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How to put numbers to array in while loop in java?

I am trying to add two binary numbers and then get their sum in binary system. I got their sum in decimal and now I am trying to turn it into binary. But there is problem that when I take their sum (in decimal) and divide by 2 and find remainders(in while loop), I need to put remainders into array in order print its reverse. However, there is an error in array part. Do you have any suggestions with my code? Thanks in advance.
Here is my code:
import java.util.Scanner;
public class ex1 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int n = scan.nextInt();
int m = scan.nextInt();
int k = dec1(n)+dec2(m);
int i=0,c;
int[] arr= {};
while(k>0) {
c = k % 2;
k = k / 2;
arr[i++]=c; //The problem is here. It shows some //error
}
while (i >= 0) {
System.out.print(arr[i--]);
}
}
public static int dec1(int n) {
int a,i=0;
int dec1 = 0;
while(n>0) {
a=n%10;
n=n/10;
dec1= dec1 + (int) (a * Math.pow(2, i));
i++;
}
return dec1;
}
public static int dec2(int m) {
int b,j=0;
int dec2 = 0;
while(m>0) {
b=m%10;
m=m/10;
dec2= dec2 + (int) (b * Math.pow(2, j));
j++;
}
return dec2;
}
}

Here:
int[] arr= {};
creates an empty array. Arrays don't grow dynamically in Java. So any attempt to access any index of arr will result in an ArrayIndexOutOfBounds exception. Because empty arrays have no "index in bounds" at all.
So:
first ask the user for the count of numbers he wants to enter
then go like: int[] arr = new int[targetCountProvidedByUser];
The "more" real answer would be to use List<Integer> numbersFromUsers = new ArrayList<>(); as such Collection classes allow for dynamic adding/removing of elements. But for a Java newbie, you better learn how to deal with arrays first.

Why are you using two different methods to do the same conversion? All you need is one.
You could have done this in the main method.
int k = dec1(n)+dec1(m);
Instead of using Math.pow which returns a double and needs to be cast, another alternative is the following:
int dec = 0;
int mult = 1;
int bin = 10110110; // 128 + 48 + 6 = 182.
while (bin > 0) {
// get the right most bit
int bit = (bin % 10);
// validate
if (bit < 0 || bit > 1) {
throw new IllegalArgumentException("Not a binary number");
}
// Sum up each product, multiplied by a running power of 2.
// this is required since bits are taken from the right.
dec = dec + mult * bit;
bin /= 10;
mult *= 2; // next power of 2
}
System.out.println(dec); // prints 182
An alternative to that is to use a String to represent the binary number and take the bits from the left (high order position).
String bin1 = "10110110";
int dec1 = 0;
// Iterate over the characters, left to right (high to low)
for (char b : bin1.toCharArray()) {
// convert to a integer by subtracting off character '0'.
int bit = b - '0';
// validate
if (bit < 0 || bit > 1) {
throw new IllegalArgumentException("Not a binary number");
}
// going left to right, first multiply by 2 and then add the bit
// Each time thru, the sum will be multiplied by 2 which shifts everything left
// one bit.
dec1 = dec1 * 2 + bit;
}
System.out.println(dec1); // prints 182
One possible way to display the result in binary is to use a StringBuilder and simply insert the converted bits to characters.
public static String toBin(int dec) {
StringBuilder sb = new StringBuilder();
while (dec > 0) {
// by inserting at 0, the bits end up in
// correct order. Adding '0' to the low order
// bit of dec converts to a character.
sb.insert(0, (char) ((dec & 1) + '0'));
// shift right for next bit to convert.
dec >>= 1;
}
return sb.toString();
}

How does Integer.toString() works internally?

I found that a similar question has been asked before here : how does Float.toString() and Integer.toString() works?
But this doesn't speak about how that function internally works. When I opened the internally source code of Integer.toString(), it is not understandable for normal junior java programmer.
Can somebody please explain what happens internally in short description ?
NOTE : This was one of the interview questions that I was asked recently. I had no idea about how to answer such question !

