Convert decimal to binary using recursion

Question

I got the following code:

public class decToBin {

public static void main(String args[]) {

    int number = 32;

    System.out.println(decToBinWrapper(number));
}

public static String decToBinWrapper(int number) {

    return decToBin(number, "");
}

public static String decToBin(int number, String bin) {
    if (number >= 1)
        return decToBin(number / 2, bin + Integer.toString(number % 2));
    else
        return "0";

}
}

which is supposed to convert a decimal to binary but it only prints "0" instead of the binary string. Could someone tell me what I'm doing wrong please?

Just in case it's useful to you: Integer.toBinaryString()

user41871
– user41871

2015-02-08 02:52:19 +00:00
Commented Feb 8, 2015 at 2:52 — user41871
– user41871, Commented Feb 8, 2015 at 2:52
There is no decimal here. number is already binary.

user207421
– user207421

2017-08-22 04:31:17 +00:00
Commented Aug 22, 2017 at 4:31 — user207421
– user207421, Commented Aug 22, 2017 at 4:31

August · Accepted Answer · 2015-02-08 02:49:34Z

3

You should return the bin variable:

else
    return bin;

You also want to prepend Integer.toString(number % 2) to the previous String, not append it:

return decToBin(number / 2, Integer.toString(number % 2) + bin);

answered Feb 8, 2015 at 2:49

August

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Comments

Louis Wasserman · Accepted Answer · 2015-02-08 02:47:48Z

1

else
    return "0";

I think you probably meant return bin, since you're accumulating into that string. You're just discarding bin in your current implementation.

answered Feb 8, 2015 at 2:47

Louis Wasserman

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1 Comment

nem035 Over a year ago

He also needs to prepend values, not append to make the digit order be correct

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