decimal to binary conversion java

Question

My method seems to only work for all int less than 2^10. If it is greater than or equal to this value then it just returns 2^32. I'm not sure how my program is even getting this. I need to make it work for all int less than or equal to 2^15. Can anybody at least see why it's returning 2^32 for int greater than or equal to 2^10? If so then let me know please.

public static int DecToBin(int num) { int binNum = 0; int divisor = num; int mod = 0; int exp =0; while(divisor != 0){ mod = divisor%2; divisor = divisor/2; if(mod==1){ binNum = (int) (binNum + (Math.pow(10, exp))); } exp++; } return binNum;

}

you can't, not this way. you are trying to represent numbers that are too large. You represent 2^10 as a 10^10 number, which is > 2^32. Use a String? — njzk2
– njzk2, Commented Nov 10, 2014 at 23:44
My method has to return an int type. I thought int was good upto 2^32. — user3508348
– user3508348, Commented Nov 11, 2014 at 0:02
Never mind I've been a fool. You're absolutely correct. I need string. Thank you very much. — user3508348
– user3508348, Commented Nov 11, 2014 at 0:13

lockstock · Accepted Answer · 2014-11-11 00:23:51Z

An always easy way to accomplish this is:

Integer.toBinaryString(int);

This is the easiest way to do this but their might be more efficiant ways to do it.

Hopefully this is what you were looking for

AdamMc331 · Accepted Answer · 2014-11-11 15:56:43Z

As Taco pointed out, you can call Integer.toBinaryString(int), but if this is for an assignment or something where you should write the method yourself, here is what I would do.

I would write a method, that will constantly divide the number by 2, and store the remainders in a String. Something like this:

String binaryString = ""; int value = 100; while(value > 0){ int remainder = value % 2; value = value/2; binaryString = remainder + binaryString; }

I tested this in a quick console run and it produced '1100100' which is correct.

EDIT I used this as a reference.

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