按位移位澄清

Question

Assume I have the variable x initialized to 425 . In binary, that is 110101001 .

Bitshifting it to the right by 2 as follows: int a = x >> 2; , the answer is: 106 . In binary that is 1101010 . This makes sense as the two right-most bits are dropped and two zero's are added to the left side.

Bitshifting it to the left by 2 as follows: int a = x << 2; , the answer is: 1700 . In binary this is 11010100100 . I don't understand how this works. Why are the two left most bits preserved? How can I drop them?

Thank you,

Answer 1

This is because int is probably 32-bits on your system. (Assuming x is type int .)

So your 425 , is actually:

0000 0000 0000 0000 0000 0001 1010 1001

When left-shifted by 2, you get:

0000 0000 0000 0000 0000 0110 1010 0100

Nothing gets shifted off until you go all the way past 32. (Strictly speaking, overflow of signed-integer is undefined behavior in C/C++.)

To drop the bits that are shifted off, you need to bitwise AND against a mask that's the original length of your number:

int a = (425 << 2) & 0x1ff;  //  0x1ff is for 9 bits as the original length of the number.

Answer 2

First off, don't shift signed integers. The bitwise operations are only universally unambiguous for unsigned integral types.

Second, why shift if you can use * 4 and / 4 ?

Third, you only drop bits on the left when you exceed the size of the type. If you want to "truncate on the left" mathematically, perform a modulo operation:

(x * 4) % 256

The bitwise equivalent is AND with a bit pattern: (x << 2) & 0xFF

(That is, the fundamental unsigned integral types in C are always implicitly "modulo 2 ⁿ ", where n is the number of bits of the type.)

Answer 3

Why would you expect them to be dropped? Your int (probably) consumes 4 bytes. You're shifting them into a space that it rightfully occupies.

The entire 4-byte space in memory is embraced during evaluation. You'd need to shift entirely out of that space in memory to "drop" them.