Java - Check if input is a positive integer, negative integer, natural number and so on.

Question

Is there any in-built method in Java where you can find the user input's type whether it is positive, or negative and so on? The code below doesn't work. I am trying to find a way to input any in-built method that can do it at the if statement.

import java.util.Scanner;

public class Compare {

    public static void main(String[] args) { 

        Scanner input = new Scanner(System.in);

        System.out.print("Enter a number: ");
        int number = input.nextInt();

        if(number == int) 
            System.out.println("Number is natural and positive.");
    }
}

Most of these are possible with insanely short pieces of code, eg variable>0, variable<0 — Richard Tingle, Nov 04 '13 at 10:47

Juned Ahsan · Accepted Answer · 2019-09-29T23:11:13.457

46

If you really have to avoid operators then use Math.signum()

Returns the signum function of the argument; zero if the argument is zero, 1.0 if the argument is greater than zero, -1.0 if the argument is less than zero.

EDIT : As per the comments, this works for only double and float values. For integer values you can use the method:

Integer.signum(int i)

edited Sep 29 '19 at 23:11

answered Nov 04 '13 at 10:49

Juned Ahsan

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3

Math.signum() works only for double or float values. The question specifically mentions Integer. – Farrukh Chishti Nov 30 '17 at 08:55

blackpanther · Answer 2 · 2016-09-20T08:37:28.867

31

What about using the following:

int number = input.nextInt();
if (number < 0) {
    // negative
} else {
   // it's a positive
}

edited Sep 20 '16 at 08:37

answered Nov 04 '13 at 10:48

blackpanther

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The second `if` is redundant... Simply use `else { /*positive*/ }` if you need the context block; or do away with the `else` and it's context block entirely if you don't need it: `if (number < 0) { /*negative*/ } /*positive*/` – CosmicGiant Sep 19 '16 at 19:27
2

0 is not positive or negative. http://mathforum.org/library/drmath/view/58735.html – devdanke Feb 09 '17 at 01:06
@devdanke Yet Java has [signed zeroes](http://stackoverflow.com/a/14771363/2908724). – bishop Feb 12 '17 at 03:28
1

@bishop Wow. Nice find! – devdanke Mar 04 '17 at 04:25

score 7 · Answer 3 · edited Jan 02 '15 at 09:23

(You should you as Else-If statement to check the for the three different state (positive, negative, 0)

Here is a simple example (excludes the possibility of non-integer values)

  import java.util.Scanner;

  public class Compare {

   public static void main(String[] args) { 

    Scanner input = new Scanner(System.in);

    System.out.print("Enter a number: ");
    int number = input.nextInt();

    if( number == 0)
    { System.out.println("Number is equal to zero"); }
    else if (number > 0)
    { System.out.println("Number is positive"); }
    else 
    { System.out.println("Number is negative"); }


  }
 }

incorrect output for two cases; this line `int number = input.nextInt();` cause the wrong output for inputs of -1 and 0 — stoi2m1, Jan 02 '17 at 20:55

score 3 · Answer 4 · answered Nov 01 '17 at 11:39

3

For integers you can use Integer.signum()

Returns the signum function of the specified int value. (The return value is -1 if the specified value is negative; 0 if the specified value is zero; and 1 if the specified value is positive.)

answered Nov 01 '17 at 11:39

Kefirchiks

280
4
13

score 0 · Answer 5 · edited Nov 04 '13 at 11:08

0

Use like below code.

if(number >=0 ) {
            System.out.println("Number is natural and positive.");
}

edited Nov 04 '13 at 11:08

STT LCU

4,348
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answered Nov 04 '13 at 10:48

Prabhakaran Ramaswamy

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score -1 · Answer 6 · answered Nov 04 '13 at 10:48

-1

You could use if(number >= 0). The fact that you use int number = input.nextInt(); makes sure that it has to be an Integer.

answered Nov 04 '13 at 10:48

Theolodis

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Java - Check if input is a positive integer, negative integer, natural number and so on.

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