How I can convert a decimal number to exactly 8 bit binary?

Question

I want to write a function to convert a number to 8 bit binary String. For example: with 20, the result I got is 10100, the expected result I want is 00010100.

public static int printBinaryform(int number) {
    int remainder;

    if (number <= 1) {
        System.out.print(number);
        return 0;
    }

    remainder = number % 2;
    printBinaryform(number >> 1);
    System.out.print(remainder);

    return remainder;
}

public static int encryptionNumber(int number, int key) {
    int encrrptionNumber = number ^ key;
    return encrrptionNumber;
}

with input is 20, I expect the output of 00010100, the actual result is 10100.

pad the output with the necessary amount of zeroes untill it reaches length of 8. Your code works simoly needs padding — Nikos M., May 03 '19 at 11:03
I wouldn't do it recursively. Just use a loop that runs 8 times. A loop is simpler. — lurker, May 03 '19 at 11:04
You return an `int` - it has no leading `0`s. Use a `String` instead (`String.format()` will be your friend). — akop, May 03 '19 at 11:06
Using `String.format("%08d", number)` to get a **minimum** of 8 characters `String` using `0` character leading — AxelH, May 03 '19 at 11:12

score 1 · Answer 1 · answered May 03 '19 at 11:13

1

You can format the integer 20 as 00010100 using this oneliner:

System.out.println(String.format("%08d", Integer.parseInt(Integer.toString(20, 2), 10)));

I'm not sure this is what you are looking for. Anyways, here's how it works:

String str = Integer.toString(20, 2); // Turn an integer into a string consisting of numbers using base 2 (ie a string of 0:s and 1:s)

int i = Integer.parseInt(str, 10); // Parse the string assuming it is a decimal value

String.format("%08d", i); // Format the integer as a string with length 8 and leading 0:s

answered May 03 '19 at 11:13

Stefan

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I can not use this function of Java. I have to write my own function, anyway, that is my homework: The XOR-"encryption" of that number with a specified key. All numbers in the range will be encrypted using the same key. All encryptable numbers must be 32 bit wide, the key must be exactly 8 bit wide. The four bytes of the number are then individually encrypted using the key. – Nguyen Ngoc Hai May 03 '19 at 12:09

How I can convert a decimal number to exactly 8 bit binary?

1 Answers1