Develop Reference

How would I add two int that are in the same array to each other and convert them into an int. In the Luhn Algorithm

I am trying to add two parts of an array together to go into an int value. I am using Luhn algorithm to figure out of a credit card is a valid credit card. We are only using 6 digit credit card's just to make sure no one enter's a real credit card number. The part I am confused on is when I go to split a number that is above 10 and add it together. Example if the algorithm was to give me 12 I would need to separate it into 1 and 2 and then add them together to equal 3. I believe I am splitting it currently in the code but when I go to add them together I get some number that makes no since. here is a section of the code with some notes about it.
I have printed out numbers in certain places to show myself what is going on in certain places. I have also added in some comments that say that either the number that is printed out is what is expected, and some comments for when there isn't something I expected
int[] cardNumber = new int[]{ 1,2,3,4,5,5};
int doubleVariablesum = 0;
int singleVariablesum = 0;
int totalSum = 0;
int cutOffVar = 0;
String temp2;
for (int i = cardNumber.length - 1; i >= 0;) {
int tempSum = 0;
int temp = cardNumber[i];
temp = temp * 2;
System.out.println("This is the temp at temp * 2: " + temp);
temp2 = Integer.toString(temp);
if (temp2.length() == 1) {
System.out.println("Temp2 char 0: "+ temp2.charAt(0));
// this prints out the correct number
// Example: if there number should be 4 it will print 4
tempSum = temp2.charAt(0);
System.out.println("This is tempSum == 1: " + tempSum);
// when this goes to add temp2.charAt(0) which should be 4 it prints out //something like 56
} else {
System.out.println("TEMP2 char 0 and char 1: " + temp2.charAt(0) + " " + temp2.charAt(1));
// this prints out the correct number successfully spited
tempSum = temp2.charAt(0) + temp2.charAt(1);
System.out.println("This is tempSum != 1: " + tempSum);
// but here it when I try to add them together it is giving me something
// like 97 which doesn't make since for the numbers I am giving it
}
doubleVariablesum = tempSum + doubleVariablesum;
System.out.println("This is the Double variable: " + doubleVariablesum);
System.out.println();
i = i - 2;
}

Since you are converting the number to a string to split the integer, and then trying to add them back together. You're essentially adding the two characters numerical values together which is giving you that odd number. You would need to convert it back to an integer, which you can do by using
Integer.parseInt(String.valueOf(temp2.charAt(0)))

When adding char symbols '0' and '1' their ASCII values are added - not numbers 0 and 1.
It is possible to use method Character::getNumericValue or just subtract '0' when converting digit symbol to int.
However, it is also possible to calculate sum of digits in a 2-digit number without any conversion to String and char manipulation like this:
int sum2digits = sum / 10 + sum % 10; // sum / 10 always returns 1 if sum is a total of 2 digits

Seems like charAt() type casts into integer value, but the ascii one. Hence for the characters '0' and '1', the numbers 48 and 49 are returned resulting in a sum of 97. To fix this, you could just assign temp2 to (temp / 10) + (temp % 10). Which actually splits a two digit integer and adds their sum.

You need to be aware of the following when dealing with char and String
Assigning the result of charAt(index) to an int will assign the ASCII value and not the actual integer value. To get the actual value you need to String.valueOf(temp2.charAt(0)).
The result of concatenating chars is the sum of the ASCII values.
eg if char c = '1'; System.out.println(c + c); will print "98" not "11".
However System.out.println("" + c + c); will print "11". Note the "" will force String concatenation.