The no arg call of integer.toString() simply calls the static method Integer.toString(int i) (using the integer variables own primitive value), which is implemented as below;
public static String toString(int i) {
if (i == Integer.MIN_VALUE)
return "-2147483648";
int size = (i < 0) ? stringSize(-i) + 1 : stringSize(i);
char[] buf = new char[size];
getChars(i, size, buf);
return new String(0, size, buf);
}
First it checks whether it's value is == the lowest possible integer, and returns that if it is equal. If not, then it checks what size the String needs to be using the stringSize() method of Integer to use as the size of an array of characters.
stringSize() implementation below;
static int stringSize(int x) {
for (int i=0; ; i++)
if (x <= sizeTable[i])
return i+1;
}
Once it has a char[] of the correct size, it then populates that array using the getChars() method, implemented below;
static void getChars(int i, int index, char[] buf) {
int q, r;
int charPos = index;
char sign = 0;
if (i < 0) {
sign = '-';
i = -i;
}
// Generate two digits per iteration
while (i >= 65536) {
q = i / 100;
// really: r = i - (q * 100);
r = i - ((q << 6) + (q << 5) + (q << 2));
i = q;
buf [--charPos] = DigitOnes[r];
buf [--charPos] = DigitTens[r];
}
// Fall thru to fast mode for smaller numbers
// assert(i <= 65536, i);
for (;;) {
q = (i * 52429) >>> (16+3);
r = i - ((q << 3) + (q << 1)); // r = i-(q*10) ...
buf [--charPos] = digits [r];
i = q;
if (i == 0) break;
}
if (sign != 0) {
buf [--charPos] = sign;
}
}
Explaining each individual step would take far too long for for a stackoverflow answer. The most pertinent section however (as pointed out in the comments) is the getChars() method which, complicated bit shifting aside, is essentially process of elimination for finding each character. I am afraid I can't go into any greater detail than that without going beyond my own understanding.

What kind of output should I see from getFft?

