[java] Determine if a String is an Integer in Java

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.

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

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

Or simply

mystring.matches("\\d+")

though it would return true for numbers larger than an int

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;
}

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

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

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.

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.

Using regular expression is better.

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*)");