x = 1010 0011

n = 2

This should produce:

x = 1110 1000

Now run through the code:

mask = ~(~(1 << n) + 1)

mask = ~(~(0000 0001 << 2) + 1)

mask = ~(~(0000 0100) + 1)

mask = ~((1111 1011) + 1)

mask = ~(1111 1100)

mask = 0000 0011 (This can be used to save the first two bits, which we want, since shifting right will delete them from x)

int right = x & mask

int right = 1010 0011 & 0000 0011

int right = 0000 0011 (first two bits saved)

x = (x >> n)

x = 1010 0011 >> 2

x = 1110 1000 (The MSB was a 1, so it filled with 1's. Don't worry, it'll be fixed later)

int a = ~((1 << 31) >> ~(~n + 1))

int a = ~((0000 0001 << 7) >> ~(~2 + 1))

int a = ~((0000 0001 << 7) >> ~(~0000 0010 + 1))

int a = ~(1000 0000 >> ~(1111 1101 + 1))

int a = ~(1000 0000 >> ~1111 1110)

int a = ~(1000 0000 >> 0000 0001)

int a = ~(1100 0000)

int a = 0011 1111 (this can be used to clear out the first n bits from x)

x = (x & a)

x = 1110 1000 & 0011 1111

x = 0010 1000 (x has now been shifted to the right, and the first n bits were cleared)

right = right << (8 + (~n + 1));

right = 0000 0011 << (8 + (~2 + 1));

right = 0000 0011 << (8 + (~0000 0010 + 1));

right = 0000 0011 << (8 + (1111 1101 + 1));

right = 0000 0011 << (8 + (1111 1110));

right = 0000 0011 << (8 - 2);

right = 1100 0000

return(right | x);

return(1100 0000 |  0010 1000);

return(1110 1000); (The answer we wanted)

See how it works?