Alright, so I am working on creating an Android audio visualization app. The problem is, what I get form the method getFft() doesn't jive with what google says it should produce. I traced the source code all the way back to C++, but I am not familiar enough with C++ or FFT to actually understand what is happening.
I will try and include everything needed here:
(Java) Visualizer.getFft(byte[] fft)
/**
* Returns a frequency capture of currently playing audio content. The capture is a 8-bit
* magnitude FFT. Note that the size of the FFT is half of the specified capture size but both
* sides of the spectrum are returned yielding in a number of bytes equal to the capture size.
* {#see #getCaptureSize()}.
* <p>This method must be called when the Visualizer is enabled.
* #param fft array of bytes where the FFT should be returned
* #return {#link #SUCCESS} in case of success,
* {#link #ERROR_NO_MEMORY}, {#link #ERROR_INVALID_OPERATION} or {#link #ERROR_DEAD_OBJECT}
* in case of failure.
* #throws IllegalStateException
*/
public int getFft(byte[] fft)
throws IllegalStateException {
synchronized (mStateLock) {
if (mState != STATE_ENABLED) {
throw(new IllegalStateException("getFft() called in wrong state: "+mState));
}
return native_getFft(fft);
}
}
(C++) Visualizer.getFft(uint8_t *fft)
status_t Visualizer::getFft(uint8_t *fft)
{
if (fft == NULL) {
return BAD_VALUE;
}
if (mCaptureSize == 0) {
return NO_INIT;
}
status_t status = NO_ERROR;
if (mEnabled) {
uint8_t buf[mCaptureSize];
status = getWaveForm(buf);
if (status == NO_ERROR) {
status = doFft(fft, buf);
}
} else {
memset(fft, 0, mCaptureSize);
}
return status;
}
(C++) Visualizer.doFft(uint8_t *fft, uint8_t *waveform)
status_t Visualizer::doFft(uint8_t *fft, uint8_t *waveform)
{
int32_t workspace[mCaptureSize >> 1];
int32_t nonzero = 0;
for (uint32_t i = 0; i < mCaptureSize; i += 2) {
workspace[i >> 1] = (waveform[i] ^ 0x80) << 23;
workspace[i >> 1] |= (waveform[i + 1] ^ 0x80) << 7;
nonzero |= workspace[i >> 1];
}
if (nonzero) {
fixed_fft_real(mCaptureSize >> 1, workspace);
}
for (uint32_t i = 0; i < mCaptureSize; i += 2) {
fft[i] = workspace[i >> 1] >> 23;
fft[i + 1] = workspace[i >> 1] >> 7;
}
return NO_ERROR;
}
(C++) fixedfft.fixed_fft_real(int n, int32_t *v)
void fixed_fft_real(int n, int32_t *v)
{
int scale = LOG_FFT_SIZE, m = n >> 1, i;
fixed_fft(n, v);
for (i = 1; i <= n; i <<= 1, --scale);
v[0] = mult(~v[0], 0x80008000);
v[m] = half(v[m]);
for (i = 1; i < n >> 1; ++i) {
int32_t x = half(v[i]);
int32_t z = half(v[n - i]);
int32_t y = z - (x ^ 0xFFFF);
x = half(x + (z ^ 0xFFFF));
y = mult(y, twiddle[i << scale]);
v[i] = x - y;
v[n - i] = (x + y) ^ 0xFFFF;
}
}
(C++) fixedfft.fixed_fft(int n, int32_t *v)
void fixed_fft(int n, int32_t *v)
{
int scale = LOG_FFT_SIZE, i, p, r;
for (r = 0, i = 1; i < n; ++i) {
for (p = n; !(p & r); p >>= 1, r ^= p);
if (i < r) {
int32_t t = v[i];
v[i] = v[r];
v[r] = t;
}
}
for (p = 1; p < n; p <<= 1) {
--scale;
for (i = 0; i < n; i += p << 1) {
int32_t x = half(v[i]);
int32_t y = half(v[i + p]);
v[i] = x + y;
v[i + p] = x - y;
}
for (r = 1; r < p; ++r) {
int32_t w = MAX_FFT_SIZE / 4 - (r << scale);
i = w >> 31;
w = twiddle[(w ^ i) - i] ^ (i << 16);
for (i = r; i < n; i += p << 1) {
int32_t x = half(v[i]);
int32_t y = mult(w, v[i + p]);
v[i] = x - y;
v[i + p] = x + y;
}
}
}
}
If you made it through all that, you are awesome! So my issue, is when I call the java method getFft() I end up with negative values, which shouldn't exist if the returned array is meant to represent magnitude. So my question is, what do I need to do to make the array represent magnitude?
EDIT: It appears my data may actually be the Fourier coefficients. I was poking around the web and found this. The applet "Start Function FFT" displays a graphed representation of coefficients and it is a spitting image of what happens when I graph the data from getFft(). So new question: Is this what my data is? and if so, how do I go from the coefficients to a spectral analysis of it?

An FFT doesn't just produce magnitude; it produces phase as well (the output for each sample is a complex number). If you want magnitude, then you need to explicitly calculate it for each output sample, as re*re + im*im, where re and im are the real and imaginary components of each complex number, respectively.
Unfortunately, I can't see anywhere in your code where you're working with complex numbers, so perhaps some rewrite is required.
UPDATE
If I had to guess (after glancing at the code), I'd say that real components were at even indices, and odd components were at odd indices. So to get magnitudes, you'd need to do something like:
uint32_t mag[N/2];
for (int i = 0; i < N/2; i++)
{
mag[i] = fft[2*i]*fft[2*i] + fft[2*i+1]*fft[2*i+1];
}

