Why does this equation return the answer it does?

Question

I am Confused as to why this returns 1;

(char)('0' + 11) = ; why?

Full code below where ending = 1;

 char[] ending;
       char a = (char)('0' + 11/10); 
        ending = new char[]{a, (char)('0' + 11)};
        System.out.println(ending);

If `ending` is an array, then this program should not print `1`. — kaya3, Mar 03 '20 at 21:03

Kosonome · Accepted Answer · 2020-03-03T20:59:27.277

2

Char value of '0' is 48.
48 + 11 = 59
Char value of 59 is ';'.

You can check char values in integer value in any ASCII Character Set in internet.

In Java, char can be use as a int, short, byte, long with values between 0 and 65535 without any casting.

A better explanation is found in: Java char is also an int?

edited Mar 03 '20 at 20:59

answered Mar 03 '20 at 17:29

Kosonome

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Internally java chars are 2 bytes so more like shorts than integers. – cquezel Mar 03 '20 at 17:37
1

Java characters are UTF-16 encodings of Unicode characters. You can look into a unicode table for characters that are up to 15 significant bits. – cquezel Mar 03 '20 at 17:46

score 1 · Answer 2 · answered Mar 03 '20 at 17:33

1

You are assigning '1' to variable a. '0' + 11/10 => '0' + 1 You are assigning a two letter string to endings. Le first letter is a ('1') the second is a semi colon. ('0' + 11).

answered Mar 03 '20 at 17:33

cquezel

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Why does this equation return the answer it does?

2 Answers2