Validate scanner user input in if statement WITHOUT variables

Question

Purpose is to reduce the number of variables so instead of making many variables I want to do something like this:

Scanner scnr = new Scanner(System.in); 

int number = 0;

scnr.nextInt();  

if (((scnr.nextInt() >= 4) && (scnr.nextInt() <=10))) 
{
   number = scnr.nextInt();
}

Instead of

Scanner scnr = new Scanner(System.in); 

int number = 0;
int validNum = 0;

number = scnr.nextInt();  

if (((number >= 4) && (number <=10))) 
{
   validNum = number;
}

You can create any number of variables, those live for short time only, so your GC will sweep all these anyway once they are out of execution scope. So don't worry about that. — varra
– varra, Commented Nov 23, 2016 at 5:11
@varra, Yeah, but it looks messy in a big application in terms of visibility. From that point, I'm saying. Any solution? — Anonymous
– Anonymous, Commented Nov 23, 2016 at 5:13
Can you please explain what are trying to get from user and what to do store — Venkat
– Venkat, Commented Nov 23, 2016 at 5:14
Do you want to keep 'number' variable value after 'if' checking ? If not assign value to same 'number' variable inside 'if' condition. — Niju
– Niju, Commented Nov 23, 2016 at 5:14
As per me, You should never spend more time on micro performance optimisation. The reason is they simply add no value considering the time invested in that work. Micro optimisation should be considered only if you are certain that a high percent of time will be spent in small cpu bound sections of your code. — FatherMathew
– FatherMathew, Commented Nov 23, 2016 at 6:05

rodridevops · Accepted Answer · 2016-11-23 05:36:59Z

5

You can use hasNext(String pattern)

Main:

import java.util.Scanner;

public class Test
{

    public static void main ( String [ ] args )
    {
        System.out.print ( "Enter number: " );
        Scanner scnr = new Scanner(System.in); 

        int number = 0;
        //Check number within range 4-10
        if (scnr.hasNext ( "^[4-9]|10" )) 
        {
           number = scnr.nextInt();
           System.out.println ( "Good Number: " + number );
        }
        else{
            System.out.println ( "Is not number or not in range" );
        }

    }
}

Tests:

Enter number: 3
Is not number or not in range
Enter number: 4
Good Number: 4
Enter number: 10
Good Number: 10
Enter number: 11
Is not number or not in range

edited Nov 23, 2016 at 5:36

answered Nov 23, 2016 at 5:30

rodridevops

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Comments

Madhawa Priyashantha · Accepted Answer · 2016-11-23 07:48:26Z

4

nextInt() will return new number on each call, so you can't do this

edited Nov 23, 2016 at 7:48

Madhawa Priyashantha

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answered Nov 23, 2016 at 5:08

Oleg Antonyan

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3 Comments

Anonymous Over a year ago

Then how can I reduce so many variables? It is like I have to create two variables for each user input, one for accepting and another one for validating.

Anonymous Over a year ago

@FastSnail, it's just a demonstration definitely. You can think of it in terms of big application.

Madhawa Priyashantha Over a year ago

@TitanicLover you can use objects,arrays ,collections depending on your application

jfo420 · Accepted Answer · 2021-03-16 23:31:31Z

0

I would be tempted to embed the condition within a do-while loop:

Scanner scnr = new Scanner(System.in); 
int number = 0;

scnr.nextInt();  

do{
   number = scnr.nextInt();
} while(number >= 4 && number <= 10);

I am new to java though, and still have some doubts about what is left in the scanner buffer, the nextInt() method parses the line until it finds an integer but is not returning the \n, or whatever it may be, as of I understood so far.

answered Mar 16, 2021 at 23:31

jfo420

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