When does a Char behave like a Char and an integer in Java?

I'm confused how a character behaves when it is printed as an integer and a character in a string or alone?

char j = 'a';
    
    for loop
    {
          String arr[i] = Character.toString(j);

			j++;

    }
print arr[i];

This prints abcde

Why in this code is j++ allowed , when i tried to print j+1 it gives an error of lossy conversion from int to string?

char a = 65, b = 97;
String str = a+b+"hi";
print(str);

this prints 162 hi;

Why didnt this print ascii(162) hi ?

char is a misnomer, really. char as a datatype in java represents an unsigned 16-bit integer value. As in, it starts at 0, it goes as high as 65535, and if you then add 1 to it, it would loop back to 0 again.

Java itself mostly only supports int and long - you do have byte, short and char but these want to upconvert to an integer as soon as they can.

Keep these two things in mind:

1. char, like byte and short are inferior and tend to get lost, as almost any mathematical operation done to char/byte/short tends to upconvert to int
2. char is a numeric type. It has nothing to do with character.

There is one exception to the rule: While char is a numeric type and has nothing to do with characters, we all know that 'char' is supposed to represent characters. Therefore, METHODS in the JVM will act accordingly.

Thus, Calling System.out.println((char) 65);, which invokes the char variant of println, prints A, because the implementation of the char variant of println will take the integral value passed into it, and looks up the number provided in the lower-plane unicode table and prints the character it finds there. There is nothing intrinsic about char (the data type), that is purely an aspect of the source code of java.lang.System.

Thus:

char a = 5; a++;

is legal, because char is a numeric type.

String str = a + b + "hi"; prints 165hi because java evaluates the + operator left-toright, so that is shorthand for:

String str = (a + b) + "hi";

thus we get to, what is:

char a = 65, b = 97;
System.out.println(a + b);

This prints 162 - as I said, chars tend to upconvert ASAP to ints, and that happens here: a+b is an int expression, thus the int variant of println is invoked.

In a + b + "hi" the same rule applies: (a + b) is an int expression with value 162, so that boils down to 162 + "hi", which is string concatenation and makes the string "162hi".

Basically, when does char behave like an integer in java? It always behaves like an integer in java. Except if you pass char expressions to methods, that tends to invoke the version of a method that treats the incoming number as a thing to look up in a unicode table.

The short answer is that that "++" operator in Java does not widen. The data type of the result is the same as the data type of the operand, which is a char in this case.

But the "+" operator may widen, that is, produce a result that is of a type with a greater range. Adding anything to a char (even another char) widens the result to integer, because the compiler knows that the result may overflow. Having widened, the result can't be assigned to a char -- it is potentially too large.

Of course, the ++ operator on a char may overflow as well. Still, the Java platform spec says (essentially) that ++ is non-widening.

While + works for concatenation of strings, it does not for chars. So when you do a + b where a and b are chars, Java converts those chars to integers and performs integer addition. And next appends hi to the obtained integer result.

If you want, you may put an empty string first to tell Java that you want concatenation, not addition:

	char a = 65;
	char b = 97;
	String str = "" + a + b + "hi";
	System.out.println(str);

This prints:

> Aahi

There are many other ways to obtain the same output. One of my favourites is format():

	System.out.format("%c%chi%n", a, b);

%c in the format string tells Java to print a char. It will do so no matter if the argument passed is a char or an integer. This in turn means that the following works:

	System.out.format("%chi%n", a + b);

> ¢hi

Here the result of a + b is an int, but %c causes it to be converted back to a char, in this case ¢.

j++

The case of j++ where j is a char is special and confusing. It basically performs the same calculation as j + 1 but converts the result back to char (compiler optimization may cause it not to be implemented in this way, but the result is guaranteed to be as though it were). So you may say you get the same result as from j = (char) (j + 1);.

General

You cannot do math on chars. Whenever you apply an arithmetic operation to a char, Java first converts it to an integer. This goes smoothly because an int can hold all the possible values of a char. The same is not true for the opposite conversion. Therefore Java generally refuses to convert from int to char on its own — you need to tell it explicitly to do that, and thereby you assume the responsibility of a correct conversion also in cases where data is lost. The ++ case being the exception where conversion back to char is implicit.