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Adding Byte and Short [duplicate]

The rules for promotion is "when operands are of different types, automatic binary numeric promotion occurs with the smaller operand type being converted to the larger". But the operands are of same type for example,
byte=byte+byte // Compile time error... found int..
So why is it so?

There's no + operator for byte. Instead, both operands are promoted to int, so you've got
byte = byte + byte
... becomes (widening to find + operator) ...
byte = int + int
... becomes (result of + operator) ...
byte = int
... which then fails because there's no implicit conversion from int to byte. You need to cast:
byte a = 1;
byte b = 2;
byte c = (byte) (a + b);
Here are the actual rules for numeric promotion, from section 5.6.2 of the JLS:
When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:
If any of the operands is of a reference type, unboxing conversion (§5.1.8) is performed. Then:
If either operand is of type double, the other is converted to double.
Otherwise, if either operand is of type float, the other is converted to float.
Otherwise, if either operand is of type long, the other is converted to long.
Otherwise, both operands are converted to type int.

You were provided with correct answer about automatic promotion to 'int'.
There is one more note about that - compound assignment operators behave as they have an implicit type case. Example:
byte b1 = 1;
byte b2 = 2;
b1 = b1 + b2; // compilation fails
b1 += b2; // compilation successful

I would like to talk about Promotion in general
Java can evaluate only arithmetic expressions in which the operands’ types are identical
For example, in an expression containing int and double values, the int values are promoted to double values for use in the expression.
in another word
double someVar = 1 / 2;// someVar = 0
but
double someVar = (double)1 / 2;// someVar = 0.5
why?
we use the (double) cast operator to create a temporary
floating-point copy of its operand "1" (it's called explicit conversion)
The calculation now consists of a floating-point value (the temporary double copy of 1) divided by the integer 2
according to the above statement, Java performs an operation called promotion (or implicit conversion), so the int values are promoted to double values for use in the expression => the integer 2 is promoted to double
the expression became double someVar = 1.0 / 2.0; // someVar= 0.5
hope this is helpful, even if it is out of the core of the question

To understand this you should refer the 2 things:
Implicit Casting:
byte (8 bit) -> short (16 bit) -> int (32 bit) -> float (32 bit) -> double (64 bit)
char (16 bit) -> int (32 bit)
Arithmetic Operator Rules
Operator Precedence Rule : This rule states that Group of (*,/, %) will be evaluated first. Then Group of (+,-) operator will be evaluated. From a same Group of Operators, calculate from the left.
Operand Promotion Rule : This rule states that Operands having data type smaller than int will be promoted to int. order of promotion (byte->short->int, char->int)
Same Type Operand Rule: This rule states that if both operands are int,long,float,double then the same type is carried to the result type.
i.e. long+long => long , float + float => float, int+int => int
Mix Type Operand Rule : Follow the order of Promotion (int->long->float->double) if any of the operand is from the above order then the smaller will be promoted to the bigger one and result will be calculated in bigger type.
i.e. long + double => double , int + long => long

From this SO question, and above answers due to + arithmetic operator, both operands are converted to type int.
byte b1 = 1;
byte b2 = 2;
byte b3 = b1 + b2; // compile time error
In above code, value of b1 and b2 will be resolved at runtime, so compiler will convert both to int before resolving the value.
But if we consider the following code,
final byte b1 = 1;
final byte b2 = 2;
int b3 = b1 + b2; // constant expression, value resolved at compile time
b1 and b2 are final variables and values will be resolved at compile time so compilation won't fail.

Basic java logic expression is incorrect [duplicate]

I have seen it discussed somewhere that the following code results in obj being a Double, but that it prints 200.0 from the left hand side.
Object obj = true ? new Integer(200) : new Double(0.0);
System.out.println(obj);
Result: 200.0
However, if you put a different object in the right hand side, e.g. BigDecimal, the type of obj is Integer as it should be.
Object obj = true ? new Integer(200) : new BigDecimal(0.0);
System.out.println(obj);
Result: 200
I presume that the reason for this is something to do with casting the left hand side to a double in the same way that it happens for integer/double comparisons and calculations, but here the left and right sides do not interact in this way.
Why does this happen?

