If input value can be in numeric form other than integer , check by
if (x == (int)x)
{
// Number is integer
}
If string value is being passed , use Integer.parseInt(string_var).
Please ensure error handling using try catch in case conversion fails.
If input value can be in numeric form other than integer , check by
if (x == (int)x)
{
// Number is integer
}
If string value is being passed , use Integer.parseInt(string_var).
Please ensure error handling using try catch in case conversion fails.
If you have a double/float/floating point number and want to see if it's an integer.
public boolean isDoubleInt(double d)
{
//select a "tolerance range" for being an integer
double TOLERANCE = 1E-5;
//do not use (int)d, due to weird floating point conversions!
return Math.abs(Math.floor(d) - d) < TOLERANCE;
}
If you have a string and want to see if it's an integer. Preferably, don't throw out the Integer.valueOf() result:
public boolean isStringInt(String s)
{
try
{
Integer.parseInt(s);
return true;
} catch (NumberFormatException ex)
{
return false;
}
}
If you want to see if something is an Integer object (and hence wraps an int):
public boolean isObjectInteger(Object o)
{
return o instanceof Integer;
}
Videos
Quick and dirty...
if (x == (int)x)
{
...
}
edit: This is assuming x is already in some other numeric form. If you're dealing with strings, look into Integer.parseInt.
One example more :)
double a = 1.00
if(floor(a) == a) {
// a is an integer
} else {
//a is not an integer.
}
In this example, ceil can be used and have the exact same effect.
If what you receive is a String, you can try to parse it into an integer, if it fails, it's because it was not an integer after all. Something like this:
public static boolean isInteger(String str) {
try {
Integer.parseInt(str);
return true;
} catch (NumberFormatException nfe) {
return false;
}
}
Object x = someApi();
if (x instanceof Integer)
Note that if someApi() returns type Integer the only possibilities of something returned are:
- an
Integer null
In which case you can:
if (x == null) {
// not an Integer
} else {
// yes an Integer
}
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;
}
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*)");