Get the Unicode from a Hexadecimal

I've been searching for a solution to my problem for days but can't get a spot-on answer when looking at previously answered questions/ blogs / tutorials etc. all over the internet.
My aim is to write a program which takes a decimal number as an input and then calculates the hexadecimal number and also prints the unicode-symbol of said hexadecimal number (\uXXXX).
My problem is I can't "convert" the hexadecimal number to unicode. (It has to be written in this format: \uXXXX)
Example:
Input:
122 (= Decimal)
Output:
Hexadecimal: 7A
Unicode: \u007A | Unicode Symbol: Latin small letter "z"
The only thing I've managed to do is print the unicode (\u007A), but I want the symbol ("z").
I thought if the unicode only has 4 numbers/letters, I would just need to "copy" the hexadecimal into the code and fill up the remaining places with 0's and it kinda worked, but as I said I need the symbol not the code. So I tried and tried, but I just couldn't get the symbol.
By my understanding, if you want the symbol you need to print it as a string.
But when trying it with a string I get the error "illegal unicode escape".
It's like you only can print pre-determined unicodes and not "random" ones generated on the spot in relation of your input.
I'm only a couple days into Java, so apologies if I have missed anything.
Thank you for reading.
My code:
int dec;
int quotient;
int rest;
int[]hex = new int[10];
char[]chars = new char[]{
'F',
'E',
'D',
'C',
'B',
'A'
};
String unicode;
// Input Number
System.out.println("Input decimal number:");
Scanner input = new Scanner(System.in);
dec = input.nextInt();
//
// "Converting to hexadecimal
quotient = dec / 16;
rest = dec % 16;
hex[0] = rest;
int j = 1;
while (quotient != 0) {
rest = quotient % 16;
quotient = quotient / 16;
hex[j] = rest;
j++;
}
//
/*if (j == 1) {
unicode = '\u000';
}
if (j == 2) {
unicode = '\u00';
}
if (j == 3) {
unicode = '\u0';
}*/
System.out.println("Your number: " + dec);
System.out.print("The corresponding Hexadecimal number: ");
for (int i = j - 1; i >= 0; i--) {
if (hex[i] > 9) {
if (j == 1) {
unicode = "\u000" + String.valueOf(chars[16 - hex[i] - 1]);
}
if (j == 2) {
unicode = "\u00" + String.valueOf(chars[16 - hex[i] - 1]);
}
if (j == 3) {
unicode = "\u0" + String.valueOf(chars[16 - hex[i] - 1]);
}
System.out.print(chars[16 - hex[i] - 1]);
} else {
if (j == 1) {
unicode = "\u000" + Character.valueOf[hex[i]);
}
if (j == 2) {
unicode = "\u00" + Character.valueOf(hex[i]);
}
if (j == 3) {
unicode = "\u0" + Character.valueOf(hex[i]);
}
System.out.print(hex[i]);
}
}
System.out.println();
System.out.print("Unicode: " + (unicode));
}
It's not an advanced code whatsoever, I wrote it exactly how I would calculate it on paper.
Dividing the number through 16 until I get a 0 and what remains while doing so is the hexadecimal equivalent.
So I put it in a while loop, since I would divide the number n-times until I got 0, the condition would be to repeat the division until the quotient equals zero.
While doing so the remains of each division would be the numbers/letters of my hexadecimal number, so I need them to be saved. I choose an integer array to do so. Rest (remains) = hex[j].
I also threw a variable in the called "j", so I would now how many times the division was repeated. So I could determine how long the hexadecimal is.
In the example it would 2 letters/numbers long (7A), so j = 2.
The variable would then be used to determine how many 0's I would need to fill up the unicode with.
If I have only 2 letters/numbers, it means there are 2 empty spots after \u, so we add two zeros, to get \u007A instead of \u7A.
Also the next if-command replaces any numbers higher than 9 with a character from the char array above. Basically just like you would do on paper.
I'm very sorry for this insanely long question.

U+007A is the 3 bytes int code pointer.
\u007A is the UTF-16 char.
A Unicode code pointer, symbol, sometimes is converted to two chars and then the hexadecimal numbers do not agree. Using code pointers hence is best. As UTF-16 is just an encoding scheme for two-bytes representation, where the surrogate pairs for 3 byte Unicode numbers do not contain / or such (high bit always 1).
int hex = 0x7A;
hex = Integer.parseUnsignedInt("007A", 16);
char ch = (char) hex;
String stringWith1CodePoint = new String(new int[] { hex }, 0, 1);
int[] codePoints = stringWith1CodePoint.codePoints().toArray();
String s = "𝄞"; // U+1D11E = "\uD834\uDD1E"

You can simply use System.out.printf or String.format to do what you want.
Example:
int decimal = 122;
System.out.printf("Hexadecimal: %X\n", decimal);
System.out.printf("Unicode: u%04X\n", decimal);
System.out.printf("Latin small letter: %c\n", (char)decimal);
Output:
Hexadecimal: 7A
Unicode: u007A
Latin small letter: z

Trying to increment counter using character comparison

I'm writing a method that is supposed to iterate through a string and increment a counter each time the letter c is found. The method uses the count to calculate the proportion of the string that was c. I added print statements to investigate why it was wrongly returning 0.0 for a string that had some c's in it.
I have tried declaring the counter as an int and then casting it to double later but that didn't help.
public double cRatio() {
String dna = "ATCCCCCCGTACCTAGCAA";
double cCount = 0.0;
for (int i = 0; i < dna.length(); i++) {
char ch = dna.charAt(i);
if (ch == 'c') {cCount++;}
}
System.out.println("C Count is : " + cCount);
System.out.println("dna.length() is : " + dna.length());
double cgRatio = (cCount / dna.length());
System.out.println("C Ratio is : " + cgRatio);
return cgRatio;
}
I expect the counter to have value 9.0 at the time it prints but instead, it is 0.0.

