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Java integer increases unintended

I have a function that takes String and a char[] type variables as parameter.
public String compressLine(String line, char[] fileChars){
String outputLine = new String();
int count = 0;
char lastChar = fileChars[0];
for(char c : line.toCharArray()){
if(c!=lastChar){
outputLine += String.format("%o,", count);
lastChar = c;
count = 1;
}else{
count += 1;
}
}
outputLine += count;
return outputLine;
}
String line is a line of ascii art and the char[] fileChars consists of the characters that are used in the string mentioned beforehand.
Example of line:
"##########---"
Example of fileChars:
[-, #]
Algorithm works in the following manner: given a row of data, the algorithm replaces each row with numbers that say how many consecutive pixels are the same character, always starting with the number of occurrences of the character that is encountered first in the fileChars.
When I ran a test using the examples given above, It should return;
[0, 10, 3]
But the output I get is;
[0, 12, 3]
When I tried to solve the bug, I saw the integer count somehow hops from 7 to 10.
What is causing this?

Your problem is with this line:
outputLine += String.format("%o,", count);
The "%o" output format specifier specifies that the value of count should be output in Octal. When the decimal value of count is 10, the Octal equivalent is 12.
I expect that you want this instead:
outputLine += String.format("%d,", count);
PS: When it appears that the number "jumps" from 7 to 10, that's because '10' immediately follows '7' in octal.

%o means octal. I'm not sure how you ended up on using %o, but, you don't want octal. Octal is how you would count if you have 8 fingers. 0, 1, 2, 3, 4, 5, 6, 7.., oh last finger, so.. 10!
In octal, '12' is how you write the tenth number. Just like you'd use '10' to do that in decimal, and '1010' in binary.
You want %d for decimal.

Getting ints array instead of characters array

I am trying a code problem to convert double to string and then insert that to an array. I tried various methods but these don't give expected output.
public int[] makePi() {
double PI = Math.PI;
String sPI = String.valueOf(PI);
int[] Arr = new int[3];
for(int i =0; i<3; i++)
{
Arr[i] = sPI.charAt(i);
}
return Arr;
}
Output should be an array with first three characters of PI as below :-
[ 3, 1, 4 ] while I am getting [51, 46, 49]
I will handle decimal character if needed.
Just a hint is needed.
Please don't provide full program that will be a spoiler. :-)

Look at the ASCII table. Do you see what are the corresponding chars for the integers you're getting? This should be a good hint for you.
Note that you're assigning the result to an int array, while you're running on characters.

you're storing chars into an int array. hence theie respective ascii values will be stored in array (you're effectively converting char to int)
3 (char) -> 51 (ASCII Value)
. (char) -> 46 (ASCII Value)
1 (char) -> 49 (ASCII Value)
your array length is 3, so only first 3 chars are converted to ascii which is 3.1, not 3.14
But now if you want to store it into an char array (which i feel you're trying to do), all you need is -
char[] charArray = sPI.toCharArray();
Plus, I dont think you want to store in int array as though you can convert ascii values int their respective int value, but what about '.' which is not a valid int.

What you get in your array are values of characters (so something like 70 for '3', I neither remember nor want to remember exact values). You must convert value of character into the number itself. Hint: characters are numbered in the following way:
'0' - n
'1' - n + 1
'2' - n + 2
and so on.

If you want to extract the numeric values of the digits, I would advise against doing explicit comparisons and arithmetic on the character values.
The Character class provides helper methods, which are less error-prone and more readable:
int outIndex = 0;
for (int i = 0; i < 3 /* && i < sPI.length() */; ++i) {
char c = sPI.charAt(i);
if (Character.isDigit(c)) {
Arr[outIndex++] = Character.getNumericValue(c);
}
}
/* assert outIndex == 3 */
return Arr;
I've commented out some code which I'd put in there for more robustness - it's not strictly necessary in this case, since we know that sPI has at least 3 digits in it. (Mind you, if we're going to hard-code that assumption, we may as well simply return new int[] { 3, 1, 4 };).

Java - Converting from unicode to a string?

I can easily create a unicode character and print it with the following lines of code
String uniChar = Character.toString((char)0000);
System.out.println(uniChar);
However, now I want to retrieve the number above, add 3, and print out the new unicode character that the numbers 0003 corresponds to. Is there a way for me to retrieve the ACTUAL string of unichar? As in "\u0000"? That way I could substring just the "0000", convert it to an int, add 3, and reverse the entire process.

