Inconsistent behaviour of primitive integer types in Java
java, primitive-types, scjp
Solution
Section 5.1.3 of the JLS talks about the behavior of the narrowing primitive conversion used by the cast
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.
(emphasis mine)
That is why `(int) (Math.pow(2, 32));` becomes `Integer.MAX_VALUE` and `(long) (Math.pow(2, 64))` becomes `Long.MAX_VALUE`.
Problem
Can someone explain to me like I'm five why I get different behaviour for two of four primitive types representing integers in Java? AFAIK all four are signed and they all use the most significant bit as a sign bit, so why do byte and short behave normally, and int and long act, well, strange? The fragment of oracle docs explaining this would be perfect. ``` byte a = (byte) (Math.pow(2, 7)-1); //127 - as expected short b = (short) (Math.pow(2, 15)-1); //32767 - as expected int c = (int) (Math.pow(2, 31)-1); //2147483647 - as expected long d = (long) (Math.pow(2, 63)-1); //9223372036854775807 - as expected a = (byte) (Math.pow(2, 7)); //-128 - as expected b = (short) (Math.pow(2, 15)); //-32768 - as expected c = (int) (Math.pow(2, 31)); //2147483647 - why not '-2147483648'? d = (long) (Math.pow(2, 63)); //9223372036854775807 - why not '-9223372036854775808'? a = (byte) (Math.pow(2, 8)); //0 - as expected b = (short) (Math.pow(2, 16)); //0 - as expected c = (int) (Math.pow(2, 32)); //2147483647 - why not '0'? d = (long) (Math.pow(2, 64)); //9223372036854775807 - why not '0'? ``` I'm using Oracle's Java SE 1.7 for Windows. OS is Windows 7 Professional SP1 ``` java version "1.7.0_45" Java(TM) SE Runtime Environment (build 1.7.0_45-b18) Java HotSpot(TM) 64-Bit Server VM (build 24.45-b08, mixed mode) ``` EDIT, after reading all the answers and tuning my code. So, to sum up, the only way I found to get expected values is the use of BigInteger. Shift operator works well for bytes, shorts and ints, but when it comes to longs, I cought it on one malfunction. ``` byte a = (byte) ((1l << 7) - 1); //127 - as expected short b = (short) ((1l << 15) - 1); //32767 - as expected int c = (int) (1l << 31) - 1; //2147483647 - as expected long d = (1l << 63) - 1; //9223372036854775807 - as expected a = (byte) (1l << 7); //-128 - as expected b = (short) (1l << 15); //-32768 - as expected c = (int) 1l << 31; //-2147483648 - as expected d = 1l << 63; //-9223372036854775808 - as expected a = (byte) (1l << 8); //0 - as expected b = (short) (1l << 16); //0 - as expected c = (int) (1l << 32); //0 - as expected d = 1l << 64; //1 instead of 0, probably because of the word length limitation ``` With BigInteger everything works flawlessly ``` byte a = (byte) (new BigInteger("2").pow(7).longValue() - 1); //127 - as expected short b = (short) (new BigInteger("2").pow(15).longValue() - 1); //32767 - as expected int c = (int) (new BigInteger("2").pow(31).longValue() - 1); //2147483647 - as expected long d = (new BigInteger("2").pow(63).longValue() - 1); //9223372036854775807 - as expected a = (byte) (new BigInteger("2").pow(7).longValue()); //-128 - as expected b = (short) (new BigInteger("2").pow(15).longValue()); //-32768 - as expected c = (int) new BigInteger("2").pow(31).longValue(); //-2147483648 - as expected d = new BigInteger("2").pow(63).longValue(); //-9223372036854775808 - as expected a = (byte) (new BigInteger("2").pow(8).longValue()); //0 - as expected b = (short) (new BigInteger("2").pow(16).longValue()); //0 - as expected c = (int) (new BigInteger("2").pow(32).longValue()); //0 - as expected d = new BigInteger("2").pow(64).longValue(); //0 - as expected ``` Thanks everyone for big help!