How does double to int cast work in Java
It's all documented in section 5.1.3 of the JLS.
In the first step, the floating-point number is converted either to a long, if T is long, or to an int, if T is byte, short, char, or int, as follows:
If the floating-point number is NaN (§4.2.3), the result of the first step of the conversion is an int or long 0.
Otherwise, if the floating-point number is not an infinity, the floating-point value is rounded to an integer value V, rounding toward zero using IEEE 754 round-toward-zero mode (§4.2.3). Then there are two cases:
If T is long, and this integer value can be represented as a long, then the result of the first step is the long value V.
Otherwise, if this integer value can be represented as an int, then the result of the first step is the int value V.
Otherwise, one of the following two cases must be true:
The value must be too small (a negative value of large magnitude or negative infinity), and the result of the first step is the smallest representable value of type int or long.
The value must be too large (a positive value of large magnitude or positive infinity), and the result of the first step is the largest representable value of type int or long.
(The second step here is irrelevant, when T
is int
.)
In most cases I'd expect this to be implemented using hardware support - converting floating point numbers to integers is something which is usually handled by CPUs.
Cast Double to Integer in Java
A Double
is not an Integer
, so the cast won't work. Note the difference between the Double
class and the double
primitive. Also note that a Double
is a Number
, so it has the method intValue
, which you can use to get the value as a primitive int
.
How to cast a double to an int in Java by rounding it down?
Casting to an int implicitly drops any decimal. No need to call Math.floor() (assuming positive numbers)
Simply typecast with (int), e.g.:
System.out.println((int)(99.9999)); // Prints 99
This being said, it does have a different behavior from Math.floor
which rounds towards negative infinity (@Chris Wong)
How do I cast a double to an int in Java?
double
is not an object, it is a primitive type.
Simply writing (int)b
will do the job.
If you really need a Double
object, you need to construct one.
Converting double to integer in Java
is there a possibility that casting a double created via
Math.round()
will still result in a truncated down number
No, round()
will always round your double to the correct value, and then, it will be cast to an long
which will truncate any decimal places. But after rounding, there will not be any fractional parts remaining.
Here are the docs from Math.round(double)
:
Returns the closest long to the argument. The result is rounded to an integer by adding 1/2, taking the floor of the result, and casting the result to type long. In other words, the result is equal to the value of the expression:
(long)Math.floor(a + 0.5d)
type casting int to double java?
The only possibilities you have are:
Type cast the
double
into anint
if you want to store your number as an integer:int a = (int)26.4 // so a will be 26
(you will obviously lose precision this way)
Store the number as a
double
to keep the precision:double a = 26.4
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