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Can anyone solve my confusion here is my code : 

byte i = 0;

i++;
System.out.println(i);

Result: 1

byte i = 0;

i = i+1;
System.out.println(i);

Generate compile time error: Type mismatch: cannot convert from int to byte

When I convert that to byte like: i = (byte) (i+1); then happily getting result 1

Performing this example i am understand i = i+1 & i++ perform can't same opearation so now i want to know what is exactally difference between them ...!!!

Saagar Elias Jacky
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Dhruv Raval
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  • Check this: http://stackoverflow.com/questions/13100019/why-can-not-i-add-two-bytes-and-get-an-int-and-i-can-add-two-final-bytes-get-a-b – Somnath Musib Apr 07 '15 at 11:47
  • implicit casting happens while doing increment.. so both are same at the end. – Selva Apr 07 '15 at 11:49
  • I think you should focus your learning on the differences between the primitive data types **byte** and **int** - and the conversions between those. – GhostCat Apr 07 '15 at 11:56

3 Answers3

6

i++ and i+=1 implicitly cast the result back to the type of i.

So if i is a byte, then i++; is not equivalent to i = i + 1; - it's actually equivalent to i = (byte)(i + 1);.

From the Java Language Specification, section 15.14.2, emphasis mine:

... the value 1 is added to the value of the variable and the sum is stored back into the variable. Before the addition, binary numeric promotion (§5.6.2) is performed on the value 1 and the value of the variable. If necessary, the sum is narrowed by a narrowing primitive conversion (§5.1.3) and/or subjected to boxing conversion (§5.1.7) to the type of the variable before it is stored.

There's no fundamental reason to it, other than "because the specification says so". The people who wrote the specification most likely wrote it this way so that ++ would be useful for all numeric types (not just int and long).

user253751
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2

internally for short, char, byte and int datatype if any arithmetic operation is performed compiler will upgrade data type to int and perform the operation. compiler will change the data type of expression result value to int

so for byte i

i = i+1; // will not work

because i+1 will result an integer data type value. so you have to typecast externally as

i = (byte)(i+1); // this is equivalent to i +=1; or you can say i++;
Prashant
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  • @Jägermeister : dude add two byte values and assign in byte variable and see. – Prashant Apr 07 '15 at 12:00
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    @Jägermeister No, *that's* not true. The result of adding two `byte`s is an `int`, [try it and see](http://ideone.com/D4WzNI). – user253751 Apr 07 '15 at 12:00
  • @Jägermeister : as per you `byte b=0;byte c=0; byte d=b+c;` this should work?? but it will not work because the result is an int. – Prashant Apr 07 '15 at 12:01
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    @Jägermeister this is called binary numeric promotion: http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.6.2 – weston Apr 07 '15 at 12:04
  • I think it's because you missed the brackets. `i = (byte)(i + 1)`; – Paul Boddington Apr 07 '15 at 12:04
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Finally i am found my problems(questions) superior understand so here describe it to understands others easily..!!!

1> i = i + 1

In java whenever any arithmetic operation performs between two or more variables then it's return value's type depends upon like this equation :

RETURN VALUE TYPE = (int, MAXIMUM RANGE OF VARIABLE-1, MAXIMUM RANGE OF VARIABLE-N )

here, Return value type : int

For eqution (int, byte, byte) so int have maximum range..

byte -> 1 byte size & int -> 4 byte size

That's why above exe i can't store that return value in byte directly it's require external type casting i = (byte) (i+1)

2> i++

In java whenever increment & decrement operation perform like these then it's return value's type depends upon this equation:

RETURN VALUE TYPE = (VARIABLE TYPE, VALUE + 1)

these equation denote internal type casting perform in this case

MUST REMEMBER:

case 1: External type cast require if needed

case 2: internal type casting perform automatically

Dhruv Raval
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