How to check whether input value is integer or float?

Question

How to check whether input value is integer or float?

Suppose 312/100=3.12 Here i need check whether 3.12 is a float value or integer value, i.e., without any decimal

Answer 1

Math.round() returns the nearest integer to your given input value. If your float already has an integer value the "nearest" integer will be that same value, so all you need to do is check whether Math.round() changes the value or not:

if (value == Math.round(value)) {
  System.out.println("Integer");
} else {
  System.out.println("Not an integer");
}

Answer 2

You should check that fractional part of the number is 0. Use

x==Math.ceil(x)

or

x==Math.round(x)

or something like that

Answer 3

You can use RoundingMode.#UNNECESSARY if you want/accept exception thrown otherwise

new BigDecimal(value).setScale(2, RoundingMode.UNNECESSARY);

If this rounding mode is specified on an operation that yields an inexact result, an ArithmeticException is thrown.

Exception if not integer value:

java.lang.ArithmeticException: Rounding necessary

Answer 4

How about this. using the modulo operator

if(a%b==0) 
{
    System.out.println("b is a factor of a. i.e. the result of a/b is going to be an integer");
}
else
{
    System.out.println("b is NOT a factor of a");
}

Answer 5

Do this to distinguish that.

If for example your number is 3.1214 and stored in num but you don't know kind of num:

num = 3.1214
// cast num to int
int x = (int)num;
if(x == num)
{
  // num is a integer
} 
else
  // num is float
}

In this example we see that num is not integer.

Answer 6

The ceil and floor methods will help you determine if the number is a whole number.

However if you want to determine if the number can be represented by an int value.

if(value == (int) value)

or a long (64-bit integer)

if(value == (long) value)

or can be safely represented by a float without a loss of precision

if(value == (float) value)

BTW: don't use a 32-bit float unless you have to. In 99% of cases a 64-bit double is a better choice.