Convert integer to integer array or binary

Question

I am trying to convert an integer to a binary number using an integer array. The first conversion is toBinaryString where I get the proper conversion "11111111" then the next step to convert to an array. This is where it goes wrong and I think its the getChar line.

int x = 255;

string=(Integer.toBinaryString(x));

int[] array = new int[string.length()];

for (int i=0; i< string.length(); i++){
array[i] = string.getChar(i);

   Log.d("TAG", " Data " + array[1] "," + array[2] + "," + array[3]);

Log displays ( Data 0,0,0 ) the results I am looking for is ( Data 1,1,1 )

Here is the final code and it works.

        // NEW
int x = 128;

string=(Integer.toBinaryString(x));

int[] array = new int[string.length()];

for (int i=0; i < string.length(); i++) {
array[i] = Integer.parseInt(string.substring(i,i+1));
}

Log.d("TAG", "Data   " + array[0] + "" + array[1]+ "" + array[2] + "" + array[3]+  " " + array[4]+ "" + array[5] + "" + array[6] + "" + array[7]);

How do you call String.getChar(), dont see it available. I wonder how your app is not crashing when doing array[3] since I see that there would be indexes as [0,1&2] — Varun
– Varun, Commented Sep 5, 2013 at 22:29
There's no String.getChar(int) method. Did you mean String.charAt(int index)? Try posting code that compiles. — Mattias Buelens
– Mattias Buelens, Commented Sep 5, 2013 at 22:30
Code did compile but I had array[i] = string.getChar(i); commented out since there was errors. Array should be 0 through 7 so log should not crash the array[3]. Mattias, I get a Type mismatch: cannot convert from Integer[] to int[] on int digits[] = new Integer[Integer.SIZE]; — Bobby
– Bobby, Commented Sep 5, 2013 at 22:47
Thank you EVERYONE, I used code from each of your ideas and got it working. Thank you again.. — Bobby
– Bobby, Commented Sep 5, 2013 at 23:37

2 revs · Accepted Answer · 2013-09-07 19:33:18Z

5

// Take your input integer
int x = 255;
// make an array of integers the size of Integers (in bits)
int[] digits = new Integer[Integer.SIZE];
// Iterate SIZE times through that array
for (int j = 0; j < Integer.SIZE; ++j) {
  // mask of the lowest bit and assign it to the next-to-last
  // Don't forget to subtract one to get indicies 0..(SIZE-1)
  digits[Integer.SIZE-j-1] = x & 0x1;
  // Shift off that bit moving the next bit into place
  x >>= 1;
}

edited Sep 7, 2013 at 19:33

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2 revs
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2 Comments

Mattias Buelens Over a year ago

I was thinking of something similar, just using a boolean[] to represent the bit array and use one of the answers to this question. However, it is unclear from the OP if the resulting should array should have minimally size (array[0] is always 1, unless x == 0) or fixed size (array.length == Integer.SIZE (i.e. 32)).

Bobby Over a year ago

I understand the array starts at 0 but I was only interested in 1,2,3 for log reference. Any ideas on why : int digits[] = new Integer[Integer.SIZE];gave errors.

Petros Koutsolampros · Accepted Answer · 2013-09-05 22:34:23Z

2

how about :

array[i] = Integer.parseInt(string.substring(i,i+1));

answered Sep 5, 2013 at 22:34

community wiki

Petros Koutsolampros

Comments

MrByte · Accepted Answer · 2013-09-05 22:35:14Z

2

First of all, an array starts from the index of 0 instead of 1, so change the following line of code:

Log.d("TAG", " Data " + array[1] "," + array[2] + "," + array[3]);

To:

Log.d("TAG", " Data " + array[0] "," + array[1] + "," + array[2]);

Next, the following line of code:

array[i] = string.getChar(i);

You are trying to get a character value to an Integer array, you may want to try the following instead and parse the value into an Integer (also, the "getChar" function does not exist):

array[i] = Integer.parseInt(String.valueOf(string.charAt(i)));

I hope I've helped.

P.S - Thanks to Hyrum Hammon for pointing out about the String.valueOf.

edited Sep 5, 2013 at 22:35

answered Sep 5, 2013 at 22:30

MrByte

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4 Comments

Hyrum Hammon Over a year ago

I'm pretty sure that'll have to be Integer.parseInt(String.valueOf(string.charAt(i)));

Mattias Buelens Over a year ago

Not sure about getChar, there's no such method on String. Also, Integer.parseInt has a lower-case "p" (conforming with the Java naming conventions).

Mattias Buelens Over a year ago

Hmm, it seems there's no parseInt(char), so that will throw a compiler error. You could make it into a string first, but Character.digit(string.charAt(i), 2) is probably faster and more straight to the point.

Hyrum Hammon Over a year ago

yeah, just checked. You do need String.valueOf

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