How does double to int cast work in Java

£可爱£侵袭症+ 提交于 2019-11-27 01:49:01
Jon Skeet

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.

Java truncates its value if you use (int) cast, as you may notice:

    double d = 2.4d;
    int i = (int) d;
    System.out.println(i);
    d = 2.6;
    i = (int) d;
    System.out.println(i);

Output:

2
2

Unless you use Math.round, Math.ceil, Math.floor...

mikera

You may want to read the Java specification:

http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.3

The relevant section is 5.1.3 - "Narrowing Primitive Conversion".

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