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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 ;

Integer arithmetic with character

I have a simple block of code, can someone explain to me why this acceptable in Java?
int a = 10;
int c = 'A' + (a -1);
System.out.println(c);
The result of this displayed in compiler is: 74.
So where exactly the values of seventies come from?
Thanks for your answers.

In Java a char can be (explicitly or implicitly) casted to int, it then uses the ASCII value associated to this character.
It your case, the seventies comes from the character 'A'. The ASCII value of this character is 65. So the system implicitly does the casting 'A' → 65. Your calculation does:
c = 'A' + (a-1)
↓
c = 65 + (10-1)
↓
c = 74

The ascii value of 'A' is 65. Check this link for complete reference. The conversion occurs through (implicit) widening the char datatype to int datatype using its unicode value, which in this case for 'A' is 65 (jls 5.6.1).

Char is implicitly converted into integer resulting 'A' = 65
so 65+9 = 74

When you assign value of one data type to another, the two types might not be compatible with each other. It should be casted to correct type:
There are 2 types:-
Widening or Automatic Type Conversion.
which casts in this sequence:
byte -> short -> int -> long -> float -> double
char -> int
example :
int i = 100; //i is 100
float f = i; // f is 100.0
Narrowing or Explicit
Conversion. which casts in this sequence:
double -> float -> long -> int ->short -> byte
example:
double d = 100.100; //d is 100.100
int i = (int)d; //i is 100
So, When a Character value is used in integer context. It is automatically casted to int.
In your case:
int a = 10;
int c = 'A' + (a -1); //c = 65 + (10-1)
'A' ascii value is 65, thus you get c = 74
Hope this helps.

change ascii to string after adding number to it

I am trying to reform a string from ascii characters. I noticed when I break down ascii it matches the number with the ascii char, this changes when I tried to do (c+=c+0) on it as seen below, I tried to revert back to the original string but it doesn't seems to work. I was wondering does the ascii number change when you append it to a string?
StringBuilder b = new StringBuilder();
char c = "e".charAt(0);
System.out.println(c); //'e'
System.out.println(c+0); //101 -ascii
b.append(c+=c+0);
char d = b.charAt(0);
System.out.println(d-=d+0); //blank

When you use + to sum a char with and int, the result will be promoted
to int.
When you use compound assignment operators +=, the result will be converted to left
operand data type(in this case, char)
This is what happens with your code:
StringBuilder b = new StringBuilder();
char c = "e".charAt(0);
System.out.println(c); //'e'
System.out.println(c+0); //101 -ascii
b.append(c+=c+0); // result of c+c+0 is int 202, it is converted to char Ê
char d = b.charAt(0); // char d = Ê
System.out.println(d-=d+0); // result of d-(d+0) is int 0, it will be converted to null

It's not clear what you're trying to accomplish, but hopefully this example helps you figure out what's going on:
public static void main(String[] args) {
StringBuilder b = new StringBuilder();
char c = "e".charAt(0);
System.out.println(c); // e
System.out.println((int) c); // 101
char doubleC = (char) (c + c);
b.append(doubleC);
char d = b.charAt(0);
System.out.println((int) d); // 202
System.out.println(d - c); // 101
System.out.println((char) (d - c)); // e
}
Whether a character is printed as a letter or its ASCII value depends on its type. You can see that (int) c casts the character to its int ASCII value. Your approach of c+0 also does this but in a less intuitive way.
You also apparently want to "undo" your modification to c, but subtracting d-d always gives you 0. You need to store your value of c and subtract that from d.

+0s don't do anything because the character is underneath a number (so it's essentially like saying 101 + 0).
d-=d+0 evaluates to d = d - d + 0, so (because we're manipulating chars) d - d = 0
If you wanted to get c from d, you'd have to do d = d - c, because 'c' is the difference between the initial and the final value.
So no, the ASCII number of a character doesn't change, you're just not doing the reverse calculation as you should be.
It's like saying: x + y = z,
and then wondering why z - z != y.
You should be doing: z - x = y
#Edit
To add to your comment under pkpnd's answer - if you have only String b to work with, you'd still need to know the difference (i.e. by how much you increased/decreased the number) to reverse to the original chars.

Convert couple of characters to int

How can I convert couple of characters to int, to keep the question simple let's assume this :
char c1 = '1';
char c2 = '4';
char c3 = '5';
What would be the efficient way to get 145 type int.

int myInt = Integer.parseInt(""+c1+c2+c3);
""+c1 gives you a String containing the char c1, to which you append the characters c2 and c3. Then you use parseInt() to convert it to an integer.
A more straightforward (but less robust) method would be this:
int myInt = 100*(c1-'0') + 10*(c2-'0') + (c3-'0');
c1-'0' gives you the number represented by c1 as opposed to the character code. Then you simply shift things to the correct decimal places and add.

int myInt = (c1 - '0') * 100 + (c2 - '0') * 10 + (c3 - '0');

Here is a more general method to convert a char array to int:
int valueOf(char[] input){
int result = 0;
for(char c : input) {
result *= 10;
result += Character.digit(c, 10);
}
return result;
}

The most correct way to convert a single digit character to an int is Character.digit(char ch, int radix), which can handle values in bases besides 10.

Strings are ultimately backed by a char[], so you could do the following:
char c1 = '1';
char c2 = '4';
char c3 = '5';
int converted = Integer.parseInt(new String(new char[]{c1, c2, c3}));
There are probably several other perfectly good ways to do this.

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.