Char array compile time error upon assign a value from array

Question

So I have this code

char [] a = {'a','b','c'};

char c = 'a' + 'b'; //works
char c2 = 98 + 97; //works
char c3 = a[0] + a[1]; //compile time error

So all of them are the same functionality but upon getting and using an array value it is giving me a compile time error. What is the cause of this??

The result of the additive operator applied two char operands is an int.

then why can I do this?

char c2 = (int)((int)98 + (int)97);

@ScaryWombat it is commented above. I think he is saying that why all of them work but not the array part — Rod_Algonquin, Aug 08 '14 at 05:53
@ScaryWombat Exception in thread "main" java.lang.RuntimeException: Uncompilable source code - incompatible types: possible lossy conversion from int to char — Kick Buttowski, Aug 08 '14 at 05:54
@Rod_Algonquin My eyes must be seriously bad - where is it commented? — Scary Wombat, Aug 08 '14 at 05:56
@Rod_Algonquin Sorry I was meaning, what is the wording of the error. I think that Kick answered me though. — Scary Wombat, Aug 08 '14 at 06:01
Possibly a duplicate of http://stackoverflow.com/questions/24608241/why-cant-you-add-an-int-and-a-char-in-some-cases and http://stackoverflow.com/questions/13951459/casting-characters-in-java and many others... — Marco13, Aug 08 '14 at 15:42

Sotirios Delimanolis · Accepted Answer · 2014-08-08T06:13:37.340

2

The result of the additive operator applied two char operands is an int.

Binary numeric promotion is performed on the operands. The type of an additive expression on numeric operands is the promoted type of its operands

The first two are constant expressions where the resulting value is an int that can be safely assigned to a char.

The third is not a constant expression and so no guarantees can be made by the compiler.

Similarly

then why can I do this?

char c2 = (int)((int)98 + (int)97);

That is also a constant expression and the result can fit in a char.

Try it with bigger values, 12345 and 55555.

edited Aug 08 '14 at 06:13

answered Aug 08 '14 at 05:54

Sotirios Delimanolis

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for the first sentence, do you have any source? I am sure you right, but I wanna read more about it for myself thank you :) – Kick Buttowski Aug 08 '14 at 05:56
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@Kick Yes. It's in the JLS, give me a second. I'm on my phone. – Sotirios Delimanolis Aug 08 '14 at 05:57
would you please tell me what is JLS as well? – Kick Buttowski Aug 08 '14 at 05:58
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@KickButtowski it is the java specification language more like the rule book of java – Rod_Algonquin Aug 08 '14 at 05:58
@kick Which one? Binary numeric promotion is explained in the link. – Sotirios Delimanolis Aug 08 '14 at 06:03
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@game That is still a constant expression where the result is guaranteed to fit in a char. – Sotirios Delimanolis Aug 08 '14 at 06:04
@SotiriosDelimanolis what Binary numeric means here? additive expression means + sing? – Kick Buttowski Aug 08 '14 at 06:05
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@Kick Yes additive op means `+` applied to numeric types. Binary numeric means that two numbers (numeric values) are involved. – Sotirios Delimanolis Aug 08 '14 at 06:07
@kick No, _binary_ refers to the fact that `+` applies to **two** values. – Sotirios Delimanolis Aug 08 '14 at 06:09
@SotiriosDelimanolis so why when we concatenate two strings, we do not end up with int? Is it because String cannot cast to int or vice versa, yet char can be cast to int and vice versa? – Kick Buttowski Aug 08 '14 at 06:15
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@kick When applied to numeric types, the `+` operator is binary numeric additive operator. When applied to at least one operand of type `String`, it is the string concatenation operator. Please go through the JLS that I've linked. – Sotirios Delimanolis Aug 08 '14 at 06:15

Char array compile time error upon assign a value from array

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