I think you're looking for String#codePointAt:
Returns the character (Unicode code point) at the specified index. The index refers to char values (Unicode code units) and ranges from 0 to length()- 1.
If the char value specified at the given index is in the high-surrogate range, the following index is less than the length of this String, and the char value at the following index is in the low-surrogate range, then the supplementary code point corresponding to this surrogate pair is returned. Otherwise, the char value at the given index is returned.
For instance (live copy):
// String containing smiling face with smiling eyes emoji
String str = "😊";
// Get the code point
int cp = str.codePointAt(0);
// Show it
System.out.println(str + ", code point = U+" + toHex(cp));
// Increase it
++cp;
// Get the updated string (from an array of code points)
String updated = new String(new int[] { cp }, 0, 1);
// Show it
System.out.println(updated + ", code point = U+" + toHex(cp));
(toHex is just return Integer.toString(n, 16).toUpperCase();)
That outputs:
😊, code point = U+1F60A
😋, code point = U+1F60B

This code will work in both cases, for codepoints from Unicode BMP and from Unicode supplemental panes which uses 4 bytes in UTF-8 to encode a character. 4 byte code point requires 2 Java char entities to be stored, so in this case string.length() = 2.
// array will contain one or two characters
char[] chars = Character.toChars(codePoint);
// string.length will be 1 or 2
String str = new String(chars);

Unicode is a numbering of "characters" - code points - upto a 3-byte int range.
The UTF-16 encoding uses a sequance of byte pairs, and a java char is such a byte pair. The (int) cast of a char is imperfect and covers only a part of the Unicode. The correct way to convert a code point to possibly more than one char:
int codePoint = 0x263B;
char[] chars = Character.chars(codePoint);
To work with Unicode code points, one can do:
int[] codePoints = {0x2639, 0x263a, 0x263b};
String s = new String(codePoints, 0, codePoints.length);
codePoints[0} += 2;
You code use an int array of 1 code point.
In java 8 one can get an IntStream of code points:
s.codePoints().forEach(cp -> {
System.out.printf("U+%X = %s%n", cp, Character.getName(cp));
};

Hash a String into fixed bit hash value

I want to hash a word into fixed bit hash value say 64 bit,32 bit (binary).
I used the following code
long murmur_hash= MurmurHash.hash64(word);
Then murmur_hash value is converted into binary by the following function
public static String intToBinary (int n, int numOfBits) {
String binary = "";
for(int i = 0; i < numOfBits; ++i) {
n/=2;
if(n%2 == 0)
{
binary="0"+binary;
}
else
binary="1"+binary;
}
return binary;
}
Is there any direct hash method to convert into binary?

Just use this
Integer.toBinaryString(int i)

If you want to convert into a fixed binary string, that is, always get a 64-character long string with zero padding, then you have a couple of options. If you have Apache's StringUtils, you can use:
StringUtils.leftPad( Long.toBinaryString(murmurHash), Long.SIZE, "0" );
If you don't, you can write a padding method yourself:
public static String paddedBinaryFromLong( long val ) {
StringBuilder sb = new StringBuilder( Long.toBinaryString(val));
char[] zeros = new char[Long.SIZE - sb.length()];
Arrays.fill(zeros, '0');
sb.insert(0, zeros);
return sb.toString();
}
This method starts by using the Long.toBinaryString(long) method, which conveniently does the bit conversion for you. The only thing it doesn't do is pad on the left if the value is shorter than 64 characters.
The next step is to create an array of 0 characters with the missing zeros needed to pad to the left.
Finally, we insert that array of zeros at the beginning of our StringBuilder, and we have a 64-character, zero-padded bit string.
Note: there is a difference between using Long.toBinaryString(long) and Long.toString(long,radix). The difference is in negative numbers. In the first, you'll get the full, two's complement value of the number. In the second, you'll get the number with a minus sign:
System.out.println(Long.toString(-15L,2));
result:
-1111
System.out.println(Long.toBinaryString(-15L));
result:
1111111111111111111111111111111111111111111111111111111111110001

Another other way is using
Integer.toString(i, radix)
you can get string representation of the first argument i in the radix ( Binary - 2, Octal - 8, Decimal - 10, Hex - 16) specified by the second argument.

BigInteger to String according to ASCII

Is there a way to convert a series of integers to a String according to the ASCII table. I want to take the ASCII value of a String and convert it back to a String. For example,
97098097=> "aba"
I really need an effective way of taking an integer and converting it to a String according to its ASCII value. This method must also take into account the fact that there is no zero in front of the '9' when the String "aba" has an ASCII value of 97098097 as 'a' has an ASCII value of 097 and a String "dee" has one of 100101101. This means that not every number will have an ASCII value that has a number of digits that is a multiple of three.
If you have any misunderstandings of what I'm trying to do please let me know.

No lookup table required.
while (string.length() % 3 != 0)
{
string = '0' + string;
}
String result = "";
for (int i = 0; i < string.length(); i += 3)
{
result += (char)(Integer.parseInt(string.substring(i, i + 3)));
}

First, I would create some sort of lookup table in your code with all the ascii values and their String equivalent. Then take the big int and convert it to a String. Then do the mod of 3 with the length of your bigint string to determine if you need to add 1, 2, or no 0's to the front of it. Then just grab every 3 integers from the front of the number, compare it to the lookup table, and append the corresponding value to your result string.
Example:
Given 97098097
You would convert it to: "97098097"
Then you do a mod with 3 resulting in a value of 1, so 1 zero needs to be added.
Append 1 zero: "097098097"
Then grab every 3 from the front and compare to look up table:
097 -> a, so result += "a"
098 -> b, so result += "b"
097 -> a, so result += "a"
You end with result being "aba"