random 4-digit number generator

The problem of random 4-digit number generator

In this blog, we will understand how to create a code for a random 4-digit number generator ranging between 0 to 9999. The code we will write will be specific to java language only.

Solution

The first will create a number in the string format. For this code, we will use an inbuilt function called random to create a random number in java. 

String.format("%04d", number);

The above code will create a string of length “0” to “0000”. Here %04 represents the number of digits we want in our code. The format function is responsible for converting the text into string type.

public static String getRandomNumberString() {
    // 6 digit random Number it can generate.
    // from 0 to 999999
    Random rnd = new Random();
    int number = rnd.nextInt(999999);

    // Here we can convert number sequence into 6 character.
    return String.format("%06d", number);
}

One may carry out the following: Make a class for BinaryNumber and include a function Object() { [native code] } that generates a char[] of 6 characters, each of which is generated using a random number generator between 0 and 1. If you want to display it, override the toString() method so that it returns the digits char[] as a string. then create a method called toInt() that multiplies the current digit by 10 to the power of I to convert the string’s characters one at a time into an integer:

char[] digits = {‘1’ , ‘0’ , ‘1’ , ‘1’ , ‘0’ , ‘1’};
//random 

int result = 0;
for( int i = 0; i < digits.length; i++) {
    result += Integer.parseInt(digits[i]) * Math.pow(10, i);
}

return result;

Also Read: What is IndexError and how to solve it?