Checking if a number is an Integer in Java

Is there any method or quick way to check whether a number is an Integer (belongs to Z field) in Java?

I thought of maybe subtracting it from the rounded number, but I didn't find any method that will help me with this.

Where should I check开发者_如何学编程? Integer Api?

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.

/**
 * Check if the passed argument is an integer value.
 *
 * @param number double
 * @return true if the passed argument is an integer value.
 */
boolean isInteger(double number) {
    return number % 1 == 0;// if the modulus(remainder of the division) of the argument(number) with 1 is 0 then return true otherwise false.
}

if you're talking floating point values, you have to be very careful due to the nature of the format.

the best way that i know of doing this is deciding on some epsilon value, say, 0.000001f, and then doing something like this:

boolean nearZero(float f)
{
    return ((-episilon < f) && (f <epsilon)); 
}

then

if(nearZero(z-(int)z))
{ 
    //do stuff
}

essentially you're checking to see if z and the integer case of z have the same magnitude within some tolerance. This is necessary because floating are inherently imprecise.

NOTE, HOWEVER: this will probably break if your floats have magnitude greater than Integer.MAX_VALUE (2147483647), and you should be aware that it is by necessity impossible to check for integral-ness on floats above that value.

With Z I assume you mean Integers , i.e 3,-5,77 not 3.14, 4.02 etc.

A regular expression may help:

Pattern isInteger = Pattern.compile("\\d+");

    double x == 2.15;

    if(Math.floor(x) == x){
        System.out.println("an integer");
    } else{
        System.out.println("not an integer");
    }

I think you can similarly use the Math.ceil() method to verify whether x is an integer or not. This works because Math.ceil or Math.floor rounds up x to the nearest integer (say y) and if x==y then our original `x' was an integer.

    if((number%1)!=0)
    {
        System.out.println("not a integer");
    }
    else
    {
        System.out.println("integer");
    }

 int x = 3;

 if(ceil(x) == x) {

  System.out.println("x is an integer");

 } else {

  System.out.println("x is not an integer");

 }

change x to 1 and output is integer, else its not an integer add to count example whole numbers, decimal numbers etc.

   double x = 1.1;
   int count = 0;
   if (x == (int)x)
    {
       System.out.println("X is an integer: " + x);
       count++; 
       System.out.println("This has been added to the count " + count);
    }else
   {
       System.out.println("X is not an integer: " + x);
       System.out.println("This has not been added to the count " + count);


   }

Check if ceil function and floor function returns the same value

static boolean isInteger(int n) 
{ 
return (int)(Math.ceil(n)) == (int)(Math.floor(n)); 
} 

All given solutions are good, however most of them can give issues with Static Code Analysis (e.g. SONAR): "Floating point numbers should not be tested for equality" (see https://jira.sonarsource.com/browse/RSPEC-1244).

I am assuming that input that should be tested is double, not string.

As a workaround, I test numbers for not being integers:

public boolean isNotInteger(double x) {
    return x - Math.floor(x) > 0;
}

You can just use x % 1 == 0 because x % 1 gives the residual value of x / 1

// in C language.. but the algo is same

#include <stdio.h>

int main(){
  float x = 77.6;

  if(x-(int) x>0)
    printf("True! it is float.");
  else
    printf("False! not float.");        

  return 0;
}