In the JLS in 5.6.2. Binary Numeric Promotion:
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:
If any operand is of a reference type, it is subjected to unboxing conversion (§5.1.8).
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
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.
This confuses me because both
http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/04/mixed.html and
https://alvinalexander.com/java/java-int-double-float-mixed-type-division-arithmetic-rules
state that
All floating point values (float and double) in an arithmetic operation (+, −, *, /) are converted to double type before the arithmetic operation in performed.
With my understanding, according to the JLS, adding 1 float and 1 integer would result in a float. However, according to the other sources, adding 1 float and 1 integer would result in a double?
When a source makes a claim contrary to the JLS, chances are good that it's wrong. In this case, the following demonstrates the behaviour that the JLS describes:
Number n = 1 + 0.1f;
System.out.println(n instanceof Float); // true
System.out.println(n instanceof Double); // false
Related
So, this university exercise asks to write a method that takes as an argument a Function<Short, Short> and result type a Function<Short, Short>. It should return the outcome of the function divided by 9.
This is the code I wrote, and while I'm quite confident that it should be working, eclipse shows an error at line 16 that I simply don't understand:
"Type mismatch: cannot convert from Function<Short,Object> to Function<Short,Short>.
It works fine if I put Function<Integer, Integer> in both, or if I write Function<Short, Object> at the return type, and leave the argument as Function<Short, Short>.
public Function<Short, Short> ulmic(Function<Short, Short> period) {
return period.andThen(a -> a / 9);
}
In order to make it work, you have to write as follows:
public Function<Short, Short> ulmic(Function<Short, Short> period) {
return period.andThen(a -> (short) (a / 9));
}
When you apply the operand, / over a and 9, Java converts the result to an integer.
From Chapter 5. Conversions and Promotions one can read:
One conversion context is the operand of a numeric operator such as +
or *. The conversion process for such operands is called numeric
promotion. Promotion is special in that, in the case of binary
operators, the conversion chosen for one operand may depend in part on
the type of the other operand expression.
and 5.6.2. Binary Numeric Promotion
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:
If any operand is of a reference type, it is subjected to unboxing
conversion (§5.1.8).
Widening primitive conversion (§5.1.2) is applied to convert either or
both operands as specified by the following rules:
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.
when exactly does the widening primitive conversion happen in the program?
If I got an expression:
long l = 3L;
double d = 5.2L + l;
are 5.2L and l converted into double and then calculated, or is the calculation happening in long, with the result being converted to double afterwards?
The statement a = b + c consists of 2 operations:
Additive Operator +
JLS §15.18.2. Additive Operators (+ and -) for Numeric Types says:
Binary numeric promotion is performed on the operands (§5.6.2).
JLS §5.6.2. Binary Numeric Promotion says:
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
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.
Assignment Operator =
JLS §5.2. Assignment Contexts says:
Assignment contexts allow the value of an expression to be assigned (§15.26) to a variable; the type of the expression must be converted to the type of the variable.
Assignment contexts allow the use of one of the following:
an identity conversion (§5.1.1)
a widening primitive conversion (§5.1.2)
...
JLS §5.1.2. Widening Primitive Conversion says:
19 specific conversions on primitive types are called the widening primitive conversions:
byte to short, int, long, float, or double
short to int, long, float, or double
char to int, long, float, or double
int to long, float, or double
long to float or double
float to double
To re-cap, first the + operator causes the two operands to be widened to int, long, float, or double, whichever first covers both operands. The result of the + operator may then be further widened to fit the variable of the assignment.
Your original code
long l = 3L;
double d = 5.2L + l;
will not compile because 5.2L is not a valid literal.
In answer to your further questions:
which case would happen, if it was 5L instead?
In that case you would be adding two longs, and the result would be a long. That long would be converted to a double if you store it in a double variable.
And what would happen if there was an integer involved?
If you add an int and a long, you get a long. If you are assigning that to a double variable, it will subsequently be converted to a double.
byte a=10;
byte b=20;
b=a+b;
In this case, I need to explicitly convert a+b to byte like this :
b=(byte)(a+b);
It's the same with short :
short x=23;
short y=24;
Otherwise it gives an error.
But in case of integers, it's not required to convert explicitly :
int p=7788;
int q=7668;
p=p+q;
This will work just fine.
Why is that?
We don't need to explicitly cast even in the case of long as well.
If you look at the JLS 4.2.2 Integer Operations, it states that the result of a numerical operation between two integral operands is an int or a long. Since there's no implicit cast from an int to byte or a short, you need an explicit cast.
