According to the reference char() returns a value of type char (Java primitive type - char) and that will be a 16 bit unicode value.

Unfortunately I can't find the Processing source code for this function so I can't see how it is implemented.

Perhaps you could provide some simple code to demonstrate what you mean.

char() cannot be a function, because char is a keyword. That's Processing's pre-processor that transforms it to parseChar() or similar.

@PhiLho - thanks you are right it is the pre-processor and indeed it does call the parseChar method so

char a = char(64);
in Processing becomes
char a = PApplet.parseChar(64);
after pre-processing.

I have looked at the source code and the parseChar method and it can accept either a byte (8 bit) or an int (32 bit) variable.

If the parameter is a byte then the char must be in the range 0x0000 to 0x00ff

If the parameter is an int then it simply casts the value so I assume it copies the lower 16 bits from the int so the char will be in the range 0x0000 to 0xffff

@greekkv it would still be useful to post some simple code to demonstrate what you mean.

Thank you very much for your reply

" The correct output should be"
And what do you see, instead?

Personally, I see the good old Ascii chars (32-127) followed by the characters from the Latin-1 Supplement Unicode plane. It happens that they correspond to my Windows' ISO-8859-1 code page (more or less), but theoretically, all systems should show the same characters, independently of their code page. Java is using Unicode internally.

An heavy workaround:

1. PFont  f;
2. int SIZE = 24;
3. void setup() 
4. {
5.   size(1000, 640);
6.   background(0);
7.   f = createFont("Arial", SIZE, true);
8.  
9.   for(int i = 32 ; i <= 255 ; i++)
10.   {
11.       print(char(i));
12.   }         
13.  
14.   int x = 10, y = 30;
15.   textFont(f);
16.   for (int i = 32; i < 255; i++)
17.   {
18.     byte[] chr = { (byte) i };
19.     try
20.     {
21.       String letter = new String(chr, "ISO-8859-7");
22.       text(letter, x, y);
23.       x += SIZE;
24.       if (x > width - SIZE)
25.       {
26.         x = 10;
27.         y += 30;
28.       }
29.     }
30.     catch (java.io.UnsupportedEncodingException e) { println("Typo!"); }
31.   }
32. }

Using Unicode:
int x = 10;
for (int i = 0x0391; i < 0x03AA; i++)
{
  char c = char(i);
  println(c);
  text(c, x, 20);
  x += 20;
}

Alternatively:
char c = '\u0391';
(but it seems fail in Processing...)