One possible explanation why you see negative values: byte is a signed data type in Java. All values, that are greater or equal 1000 00002 are interpreted as negative integers.
If we know that all values should are expected to be in the range [0..255], then we have map the values to a larger type and filter the upper bits:
byte signedByte = 0xff; // = -1
short unsignedByte = ((short) signedByte) & 0xff; // = 255

"The capture is a 8-bit magnitude FFT" probably means that the return values have an 8-bit magnitude, not that they are magnitudes themselves.
According to Jason
For real-valued signals, like the ones
you have in audio processing, the
negative frequency output will be a
mirror image of the positive
frequencies.
Android 2.3 Visualizer - Trouble understanding getFft()

how to find 2 to the power of n . n ranges from 0 to 200

Assume my system as 32 bit machine. Considering this if I use long int for n>63 I will get my value as 0. How to solve it?

double is perfectly capable of storing powers of two up to 1023 exactly. Don't let someone tell you that floating point numbers are somehow always inexact. This is a special case where they aren't!
double x = 1.0;
for (int n = 0; n <= 200; ++n)
{
printf("2^%d = %.0f\n", n, x);
x *= 2.0;
}
Some output of the program:
2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8
2^4 = 16
...
2^196 = 100433627766186892221372630771322662657637687111424552206336
2^197 = 200867255532373784442745261542645325315275374222849104412672
2^198 = 401734511064747568885490523085290650630550748445698208825344
2^199 = 803469022129495137770981046170581301261101496891396417650688
2^200 = 1606938044258990275541962092341162602522202993782792835301376

Just wait around for a 256-bit compiler, then use int :-)
No, seriously, since you just want to start with 1 and keep doubling, your best bet is to get a big integer library like GNU MP.
You would do that with a piece of code like (untested):
#include <stdio.h>
#include "gmp.h"
int main (void) {
int i;
mpz_t num;
mpz_init_set_ui (num, 1);
for (i = 0; i <= 200; i++) {
printf ("2^%d = ", i);
mpz_out_str (NULL, 10, num);
printf ("\n");
mpz_mul_ui (num, num, 2);
}
return 0;
}
You could code up your own data structure of an array of longs with only two operations, double and print but I think it would be far easier to just use GMP.
If you do want to roll your own, have a look at this. It's a variation/simplification of some big integer libraries I've developed in the past:
#include <stdio.h>
#include <stdlib.h>
// Use 16-bit integer for maximum portability. You could adjust
// these values for larger (or smaller) data types. SZ is the
// number of segments in a number, ROLLOVER is the maximum
// value of a segment plus one (need to be less than the
// maximum value of your datatype divided by two. WIDTH is
// the width for printing (number of "0" characters in
// ROLLOVER).
#define SZ 20
#define ROLLOVER 10000
#define WIDTH 4
typedef struct {
int data[SZ];
} tNum;
// Create a number based on an integer. It allocates the segments
// then initialises all to zero except the last - that one is
// set to the passed-in integer.
static tNum *tNumCreate (int val) {
int i;
tNum *num = malloc (sizeof (tNum));
if (num == NULL) {
printf ("MEMORY ERROR\n");
exit (1);
}
for (i = 0; i < SZ - 1; i++) {
num->data[i] = 0;
}
num->data[SZ-1] = val;
}
// Destroy the number. Simple free operation.
static void tNumDestroy (tNum *num) {
free (num);
}
// Print the number. Ignores segments until the first non-zero
// one then prints it normally. All following segments are
// padded with zeros on the left to ensure number is correct.
// If no segments were printed, the number is zero so we just
// output "0". Then, no matter what, we output newline.
static void tNumPrint (tNum *num) {
int i, first;
for (first = 1, i = 0; i < SZ; i++) {
if (first) {
if (num->data[i] != 0) {
printf ("%d", num->data[i]);
first = 0;
}
} else {
printf ("%0*d", WIDTH, num->data[i]);
}
}
if (first) {
printf ("0");
}
printf ("\n");
}
// Double a number. Simplified form of add with carry. Carry is
// initialised to zero then we work with the segments from right
// to left. We double each one and add the current carry. If
// there's overflow, we adjust for it and set carry to 1, else
// carry is set to 0. If there's carry at the end, then we have
// arithmetic overflow.
static void tNumDouble (tNum *num) {
int i, carry;
for (carry = 0, i = SZ - 1; i >= 0; i--) {
num->data[i] = num->data[i] * 2 + carry;
if (num->data[i] >= ROLLOVER) {
num->data[i] -= ROLLOVER;
carry = 1;
} else {
carry = 0;
}
}
if (carry == 1) {
printf ("OVERFLOW ERROR\n");
exit (1);
}
}
// Test program to output all powers of 2^n where n is in
// the range 0 to 200 inclusive.
int main (void) {
int i;
tNum *num = tNumCreate (1);
printf ("2^ 0 = ");
tNumPrint (num);
for (i = 1; i <= 200; i++) {
tNumDouble (num);
printf ("2^%3d = ", i);
tNumPrint (num);
}
tNumDestroy (num);
return 0;
}
and its associated output:
2^ 0 = 1
2^ 1 = 2
2^ 2 = 4
2^ 3 = 8
2^ 4 = 16
2^ 5 = 32
2^ 6 = 64
2^ 7 = 128
2^ 8 = 256
2^ 9 = 512
: : : : :
2^191 = 3138550867693340381917894711603833208051177722232017256448
2^192 = 6277101735386680763835789423207666416102355444464034512896
2^193 = 12554203470773361527671578846415332832204710888928069025792
2^194 = 25108406941546723055343157692830665664409421777856138051584
2^195 = 50216813883093446110686315385661331328818843555712276103168
2^196 = 100433627766186892221372630771322662657637687111424552206336
2^197 = 200867255532373784442745261542645325315275374222849104412672
2^198 = 401734511064747568885490523085290650630550748445698208825344
2^199 = 803469022129495137770981046170581301261101496891396417650688
2^200 = 1606938044258990275541962092341162602522202993782792835301376

