Question

Alright, guys, I’m trying to get a basic hang on Java and I figured I would make something a bit more complicated than usual. What I have in this method is an attempt to first check if the first character can be an int, and if it is, I want to see how far that number will go in the input (so if the user inputted 455+60, I’d want to pull out the 455 and stop at the +).

For some reason, if the input is 500t, it will work fine. But if the index ends after the number (say the input is just 500), it has a problem. I am thinking the for loop keeps going for some reason. Can someone tell me where I messed up? I added an if statement to see if I could stop it, but to no avail.

public void init(){
    int chars = 0;
    /*
     * Input fancy intro here.
     */
    clearArray();
    String uInput = getInput();
    rawInput=uInput.toCharArray();
    cleanArray(rawInput);
    try{Integer.parseInt(Character.toString(rawInput[0]));
    boolean isNum = true;
    for(int i=0;isNum==true;i++){
        if(i != rawInput.length){
        try{Integer.parseInt(Character.toString(rawInput[i]));}
        catch (NumberFormatException s){
            isNum = false;
            chars = i;
            System.out.println("The number has: "+chars+" characters.");
        }
        }
        }       
    }
    catch (NumberFormatException s){
        System.out.println("What is this madness?");

    }



}

Answer 1

Here, you avoid asking for rawInput[rawInput.length]:

if(i != rawInput.length)

… but you do keep going for the next iteration.

Is there any reason you can’t just do:

for (int i = 0; i < rawInput.length && isNum; i++)

? I would also suggest using Character.isDigit or a custom check rather than converting each character to a string, then parsing it and catching an exception.

Note that Character.isDigit will return true for non-ASCII digits, which you may not want. You quite possibly just want a simple check of:

char c = rawInput[i];
if (c < '0' || c > '9') {
    isNum = false;
}

Also note that a regular expression may well be a better solution for this…

Oh, and if you’ve got an empty string, that will count as a number at the moment…