Adding and subtracting chars, why does this work? [duplicate]

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From the Docs

The char data type is a single 16-bit Unicode character.

A char is represented by its code point value:

  • min '\u0000' (or 0)
  • max: '\uffff' (or 65,535)

You can see all of the English alphabetic code points on an ASCII table.

Note that 0 == \u0000 and 65,535 == \uffff, as well as everything in between. They are corresponding values.

A char is actually just stored as a number (its code point value). We have syntax to represent characters like char c="A";, but it’s equivalent to char c = 65; and 'A' == 65 is true.

So in your code, the chars are being represented by their decimal values to do arithmetic (whole numbers from 0 to 65,535).

For example, the char 'A' is represented by its code point 65 (decimal value in ASCII table):

System.out.print('A'); // prints A
System.out.print((int)('A')); // prints 65 because you casted it to an int

As a note, a short is a 16-bit signed integer, so even though a char is also 16-bits, the maximum integer value of a char (65,535) exceeds the maximum integer value of a short (32,767). Therefore, a cast to (short) from a char cannot always work. And the minimum integer value of a char is 0, whereas the minimum integer value of a short is -32,768.

For your code, let’s say that the char was 'D'. Note that 'D' == 68 since its code point is 68.

return 10 + ch - 'A';

This returns 10 + 68 - 65, so it will return 13.

Now let’s say the char was 'Q' == 81.

if (ch >= 'A' && ch <= 'F')

This is false since 'Q' > 'F' (81 > 70), so it would go into the else block and execute:

return ch - '0';

This returns 81 - 48 so it will return 33.

Your function returns an int type, but if it were to instead return a char or have the int casted to a char afterward, then the value 33 returned would represent the '!' character, since 33 is its code point value. Look up the character in ASCII table or Unicode table to verify that '!' == 33 (compare decimal values).

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