python supports big integers out of the box. At any linux prompt, run this:
$ python -c "for power in range(201): print power, 2**power"
0 1
1 2
2 4
3 8
4 16
5 32
6 64
<snip>
196 100433627766186892221372630771322662657637687111424552206336
197 200867255532373784442745261542645325315275374222849104412672
198 401734511064747568885490523085290650630550748445698208825344
199 803469022129495137770981046170581301261101496891396417650688
200 1606938044258990275541962092341162602522202993782792835301376
This can be easily made into a script if necessary. See any python tutorial.

It's been ages since I've used Java seriously, but: BigInteger class? It has all the usual mathematical (multiply, pow) and bitwise (shiftLeft) operations.
Your tagging is a little confusing though, which language did you prefer?

Use java.math.BigInteger.shiftLeft.
for (int i = 0; i <= 200; i++) {
System.out.format("%d = %s%n", i, BigInteger.ONE.shiftLeft(i));
}
Excerpt of output:
0 = 1
1 = 2
2 = 4
3 = 8
4 = 16
:
197 = 200867255532373784442745261542645325315275374222849104412672
198 = 401734511064747568885490523085290650630550748445698208825344
199 = 803469022129495137770981046170581301261101496891396417650688
200 = 1606938044258990275541962092341162602522202993782792835301376
If BigInteger is unavailable, you can also just manually do the multiplication and store it in a String.
String s = "1";
for (int i = 0; i < 200; i++) {
StringBuilder sb = new StringBuilder();
int carry = 0;
for (char ch : s.toCharArray()) {
int d = Character.digit(ch, 10) * 2 + carry;
sb.append(d % 10);
carry = d / 10;
}
if (carry != 0) sb.append(carry);
s = sb.toString();
System.out.format("%d = %s%n", i + 1, sb.reverse());
}
(see full output)