Your code does not count the letter c because you are checking for the letter C. The small c does not equal the upper C (different ASCII code).
if (ch == 'C') {cCount++;}
This will fix your problem.

Why doesn't my compare work between char and int in Java?

char c = '0';
int i = 0;
System.out.println(c == i);
Why does this always returns false?

Although this question is very unclear, I am pretty sure the poster wants to know why this prints false:
char c = '0';
int i = 0;
System.out.println(c == i);
The answer is because every printable character is assigned a unique code number, and that's the value that a char has when treated as an int. The code number for the character 0 is decimal 48, and obviously 48 is not equal to 0.
Why aren't the character codes for the digits equal to the digits themselves? Mostly because the first few codes, especially 0, are too special to be used for such a mundane purpose.

The char c = '0' has the ascii code 48. This number is compared to s, not '0'. If you want to compare c with s you can either do:
if(c == s) // compare ascii code of c with s
This will be true if c = '0' and s = 48.
or
if(c == s + '0') // compare the digit represented by c
// with the digit represented by s
This will be true if c = '0' and s = 0.

The char and int value can not we directly compare we need to apply casting. So need to casting char to string and after string will pars into integer
char c='0';
int i=0;
Answer is like
String c = String.valueOf(c);
System.out.println(Integer.parseInt(c) == i)
It will return true;
Hope it will help you
Thanks

You're saying that s is an Integer and c (from what I see) is a Char.. so there you, that's the problem: Integer vs. Char comparation.

Conversion from ASCII values to Char

String source = "WEDGEZ"
char letter = source.charAt(i);
shift=5;
for (int i=0;i<source.length();i++){
if (source.charAt(i) >=65 && source.charAt(i) <=90 )
letterMix =(char)(('D' + (letter - 'D' + shift) % 26));
}
Ok what I'm trying to do is take the string WEDGEZ, and shift each letter by 5, so W becomes B and E becomes J, etc. However I feel like there is some inconsistency with the numbers I'm using.
For the if statement, I'm using ASCII values, and for the
letterMix= statement, I'm using the numbers from 1-26 (I think). Well actually, the question is about that too:
What does
(char)(('D' + (letter - 'D' + shift) % 26)); return anyway? It returns a char right, but converted from an int. I found that statement online somewhere I didn't compose it entirely myself so what exactly does that statement return.
The general problem with this code is that for W it returns '/' and for Z it returns _, which I'm guessing means it's using the ASCII values. I really dont know how to approach this.
Edit: New code
for (int i=0;i<source.length();i++)
{
char letter = source.charAt(i);
letterMix=source.charAt(i);
if (source.charAt(i) >=65 && source.charAt(i) <=90 ){
letterMix=(char)('A' + ( ( (letter - 'A') + input ) % 26));
}
}

Well I'm not sure if this homework, so i'll be stingy with the Code.
You're Writing a Caesar Cipher with a shift of 5.
To address your Z -> _ problem...I'm Assuming you want all the letters to be changed into encoded letters (and not weird Symbols). The problem is ASCII values of A-Z lie between 65 and 90.
When coding Z (for eg), you end up adding 5 to it, which gives u the value 95 (_).
What you need to do is Wrap around the available alphabets. First isolate, the relative position of the character in the alphabets (ie A = 0, B = 1 ...) You Need to subtract 65 (which is ASCII of A. Add your Shift and then apply modulus 26. This will cause your value to wrap around.
eg, it your encoding Z, (ASCII=90), so relative position is 25 (= 90 - 65).
now, 25 + 5 = 30, but you need the value to be within 26. so you take modulus 26
so 30 % 26 is 4 which is E.
So here it is
char letter = message(i);
int relativePosition = letter - 'A'; // 0-25
int encode = (relativePosition + shift) % 26
char encodedChar = encode + 'A' // convert it back to ASCII.
So in one line,
char encodedChar = 'A' + ( ( (letter - 'A') + shift ) % 26)
Note, This will work only for upper case, if your planning to use lower case, you'll need some extra processing.
You can use Character.isUpperCase() to check for upper case.

You can try this code for convert ASCII values to Char
class Ascii {
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
char ch=sc.next().charAt(0);
if(ch==' ') {
int in=ch;
System.out.println(in);
}
}
}