问题
Consider the following Java code:
byte a = -64;
System.out.println(a << 1);
The output of this code is -128
I tried as follows to figure out why this is the output:
64 = 0 1000000 (the MSB is the sign bit)
-64= 1 1000000 (Tow's complement format)
Expected output after shifting: 1 0000000 (This is equal to 0, because the MSB is just a sign bit)
Please anyone explain what I am missing.
回答1:
The two's complement representation of -128 is 10000000, thus your results are correct.
回答2:
10000000 is -128
10000001 is -127
10000010 is -126
...
So 10000000 is not 0. It is -128 which was your output.
回答3:
This program
System.out.println(Integer.toBinaryString(-64));
System.out.println(Integer.toBinaryString(-64 << 1));
System.out.println("-64 << 1 = " + (-64 << 1));
prints
11111111111111111111111111000000
11111111111111111111111110000000
-64 << 1 = -128
You can see that -64 << 1 is the same as -64 except all the bits have been shift left by 1 (the lowest bit becomes a 0)
回答4:
In two's complement, the MSB is not just a sign bit, you're thinking ones'-complement maybe? In 8 bit two complement,
10000000 = 0x80 = -128
回答5:
In shift operators sign bit is ignored. So 1 1000000 << 1 is 10000000 which is -128.what's the problem?
Our machines are using two's complement to represent numbers (signed and unsigned). For representing a negative number machine negates it's positive and adds 1.
-128 is !10000000 + 1 = 01111111 + 1 = 10000000
EDIT:
I was wrong, only right shift operator's ignoring the sign bit. 10100000 << 1 == 01000000
For unsigned right shifting there's an operator >>> which shifts sign bit too.
11000000>>1 == 10100000 and 11000000>>>1 == 01100000
回答6:
<<means multiply by 2
>>means divide by 2
And, during shift operations don't consider signed bit.
回答7:
I’m wondering. << 1 (ignoring all details) is “multiply with 2.” -64 * 2 = -128. So why are you wondering that it indeed is -128?
来源：https://stackoverflow.com/questions/9207007/java-shift-operator