Java 中原始整数类型的行为不一致

有人可以向我解释一下，就像我五岁一样，为什么我在 Java 中表示整数的四种基本类型中的两种会得到不同的行为？ AFAIK 所有四个都是有符号的，并且它们都使用最高有效位作为符号位，那么为什么 byte 和 Short 表现正常，而 int 和 long 表现，嗯，奇怪？解释这一点的 Oracle 文档片段将是完美的。

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'?

我正在使用适用于 Windows 的 Oracle Java SE 1.7。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)

阅读所有答案并调整我的代码后进行编辑。
因此，总而言之，我发现获得期望值的唯一方法是使用 BigInteger。移位运算符适用于字节、短整型和整数，但当涉及长整型时，我因一个故障而放弃了它。

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

感谢大家的大力帮助！

JLS 第 5.1.3 节谈论强制转换使用的缩小原始转换的行为

否则，必须满足以下两种情况之一：

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

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

（强调我的）

因此(int) (Math.pow(2, 32));变成Integer.MAX_VALUE and (long) (Math.pow(2, 64))变成Long.MAX_VALUE.