Why am I getting this Result?

Question

char ch = '3';
result = cube(ch);
// Call the method that accepts a char
System.out.println("Cubed 'char' = " + result);

The method cube

private static char cube(char c) {
    char result;
    result = (char) (c *c *c);
    return result;
}

When I run this this code i get cubed 'char' = 1579, I would like to know where this number comes from within my code. Help!

`'3' == 51`. `51^3 == 132651`, which is larger than `Character.MAX_VALUE`; so it overflows. — Andy Turner, Oct 18 '16 at 14:53
Take a look at [this](http://stackoverflow.com/questions/8688668/in-java-is-the-result-of-the-addition-of-two-chars-an-int-or-a-char) related post. — Sotirios Delimanolis, Oct 18 '16 at 16:15

Andy Turner · Answer 1 · 2016-10-18T15:18:03.727

When you multiple chars together, they are promoted to int.

The int value of '3' is 51 (the ASCII value; as stated in JLS Sec 3.1, the first 128 characters of UTF-16 encoding are the same as ASCII characters).

So '3' * '3' * '3' == 51 * 51 * 51 == 132651.

But 132651 is too large to fit into a char when you cast it; so it overflows (twice, since Character.MAX_VALUE == 65535), and you get the value

   (char) (c * c * c)
== (char) 132651
== (char) (132651 % 65536)
== (char) 1579
== 'ث'

But then you are assigning the result of the call to cube to an int (or long) variable called result; so it is promoted again to int, and then concatenated to the string: this means that the integer value is appended to the string, not the char. Hence:

Cubed 'char' = 1579

in the output.

Notice that if you call without assigning to result:

System.out.println("Cubed 'char' = " + cube(ch));

then the output is

Cubed 'char' = ث

Ideone demo

Wouldn't it be simpler to say that '3' is a literal value for a `char` (a UTF-16 code unit) with the decimal value of 51? While term "ASCII" might be familiar to some, what the value for "3" is in ASCII has no bearing. — Tom Blodget, Oct 18 '16 at 16:25

Bathsheba · Answer 2 · 2016-10-18T15:10:01.160

A char in Java is a 16 bit unsigned type with a range 0 to 65535.

Your code char ch = '3'; is actually assigning the value 51 to ch. (Java mandates ASCII for the first 7 bits of the char type and 51 is the ASCII value of the number 3).

Formally the multiplication happens just after the arguments to the muliplication are converted to int. The result, 51 * 51 * 51 is 132651 and is an int type. When you assign this to a char, Java will subtract as many 65536 values as are necessary to pull the result into the accepted range. So your answer is

51 * 51 * 51 - 65536 - 65536 which is 1579

score 1 · Answer 3 · answered Oct 18 '16 at 14:54

1

The number is the result of: (51^3) % (2^16)

Because, as andy says in a comment, 51 (ascii value of '3') ^ 3 overflows, so every multiple of 2^16 (character size in java) is discarded and only what is left is shown.

answered Oct 18 '16 at 14:54

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Why am I getting this Result?

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