If you refer to JLS Sec 15.18.2, which deals with addition of numeric types, it says:
Binary numeric promotion is performed on the operands (§5.6.2).
...
The type of an additive expression on numeric operands is the promoted type of its operands.
JLS Sec 5.6.2 describes binary numeric promotion:
If any operand is of a reference type, it is subjected to unboxing conversion (§5.1.8).
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
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.
So, in the case of int and long (where both operands are of that type), binary numeric promotion is a no-op: the operands remain int and long respectively, and the result of the addition is int and long respectively, meaning the result can be assigned to variables of that type.
In the case of byte and short, binary numeric promotion causes both of those to be widened to int to perform the addition, and the result of the addition is int; you have to explicitly cast back again to the narrower type, because not all int values fit into a byte or short.
There are 2 exceptions to this requirement to do an explicit narrowing cast.
Firstly, compound assignments: this would have worked:
b += a;
because, as stated in JLS Sec 15.26.2:
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.
In other words: the compiler inserts the cast for you:
b = (byte) ((b) + (a));
Secondly, if the operands have constant values, and the result of the addition is known to fit into the range of the narrower type, and you are doing the assignment at the variable declaration.
For example:
final byte a=10; // final is necessary for a and b to be constant expressions.
final byte b=20;
byte c = a + b;
This requires no cast.
With addition in java, java will promote the smaller data type to the larger one. And when the data type is smaller than int it will promote both the operant to int. Take a look at: Promotion in Java?
This question already has answers here:
Unexpected type resulting from the ternary operator
(4 answers)
Closed 8 years ago.
Object myObject = true ? new Integer(25) : new Double(25.0);
System.out.println(myObject);
Strangely, it outputs 25.0 instead of 25
Whats going on?
Your code returns the second operand (new Integer(25)) as you expected, but it converts it to a Double due to the following rules.
Here's what JLS 15.25 says :
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 (§15.28) 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 T, where T is Byte, Short, or Character, and the other operand is a constant expression (§15.28) of type int whose value is representable in the type U which is the result of applying unboxing conversion to T, then the type of the conditional expression is U.
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 value set conversion (§5.1.13) and may perform unboxing conversion (§5.1.8).
And the numeric promotion :
5.6.2. Binary Numeric Promotion
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:
If any operand is of a reference type, it is subjected to unboxing conversion (§5.1.8).
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
If either operand is of type double, the other is converted to double.
In your example, you have an Integer and a Double. They are unboxed to int and double and then the int is converted to a double.
There is some wierd autoboxing stuff going on - you can see it better if you use different numbers:
Object myObject = true ? new Integer(25) : new Double(22.0);
Now, myObject will still be assigned a Double(25.0), not the 22.0 you would expect if the conditional didn't work. Basically, because Java thinks you are doing some sort of calculation involving an int and a double it returns the result of the iif as a "double" primative and then autoboxes it back to a Double().
You could also get it to behave as expected by forcing it to treat the values as type Object():
Object myObject = true ? (Object) new Integer(25) : (Object) new Double(22.0);
After compilation
Object myObject = true ? new Integer(25) : new Double(25.0);
will be something as below
Object myObject = (double) new Integer(25);
So, it returns new Integer(25) as expected but it gets converted to double.
As Eran mentioned, as per JLS 5.6.2:
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.
If any operand is of a reference type, it is subjected to unboxing conversion (§5.1.8).
Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:
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.
I have two variables that should be compared:
Double a = 1D;
Double b = 2D;
if (a > b) {
System.out.print("Ok");
}
In this case java will use autoboxing or compare two object's references?
From section 15.20.1 of the JLS:
The type of each of the operands of a numerical comparison operator must be a type that is convertible (§5.1.8) to a primitive numeric type, or a compile-time error occurs. Binary numeric promotion is performed on the operands (§5.6.2). If the promoted type of the operands is int or long, then signed integer comparison is performed; if this promoted type is float or double, then floating-point comparison is performed.
Section 5.6.2 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.
So yes, unboxing is performed. > has no meaning for references themselves.
More interesting is the == case where both options would be possible - and in that case, if either operand is a primitive and the other can be converted via numeric promotion, then that happens... but if both are reference types, the reference comparison is performed. For example:
Double d1 = new Double(1.0);
Double d2 = new Double(1.0);
System.out.println(d1 == d2); // Prints false due to reference comparison
It will use autoboxing. You cannot do greater than on references