How does char c = (char) -98; works?

Question

I want to know how does following line of code works?

char c = (char) -98;

As per my knowledge all signed numbers are stored in 2's complement form. So -98 will be stored in 2's complement form. So if you type cast it into char. How does this type casting is done by JVM?

Please correct me if I am wrong.

Answer 1

When you write:

char c = (char) -98;

It's the same like writing¹:

char c = 65438;

[Because 65438 = 2^16 - 98]

When explicitly converting an int to char, the first 16 bit will be removed.

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¹ -98 in 2's complement is

11111111111111111111111110011110.

The casting to char keeps only 16-bits:

1111111110011110

This value represents 65438..

More reading:

• JLS
• 2's complement

Answer 2

From the documentation:

char: The char data type is a single 16-bit Unicode character. It has a minimum value of '\u0000' (or 0) and a maximum value of '\uffff' (or 65,535 inclusive).

Source

They are just 16-bit unsigned integers.

People have reported that if char > 65535, the result was char % 65536, so I suppose your char c will be -98 % 65536, which would result in 65536 - 98 = 65438.

Anyway to be 100% sure, why don't you just try it?

UPDATE:

I see that you want to know what the output of System.out.println(char) (for example) is.

Literals of types char and String may contain any Unicode (UTF-16) characters

Source

So System.out.println((char)65438) is then equivalent to System.out.println('\uFF9E'), which by doing a lookup on the UTF-16 encoding table (source) is a HALFWIDTH KATAKANA VOICED SOUND MARK. It will only be printed though if the font supports this character, one of such fonts is Arial Unicode MS.