a number. And if it is a number, combine that plus the one it found, to be a single number. Like if it sees 1, and checks to the right that there is a 0 there, it give you 10. Here is the code for this function

        Find = equation.charAt(i);

        if (Character.isDigit(Find))
        {
            if(Found == 0)
            {
                if(i < equation.length() - 1)
                {
                    FindNext = equation.charAt(i + 1);
                    if (Character.isDigit(FindNext))
                    {
                        one = (Character.toString(Find) + Character.toString(FindNext));
                        Found = 2;
                        One = Double.parseDouble(one);
                    }
                }
                else
                {
                    one = (Character.toString(Find));
                    Found = 2;
                    One = Double.parseDouble(one);
                }
            }
            else
            {
                if(i + 1 < equation.length() - 1)
                {
                    FindNext = equation.charAt(i + 1);
                    if (Character.isDigit(FindNext))
                    {
                        two = (Character.toString(Find) + Character.toString(FindNext));
                        Two = Double.parseDouble(one);
                    }
                }
                else
                {
                    two = (Character.toString(Find));
                    Two = Double.parseDouble(two);
                }
            }
        }

Does anyone see what I’m doing wrong?

Answer 1

And if the number is 100? 1000? 10000? …you get the point. You cannot simply check one character ahead. You should instead separate your parsing by operations you support (i.e. +, -, *, etc) and it will be much easier to obtain the numbers.