Develop Reference

How to get ASCII representation of a character?

I have very basic question.
How is possible for int a = 'a' to give 97 in output.
Below is my code:
class myClass {
int last = 'a' ;
myClass () {
System.out.println(last );
}
}

You can have a look at this: Why are we allowed to assign char to a int in java?
Basically, you are assigning a char to your int. A char is technically an unsigned 16-bit character. That's why you can assign it to an int.
Hope this helps.

You can basically cast the char to the int and store it as int:
int a = (int)'a';
System.out.println(a); //prints 97
Since Java can do the basic castings from your type specifications, you do not need to explicity write casting even.
int a = 'a';
System.out.println(a); // prints 97

The output is the ASCII value of the character stored in last. The ASCII value of the character 'a' is 97, and hence the output on the console is 97.

you have to take 'a' as a char
char char1 = 'a';
then cast it to int
int num = (int) char1 ;

How do I get the numerical value/position of a character in the alphabet (1-26) in constant time (O(1)) without using any built in method or function?

How do I get the numerical value/position of a character in the alphabet (1-26) in constant time (O(1)) without using any built in method or function and without caring about the case of the character?

If your compiler supports binary literals you can use
int value = 0b00011111 & character;
If it does not, you can use 31 instead of 0b00011111 since they are equivalent.
int value = 31 & character;
or if you want to use hex
int value = 0x1F & character;
or in octal
int value = 037 & character;
You can use any way to represent the value 31.
This works because in ASCII, undercase values are prefixed with 011, and uppercase 010 and then the binary equivalent of 1-26.
By using the bitmask of 00011111 and the AND operand, we covert the 3 most significant bits to zeros. This leaves us with 00001 to 11010, 1 to 26.

Adding to the very good (self) answer of Charles Staal.
Assuming ascii encoding following will work. Updated from the kind comment of Yves Daoust
int Get1BasedIndex(char ch) {
return ( ch | ('a' ^ 'A') ) - 'a' + 1;
}
This will make the character uppercase and change the index.
However a more readable solution (O(1)) is:
int Get1BasedIndex(char ch) {
return ('a' <= ch && ch <= 'z') ? ch - 'a' + 1 : ch - 'A' + 1;
}
One more solution that is constant time but requires some extra memory is:
static int cha[256];
static void init() {
int code = -1;
fill_n (&cha[0], &cha[256], code);
code = 1;
for(char s = 'a', l = 'A'; s <= 'z'; ++s, ++l) {
cha[s] = cha[l] = code++;
}
}
int Get1BasedIndex(char ch) {
return cha[ch];
}

We can get their ASCII values and then subtract from the starting character ASCII(a - 97, A - 65)
char ch = 'a';
if(ch >=65 && ch <= 90)//if capital letter
System.out.println((int)ch - 65);
else if(ch >=97 && ch <= 122)//if small letters
System.out.println((int)ch - 97);

Strictly speaking it is not possible to do it portably in C/C++ because there is no guarantee on the ordering of the characters.
This said, with a contiguous sequence, Char - 'a' and Char - 'A' obviously give you the position of a lowercase or uppercase letter, and you could write
Ord= 'a' <= Char && Char <= 'z' ? Char - 'a' :
('A' <= Char && Char <= 'Z' ? Char - 'A' : -1);
If you want to favor efficiency over safety, exploit the binary representation of ASCII codes and use the branchless
#define ToUpper(Char) (Char | 0x20)
Ord= ToUpper(Char) - 'a';
(the output for non-letter character is considered unspecified).
Contrary to the specs, these snippets return the position in range [0, 25], more natural with zero-based indexing languages.

charvariable=(char)(charvariable+3) what does this syntax mean?

I have been looking around on the internet at caesar ciphers and while I understand the loop I don't understand why this line of code is able to shift a char to another char? I don't understand this line here:
letter = (char)(letter - 26);
When I take (char) out it doesn't work and I have never seen it with the type being in parentheses followed by an operation.
Hopefully this is an easy question and thanks for the help.
for (int i = 0; i < buffer.Length; i++)
{
// Letter.
char letter = buffer[i];
// Add shift to all.
letter = (char)(letter + shift);
// Subtract 26 on overflow.
// Add 26 on underflow.
if (letter > 'z')
{
//The following line is the line I don't understand. Why char in parentheses then another parentheses?
letter = (char)(letter - 26);
}
else if (letter < 'a')
{
letter = (char)(letter + 26);
}
// Store.
buffer[i] = letter;
}

(char) is a cast. That means that it takes a value which is of one type, and converts it to a value of another type. Thus, if x is an int, (double)x yields a double whose value is the same value as the integer value.
The reason (char) is necessary in this expression is that Java does all its integer arithmetic on values of type int or long. So even though letter is a char, in the expression letter + 26, letter will be automatically converted to an int, and then 26 is added to the integer. (char) converts it back to a char type (which is an integer value from 0 to 65535). Java will not automatically convert a larger integer type (int, whose values are from -2147483648 to 2147483647) to a shorter integer type (char), therefore it's necessary to use a cast.
However, Java does allow this:
letter += 26;
which has the same effect, and does not require a cast.

There are 26 letters in the english alphabet, and char is an integral type
char ch = 'Z' - 25;
System.out.println(ch); // <-- A
JLS-4.2.1 - Integral Types and Values says (in part),
For char, from '\u0000' to '\uffff' inclusive, that is, from 0 to 65535

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);
}
}
}