问题
I have this behavior using Java:
int b=16;
System.out.println(b<<30);
System.out.println(b<<31);
System.out.println(b<<32);
System.out.println(b<<33);
output: 0 0 16 32
Is java bit shift circular? IF not, why I get 0 when b<<30 and 16 when b<<32?
回答1:
Bit shifting is not circular; for bit-shifting ints, Java only uses the 5 least-significant bits, so that (b << 0) is equivalent to (b << 32) (is equivalent to (b << 64), etc.). You can simply take the bit-shifting amount and take the remainder when dividing by 32.
Something similar occurs for bit-shifting longs, where Java only uses the 6 least-significant bits, so that (aLong << 0) is equivalent to (aLong << 64).
Section 15.19 of the JLS talks about this:
If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive.
If the promoted type of the left-hand operand is long, then only the six lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x3f (0b111111). The shift distance actually used is therefore always in the range 0 to 63, inclusive.
(emphasis mine)
(You can't bit-shift floats or doubles, and attempting to bit-shift a short or a byte would be subject the value to unary numeric promotion to an int anyway.)
You get 0 from 16 << 30, because the 1-bit from 16
00000000 00000000 00000000 00010000
gets shifted off the end of the int and gets discarded.
// Discarded - Result-----------------------------
(00000100) 00000000 00000000 00000000 00000000
回答2:
No, it's not circular shift. It's normal left-shift. It's just that, for int type left side operand, Java uses just 5 lower order bits of the right operand for shifting. This is as per JLS §15.9:
If the promoted type of the left-hand operand is int, only the five lowest-order bits of the right-hand operand are used as the shift distance. It is as if the right-hand operand were subjected to a bitwise logical AND operator & (§15.22.1) with the mask value 0x1f (0b11111). The shift distance actually used is therefore always in the range 0 to 31, inclusive
So, for 16 << 32, considering only 5 lower order bits of 32, the expression is equivalent to:
16 << 32 & 0x1f
which is equal to 16.
来源：https://stackoverflow.com/questions/21685632/is-java-bit-shifting-circular