Bit shifting, done by the << and >> operators, allows in C languages to express memory and storage access, which is quite important to read and write exchangeable data. But:

Question: where is the bit when moved shifted left?

Answere: it depends

Long Answere:

// On LSB/intel a left shift makes the bit moving 
// to the opposite, the right. Omg 
// The shift(<<,>>) operators follow the MSB scheme
// with the highest value left. 
// shift math expresses our written order.
// 10 is more than 01
// x left shift by n == x << n == x * pow(2,n)
#include <stdio.h> // printf
#include <stdint.h> // uint16_t
int main(int argc, char **argv)
{
  uint16_t u16, i, n;
  uint8_t * u8p = (uint8_t*) &u16; // uint16_t as 2 bytes
  // iterate over all bit positions
  for(n = 0; n < 16; ++n)
  {
    // left shift operation
    u16 = 0x01 << n;
    // show the mathematical result
    printf("0x01 << %u:\t%d\n", n, u16);
    // show the bit position
    for(i = 0; i < 16; ++i) printf( "%u", u16 >> i & 0x01);
    // show the bit location in the actual byte
    for(i = 0; i < 2; ++i)
      if(u8p[i]) 
        printf(" byte[%d]", i); 
    printf("\n");
  }
  return 0;
}

Result on a LSB/intel machine:

0x01 << 0:      1 
1000000000000000 byte[0] 
0x01 << 1:      2 
0100000000000000 byte[0] 
0x01 << 2:      4 
0010000000000000 byte[0] 
0x01 << 3:      8 
0001000000000000 byte[0]
...

<< MSB Moves bits to the left, while << LSB is a Lie and moves to the right. For directional shifts I would like to use a separate operator, e.g. <<| and |>>.