Implementing Logical Right Shift in C

int lsr(int x, int n)
{
  return (int)((unsigned int)x >> n);
}

This is what you need:

int logical_right_shift(int x, int n)
{
    int size = sizeof(int) * 8; // usually sizeof(int) is 4 bytes (32 bits)
    return (x >> n) & ~(((0x1 << size) >> n) << 1);
}

Explain

x >> n shifts n bits right. However, if x is negative, the sign bit (left-most bit) will be copied to its right, for example:

Assume every int is 32 bits here, let
x　　　　 =　-2147483648　(10000000 00000000 00000000 00000000), then
x　>>　1　=　-1073741824　(11000000 00000000 00000000 00000000)
x　>>　2　=　-536870912　 (11100000 00000000 00000000 00000000)
and so on.

So we need to erase out those sign extra sign bits when n is negative.

Assume n = 5 here:

0x1 << size moves 1 to the left-most position:

(10000000 00000000 00000000 00000000)

((0x1 << size) >> n) << 1 copies 1 to its n-1 neighbors:

(11111000 00000000 00000000 00000000)

~((0x1 << size) >> n) << 1! reverses all bits:

(00000111 11111111 11111111 11111111)

so we finally obtain a mask to extract what really need from x >> n:

(x >> n) & ~(((0x1 << size) >> n) << 1)

the & operation does the trick.

And the total cost of this function is 6 operations.

Just store your int in an unsigned int, and perform >> upon it.

(The sign is not extended or preserved if you use unsigned int)

http://en.wikipedia.org/wiki/Logical_shift