You need to read section 15.25 of the Java Language Specification.
In particular:
Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:
If one of the operands is of type byte or Byte and the other is of type short or Short, then the type of the conditional expression is short.
If one of the operands is of type T where T is byte, short, or char, and the other operand is a constant expression of type int whose value is representable in type T, then > - the type of the conditional expression is T.
If one of the operands is of type Byte and the other operand is a constant expression of type int whose value is representable in type byte, then the type of the conditional expression is byte.
If one of the operands is of type Short and the other operand is a constant expression of type int whose value is representable in type short, then the type of the conditional expression is short.
If one of the operands is of type; Character and the other operand is a constant expression of type int whose value is representable in type char, then the type of the conditional expression is char.
Otherwise, binary numeric promotion (§5.6.2) is applied to the operand types, and the type of the conditional expression is the promoted type of the second and third operands. Note that binary numeric promotion performs unboxing conversion (§5.1.8) and value set conversion (§5.1.13).
So binary numeric promotion is applied, which starts with:
When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:
If any of the operands is of a reference type, unboxing conversion (§5.1.8) is performed. Then:
If either operand is of type double, the other is converted to double.
That's exactly what happens here - the parameter types are converted to int and double respectively, the second operand (the third in the original expression) is then of type double, so the overall result type is double.

Numeric conversion in the conditional operator ? :
In the conditional operator a?b:c, if both b and c are different numeric types, the following conversion rules are applied at compile time to make their types equal, in order:
The types are converted to their corresponding primitive ones, which is called unboxing.
If one operand were a constant int (not Integer before unboxing) whose value is representable in the other type, the int operand is converted into the other type.
Otherwise the smaller type is converted into the next greater one until both operands have the same type. The conversion orders are:
byte -> short -> int -> long -> float -> double
char -> int -> long -> float -> double
Eventually the whole conditional expression gets the type of its second and third operands.
Examples:
If you combine char with short, the expression becomes int.
If you combine Integer with Integer, the expression becomes Integer.
If you combine final int i = 5 with a Character, the expression becomes char.
If you combine short with float, the expression becomes float.
In the question's example, 200 is converted from Integer into double, 0.0 is unboxed from Double into double and the whole conditional expression becomes becomes double which is eventually boxed into Double because obj is of type Object.

Example:
public static void main(String[] args) {
int i = 10;
int i2 = 10;
long l = 100;
byte b = 10;
char c = 'A';
Long result;
// combine int with int result is int compiler error
// result = true ? i : i2; // combine int with int, the expression becomes int
//result = true ? b : c; // combine byte with char, the expression becomes int
//combine int with long result will be long
result = true ? l : i; // success - > combine long with int, the expression becomes long
result = true ? i : l; // success - > combine int with long, the expression becomes long
result = true ? b : l; // success - > combine byte with long, the expression becomes long
result = true ? c : l; // success - > char long with long, the expression becomes long
Integer intResult;
intResult = true ? b : c; // combine char with byte, the expression becomes int.
// intResult = true ? l : c; // fail combine long with char, the expression becomes long.
}

unable to increment byte value with expression but with increment operator [duplicate]

This question already has answers here:
why byte += 1 compile but byte = byte + 1 not?
(8 answers)
Closed 9 years ago.
why
the following is valid
byte i=0;
i++;
where as the following is invalid
byte i=0;
i=i+1;
what would be the reason?
I know in the case of i+1 the entire value is explicitly incremented by 1 (which will be int value) so compilation problem will occur, but in case of i++ it does the same but did not get any error.

Whenever you perform a binary operation between two operands of different types, one of the operands is promoted to the higher type. And then the result of the operation is of that type.
So, in your case, the byte type a is first promoted to an int, since 1 is an int type. And then after the addition operation, the result is of type int. Now, since you cannot assign an int to a byte, you need to do a typecast to remove the compiler error:
byte a = 2;
a = a + 1; // Error: Cannot assign an int value to byte
a = (byte)(a + 1); // OK
Now, in case of Compound Assignment Operator, the typecasting is done implicitly for you. The expression:
a += 1
is internally converted to:
a = (byte)(a + 1);
This is specified in JLS - §15.26.2 Compound Assignment Operator:
A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.
Similar is the case with prefix increment operators, and postfix increment operators.
According to JLS - §15.15 Unary Operators:
The type of the prefix increment expression is the type of the variable.

add/subtract/multiply etc.. these operations are done by converting the operands in RHS to integers... So here, you are trying to reassign the integer (RHS value generated after completion of operation) to a byte... BOOM. "CANNOT CONVERT FROM INT TO BYTE" - ERROR says it all.

This is because, when you say i = i + 1, i is of type byte and 1 is of type int in java. The result of i + 1 is therefore an int. Assigning it to a byte causes loss of precision.

