Java中原始整数类型的不一致行为

Question

有人可以解释一下,为什么我在Java中为表示整数的四种基本类型中的两种获得不同的行为？AFAIK全部四个都是签名的,它们都使用最重要的位作为符号位,那么为什么字节和短行为正常,而int和long行为,好吧,奇怪？oracle文档的片段解释这将是完美的.

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'?

我正在使用Oracle的Java SE 1.7 for Windows.操作系统是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)

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在阅读完所有答案并调整我的代码后编辑.
总而言之,我发现获得预期值的唯一方法是使用BigInteger.Shift运算符适用于字节,短路和整数,但是当涉及到longs时,我会在一次故障时使用它.

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      

使用BigInteger,一切都可以完美运行

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

谢谢大家的帮助!

Answer 1

JLS的5.1.3节讨论了演员使用的缩小基元转换的行为

否则,以下两种情况之一必须为真:

该值必须太小(大幅度或负无穷大的负值),第一步的结果是int或long类型的最小可表示值.

该值必须太大(大幅度或正无穷大的正值),第一步的结果是int或long类型的最大可表示值.

(强调我的)

这就是(int) (Math.pow(2, 32));成为Integer.MAX_VALUE和(long) (Math.pow(2, 64))成为的原因Long.MAX_VALUE.