签署左移行为 [英] Signed left shift behaviour
问题描述
public class Shift {
public static void main(String[] args) {
for(int i = 0; i < 32; ++i){
System.out.println(-0x55555555 << i);
}
}
}
运行上面的代码给出了以下输出
Running the above code gives the following output
-1431655765
1431655766
-1431655764
1431655768
-1431655760
1431655776
-1431655744
1431655808
-1431655680
1431655936
-1431655424
1431656448
-1431654400
1431658496
-1431650304
1431666688
-1431633920
1431699456
-1431568384
1431830528
-1431306240
1432354816
-1430257664
1434451968
-1426063360
1442840576
-1409286144
1476395008
-1342177280
1610612736
-1073741824
-2147483648
使用其他值（64）进行测试可获得更预期的结果
While testing with an other value (64) gives a more expected result
public class Shift {
public static void main(String[] args) {
for(int i = 0; i < 32; ++i){
System.out.println(-0x40 << i);
}
}
}
输出
-64
-128
-256
-512
-1024
-2048
-4096
-8192
-16384
-32768
-65536
-131072
-262144
-524288
-1048576
-2097152
-4194304
-8388608
-16777216
-33554432
-67108864
-134217728
-268435456
-536870912
-1073741824
-2147483648
0
0
0
0
0
0
在规格（§15.19）据说：
n<<< s是n个左移位位;这相当于（即使发生溢出）乘以2乘以幂。
有人可以解释第一个输出的原因？
Can someone explain the why's of the first output?
推荐答案
n<<< s是n个左移位位;这相当于（即使发生溢出）乘以2乘以幂。
The value of n << s is n left-shifted s bit positions; this is equivalent (even if overflow occurs) to multiplication by two to the power s.
这是真的：这正是结果如果由于溢出而将 -0x55555555 乘以 s 的幂，你会得到。特别是， 0x55555555 具有交替的0和1，因此您将0和1交替转换为符号位，因此符号每次都会翻转。
And it's true: this is exactly the result you'd get if you multiplied -0x55555555 by two to the power of s, due to overflow. In particular, 0x55555555 has alternating 0s and 1s, so you're shifting alternating 0s and 1s into the sign bit, so the sign is flipping every time.
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