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.