Determine if a String is an Integer in Java

Question

I'm trying to determine if a particular item in an Array of strings is an integer or not.

I am .split(" ")'ing an infix expression in String form, and then trying to split the resultant array into two arrays; one for integers, one for operators, whilst discarding parentheses, and other miscellaneous items. What would be the best way to accomplish this?

I thought I might be able to find a Integer.isInteger(String arg) method or something, but no such luck.

Answer 1

The most naive way would be to iterate over the String and make sure all the elements are valid digits for the given radix. This is about as efficient as it could possibly get, since you must look at each element at least once. I suppose we could micro-optimize it based on the radix, but for all intents and purposes this is as good as you can expect to get.

public static boolean isInteger(String s) {
    return isInteger(s,10);
}

public static boolean isInteger(String s, int radix) {
    if(s.isEmpty()) return false;
    for(int i = 0; i < s.length(); i++) {
        if(i == 0 && s.charAt(i) == '-') {
            if(s.length() == 1) return false;
            else continue;
        }
        if(Character.digit(s.charAt(i),radix) < 0) return false;
    }
    return true;
}

Alternatively, you can rely on the Java library to have this. It's not exception based, and will catch just about every error condition you can think of. It will be a little more expensive (you have to create a Scanner object, which in a critically-tight loop you don't want to do. But it generally shouldn't be too much more expensive, so for day-to-day operations it should be pretty reliable.

public static boolean isInteger(String s, int radix) {
    Scanner sc = new Scanner(s.trim());
    if(!sc.hasNextInt(radix)) return false;
    // we know it starts with a valid int, now make sure
    // there's nothing left!
    sc.nextInt(radix);
    return !sc.hasNext();
}

If best practices don't matter to you, or you want to troll the guy who does your code reviews, try this on for size:

public static boolean isInteger(String s) {
    try { 
        Integer.parseInt(s); 
    } catch(NumberFormatException e) { 
        return false; 
    } catch(NullPointerException e) {
        return false;
    }
    // only got here if we didn't return false
    return true;
}

Answer 2

It's better to use regular expression like this:

str.matches("-?\\d+");

---

-?     --> negative sign, could have none or one
\\d+   --> one or more digits

It is not good to use NumberFormatException here if you can use if-statement instead.

If you don't want leading zero's, you can just use the regular expression as follow:

str.matches("-?(0|[1-9]\\d*)");

Answer 3

Or you can enlist a little help from our good friends at Apache Commons : StringUtils.isNumeric(String str)

Answer 4

You want to use the Integer.parseInt(String) method.

try{
  int num = Integer.parseInt(str);
  // is an integer!
} catch (NumberFormatException e) {
  // not an integer!
}

Answer 5

Or simply

mystring.matches("\\d+")

though it would return true for numbers larger than an int

Answer 6

As an alternative approach to trying to parse the string and catching NumberFormatException, you could use a regex; e.g.

if (Pattern.compile("-?[0-9]+").matches(str)) {
    // its an integer
}

This is likely to be faster, especially if you precompile and reuse the regex.

However, the problem with this approach is that Integer.parseInt(str) will also fail if str represents a number that is outside range of legal int values. While it is possible to craft a regex that only matches integers in the range Integer.MIN_INT to Integer.MAX_INT, it is not a pretty sight. (And I am not going to try it ...)

On the other hand ... it may be acceptable to treat "not an integer" and "integer too large" separately for validation purposes.

Answer 7

You can use Integer.parseInt(str) and catch the NumberFormatException if the string is not a valid integer, in the following fashion (as pointed out by all answers):

static boolean isInt(String s)
{
 try
  { int i = Integer.parseInt(s); return true; }

 catch(NumberFormatException er)
  { return false; }
}

However, note here that if the evaluated integer overflows, the same exception will be thrown. Your purpose was to find out whether or not, it was a valid integer. So its safer to make your own method to check for validity:

static boolean isInt(String s)  // assuming integer is in decimal number system
{
 for(int a=0;a<s.length();a++)
 {
    if(a==0 && s.charAt(a) == '-') continue;
    if( !Character.isDigit(s.charAt(a)) ) return false;
 }
 return true;
}

Answer 8

You can use Integer.parseInt() or Integer.valueOf() to get the integer from the string, and catch the exception if it is not a parsable int. You want to be sure to catch the NumberFormatException it can throw.

It may be helpful to note that valueOf() will return an Integer object, not the primitive int.

Answer 9

public boolean isInt(String str){
    return (str.lastIndexOf("-") == 0 && !str.equals("-0")) ? str.substring(1).matches(
            "\\d+") : str.matches("\\d+");
}