java assignment in char

Question

I am learning java and the answer of one question seems to be wrong: question: Which of the following are legal?

char c = 0x1234;//A
char c = \u1234;//B
char c = '\u1234';//C

in the book the answer is C, but I think it should be both A and C. Anyone please verify that for me?

Why don't you try it?

Sotirios Delimanolis
– Sotirios Delimanolis

2013-10-16 15:48:33 +00:00
Commented Oct 16, 2013 at 15:48 — Sotirios Delimanolis
– Sotirios Delimanolis, Commented Oct 16, 2013 at 15:48
(A) is perfectly legal.

arshajii
– arshajii

2013-10-16 15:49:35 +00:00
Commented Oct 16, 2013 at 15:49 — arshajii
– arshajii, Commented Oct 16, 2013 at 15:49
Numbers aren't characters.

SLaks
– SLaks

2013-10-16 15:49:59 +00:00
Commented Oct 16, 2013 at 15:49 — SLaks
– SLaks, Commented Oct 16, 2013 at 15:49
Just B is incorrect, missing single quotes.

bucky
– bucky

2015-11-15 17:50:09 +00:00
Commented Nov 15, 2015 at 17:50 — bucky
– bucky, Commented Nov 15, 2015 at 17:50

Sotirios Delimanolis · Accepted Answer · 2013-10-16 16:00:36Z

6

Both A and C are correct

char a = 0x1234;

The literal 0x1234 is a hex integer literal. Its value fits in the char primitive type, so it is valid;

char b = \u1234;//B

Is not valid notation and so won't compile.

char c = '\u1234';

char is

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).

So it's valid.

edited Oct 16, 2013 at 16:00

answered Oct 16, 2013 at 15:55

Sotirios Delimanolis

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Comments

Dark Knight · Accepted Answer · 2013-10-16 15:59:14Z

1

char c = 0x1234;//A -- Correct

as it's value begins with 0X it is a valid hexadecimal value. Hence Correct.

char c = \u1234;//B --- Incorrect

Not a valid u code, as it's not surrounded with''.

char c = '\u1234';//C -- Correct

it is valid u code and is correctly surrounded by '', hence it is correct too.

edited Oct 16, 2013 at 15:59

answered Oct 16, 2013 at 15:50

Dark Knight

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3 Comments

Robert H Over a year ago

Why? Explain your answer

Dark Knight Over a year ago

char c = 0x1234; , as it's value begins with 0X it is a valid hexadecimal value. Hence Correct. char c = '\u1234'; it is valid u code and is correctly surrounded by '', hence it is correct too.

Robert H Over a year ago

You should explain it in the answer itself - not everyone will read the comments

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