public class Shift {
public static void main(String[] args) {
for(int i = 0; i < 32; ++i){
System.out.println(-0x55555555 << i);
}
}
}
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
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);
}
}
}
Output
-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
In the specs (§15.19) it's said that :
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.
Can someone explain the why's of the first output?
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.
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.
来源：https://stackoverflow.com/questions/16049720/signed-left-shift-behaviour