Java data types

Question

int a = 'a'; int b = ('a' + 'b');

int declared as a char data type, what do these lines mean?

Accepted Answer · 2016-09-02T19:42:40

See it. Your variables are declared as int , but the value is assigned to them using char constants.

In Java, unlike C ++, char is a 16-bit numeric unsigned type * (as opposed to an 8-bit signed byte ).

When chars are added, however, integer promotion occurs: char converted to a 32-bit int type, and added together. In this case, obviously, there will be no overflow.

The result of the addition is also int , it is assigned to the variable b .

* although its constants ( literals ) are specified by characters like '\u1f2d' , not by numbers

@Roman: According to §4.2.1, char values are the characters '\u0000' .. '\uffff' .
It seems to me that this means that in fact the type is non-numeric, the values are not numbers.
Slightly above (4.2) it says: "The integral types are long, our values <...>
Values like all the same numbers corresponding to character codes.
I am still confused that char literals are not written as numbers, but as something-in-single-quotes.

Answer 2 · 2016-09-02T19:09:48

In fact, char is an integer type, byte sometimes unsigned. It can be implicitly cast into an int (by character code) and folded. The code is not very beautiful and confuses those who came from C ++.

UPD. Thanks for the clarification, was not quite right, see the answer VladD.

The result of adding char + char is char and to int later.
The last "character" in the alphabet is \uFFFF and even with such an overflow value there will be no.

Java data types

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