In C/C++ I don't know of a standard way you can store integers that big, pax's solution is the rightway to go.
However for Java, you do have a way out, BigInteger

Use scheme!
1 => (expt 2 200)
1606938044258990275541962092341162602522202993782792835301376

in kotlin :
var x= readLine()!!.toInt()
var y=BigDecimal(1)
for (i in 1..x)
{
y *= BigDecimal(2)
}
println(DecimalFormat().format(y))

If unsigned long int is 64 bits then the largest value for 2^n that you can represent is 2^63 (i.e. n = 63):
unsigned long int x = (1UL << n); // n = 0..63

Storing int value of bitmask - extract 1 valued bits

I am calculating the int equivalent of a given set of bits and storing that in memory. From there, I would like to determine all 1 value bits from the original bitmask. Example:
33 --> [1,6]
97 --> [1,6,7]
Ideas for an implementation in Java?

On BitSet
Use java.util.BitSet to store, well, a set of bits.
Here's how you can convert from an int to a BitSet, based on which bits in the int is set:
static BitSet fromInt(int num) {
BitSet bs = new BitSet();
for (int k = 0; k < Integer.SIZE; k++) {
if (((num >> k) & 1) == 1) {
bs.set(k);
}
}
return bs;
}
So now you can do the following:
System.out.println(fromInt(33)); // prints "{0, 5}"
System.out.println(fromInt(97)); // prints "{0, 5, 6}"
And just for completeness, here's the reverse transformation:
static int toInt(BitSet bs) {
int num = 0;
for (int k = -1; (k = bs.nextSetBit(k + 1)) != -1; ) {
num |= (1 << k);
}
return num;
}
So composing both together, we always get back the original number:
System.out.println(toInt(fromInt(33))); // prints "33"
System.out.println(toInt(fromInt(97))); // prints "97"
On 0-based indexing
Note that this uses 0-based indexing, which is the more commonly used indexing for bits (and most everything else in Java). This is also more correct. In the following, ^ denotes exponentiation:
33 = 2^0 + 2^5 = 1 + 32 97 = 2^0 + 2^5 + 2^6 = 1 + 32 + 64
33 -> {0, 5} 97 -> {0, 5, 6}
If you insist on using 1-based indexing, however, you can use bs.set(k+1); and (1 << (k-1)) in the above snippets. I would advise strongly against this recommendation, however.
Related questions
What does the ^ operator do in Java? -- it's actually not exponentiation

For bit fiddling, java.lang.Integer has some very helpful static methods. Try this code as a starting base for your problem:
public int[] extractBitNumbers(int value) {
// determine how many ones are in value
int bitCount = Integer.bitCount(value);
// allocate storage
int[] oneBits = new int[bitCount];
int putIndex = 0;
// loop until no more bits are set
while (value != 0) {
// find the number of the lowest set bit
int bitNo = Integer.numberOfTrailingZeros(value);
// store the bit number in array
oneBits[putIndex++] = bitNo+1;
// clear the bit we just processed from the value
value &= ~(1 << bitNo);
}
return oneBits;
}

I can show you C# implementation, Java should be very similar.
int value = 33;
int index = 1;
while (value > 0)
{
if ((value % 2) == 1)
Console.WriteLine(index);
index++;
value /= 2;
}

If you want to get an array like that you'll likely need to loop the number of bits you want to check & the integer with a bit shifted 1 for each step.
Something like (pseudo):
Init array
mask = 1
for (0 to BitCount):
if Integer & mask
array[] = pos
mask << 1

A bit-crunching variation would be something like:
int[] getBits(int value) {
int bitValue = 1;
int index = 1;
int[] bits = new int[33];
while (value >= bitValue)
{
bits[index++] = (value & bitValue);
bitValue << 1; // or: bitValue *= 2;
}
return bits;
}
Note that since the bits are indexed from 1 as you requested, bits[0] is left unused.