Bit Shifting

Left Shift

A left shift of bits can be performed with the << operator. Note that the operator points toward the left.

To shift a bit vector x over by y positions, you would write x << y.

The leftmost bits will be thrown away as you shift.

The right side of the bit array will be filled with 0s.

Right Shift

A right shift of bits can be performed with the >> operator. Note that the operator points toward the right.

To shift a bit vector x over by y positions, you would write x >> y.

The rightmost bits will be thrown away as you shift.

In the case of a Logical Shift, the left side of the bit array will be filled with 0s. This happens if x is an unsigned value.

In the case of an Arithmetic Shift, the left side of the bit array will be filled with 0s or 1s depending on the parity of the value x. That is, if x is a negative signed number, then it will fill with 1s.

Example 1

Let x = 0x62 be an unsigned int. Find the binary representation of x << 3.

Solution:

  • 0x62 in binary is 1100010

  • However, recall that ints are 4 bytes, or 32-bits.

  • Thus, 0x62 in memory is as follows

    00000000 00000000 00000000 01100010

    Note that we have space separated the bytes for visual purposes.

  • Now if we shift the bits 3 places to the left, we get the following:

    00000000 00000000 00000011 00010000

Example 2

Let x = 0x62 be an unsigned char. Find the binary representation of x << 3.

Solution:

  • 0x62 in binary is 1100010

  • However, recall that ints are 1 byte, or 8-bits.

  • Thus, 0x62 in memory is as follows

    01100010

    Note that we only added a leading zero.

  • Now if we shift the bits 3 places to the left, we get the following:

    00010000

  • Notice how we lost several bits in the shift.

Example 3

Let x = -8 (base 10) be an signed char. Find the binary representation of x >> 2.

Solution:

  • -8 in binary, as an unsigned char is 11111000

  • Recall that this uses two’s complement notation.

  • If we shift the bits 2 places to the right as an arithmetic shift, we get the following:

    11111110

  • Notice how the left side was filled with 1s to maintain the sign.

  • The resulting base 10 value is -2.