Primitive type casting

I have been working with Java's byte primitive lately, and I came across a silly problem:
byte a = 10;
byte b = 9;
byte c = 8;
b += b*c; // how come this statement is correct without any explicit type casting
b = b*c; // while this statement is incorrect; it requires explicit cast, of course
b = b+(b*c); // this is wrong too.
So my question is, does += specify any assignment other than just add and assign, or is this a bug in Java (which I am almost sure is not)?

Because b += b*c is equivalent to b += (byte) ((b) + (b*c)).
From the Java language specification on compound assignment operators:
A compound assignment expression of the form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is evaluated only once.

All compound assignment operators not only performs the operation, but also cast automatically their result to the type of the left-hand side variable.
So your += not only adds variables and assign the result - it also cast the result to the right type.

Why does the ternary operator unexpectedly cast integers?

I have seen it discussed somewhere that the following code results in obj being a Double, but that it prints 200.0 from the left hand side.
Object obj = true ? new Integer(200) : new Double(0.0);
System.out.println(obj);
Result: 200.0
However, if you put a different object in the right hand side, e.g. BigDecimal, the type of obj is Integer as it should be.
Object obj = true ? new Integer(200) : new BigDecimal(0.0);
System.out.println(obj);
Result: 200
I presume that the reason for this is something to do with casting the left hand side to a double in the same way that it happens for integer/double comparisons and calculations, but here the left and right sides do not interact in this way.
Why does this happen?

You need to read section 15.25 of the Java Language Specification.
In particular:
Otherwise, if the second and third operands have types that are convertible (§5.1.8) to numeric types, then there are several cases:
If one of the operands is of type byte or Byte and the other is of type short or Short, then the type of the conditional expression is short.
If one of the operands is of type T where T is byte, short, or char, and the other operand is a constant expression of type int whose value is representable in type T, then > - the type of the conditional expression is T.
If one of the operands is of type Byte and the other operand is a constant expression of type int whose value is representable in type byte, then the type of the conditional expression is byte.
If one of the operands is of type Short and the other operand is a constant expression of type int whose value is representable in type short, then the type of the conditional expression is short.
If one of the operands is of type; Character and the other operand is a constant expression of type int whose value is representable in type char, then the type of the conditional expression is char.
Otherwise, binary numeric promotion (§5.6.2) is applied to the operand types, and the type of the conditional expression is the promoted type of the second and third operands. Note that binary numeric promotion performs unboxing conversion (§5.1.8) and value set conversion (§5.1.13).
So binary numeric promotion is applied, which starts with:
When an operator applies binary numeric promotion to a pair of operands, each of which must denote a value that is convertible to a numeric type, the following rules apply, in order, using widening conversion (§5.1.2) to convert operands as necessary:
If any of the operands is of a reference type, unboxing conversion (§5.1.8) is performed. Then:
If either operand is of type double, the other is converted to double.
That's exactly what happens here - the parameter types are converted to int and double respectively, the second operand (the third in the original expression) is then of type double, so the overall result type is double.

Numeric conversion in the conditional operator ? :
In the conditional operator a?b:c, if both b and c are different numeric types, the following conversion rules are applied at compile time to make their types equal, in order:
The types are converted to their corresponding primitive ones, which is called unboxing.
If one operand were a constant int (not Integer before unboxing) whose value is representable in the other type, the int operand is converted into the other type.
Otherwise the smaller type is converted into the next greater one until both operands have the same type. The conversion orders are:
byte -> short -> int -> long -> float -> double
char -> int -> long -> float -> double
Eventually the whole conditional expression gets the type of its second and third operands.
Examples:
If you combine char with short, the expression becomes int.
If you combine Integer with Integer, the expression becomes Integer.
If you combine final int i = 5 with a Character, the expression becomes char.
If you combine short with float, the expression becomes float.
In the question's example, 200 is converted from Integer into double, 0.0 is unboxed from Double into double and the whole conditional expression becomes becomes double which is eventually boxed into Double because obj is of type Object.

Example:
public static void main(String[] args) {
int i = 10;
int i2 = 10;
long l = 100;
byte b = 10;
char c = 'A';
Long result;
// combine int with int result is int compiler error
// result = true ? i : i2; // combine int with int, the expression becomes int
//result = true ? b : c; // combine byte with char, the expression becomes int
//combine int with long result will be long
result = true ? l : i; // success - > combine long with int, the expression becomes long
result = true ? i : l; // success - > combine int with long, the expression becomes long
result = true ? b : l; // success - > combine byte with long, the expression becomes long
result = true ? c : l; // success - > char long with long, the expression becomes long
Integer intResult;
intResult = true ? b : c; // combine char with byte, the expression becomes int.
// intResult = true ? l : c; // fail combine long with char, the expression becomes long.
}