Java Programming - Language Fundamentals - Discussion

How can a char variable carry integer value? Anyone, please explain.

064770 is octal, but it's outside the valid range for char (greater than 65535).

So, 1 is not valid.

I think its option 1 and 2. You can initialize an integer directly in a char as it represents a number in hexadecimal form. And 2 is correct too as in single quotes we can initialize any character.

064770 is an octal representation of the integer value of 27128 then how can we store it in char literal?

Anyone can explain about these two differences?

char c5 = '\iface'; //illegal
char c6 = '\uface'; //legal

Can anyone say how c3 is correct? Anyone explain this?

A char literal is represented by a single character in single quotes.

char a = 'a';
char b = '@';

You can also type in the Unicode value of the character, using the Unicode.

Notation of prefixing the value with \u as follows:

char letterN = '\u004E'; // The letter 'N'.

Remember, characters are just 16-bit unsigned integers under the hood. That means you can assign a number literal, assuming it will fit into the unsigned 16-bit range (65535 or less). For example, the following are all legal:

char a = 0x892; // hexadecimal literal.
char b = 982; // int literal.
char c = (char) 70000; // The cast is required; 70000 is out of char range.
char d = (char) -98; // Ridiculous, but legal.

And the following are not legal and produce compiler errors:

char e = -29; // Possible loss of precision; needs a cast.
char f = 70000 // Possible loss of precision; needs a cast.

You can also use an escape code if you want to represent a character that can't be. Typed in as a literal, including the characters for linefeed, newline, horizontal tab, Backspace, and single quotes.

char c = '\"'; // A double quote.
char d = '\n'; // A newline.

How to its possible for 6th one?

Java doesn't support unsigned integer. char c1 = 064770; should give an error.

What are the differences between NULL, ' ', '\b' and '\0'?