embedded - understanding bit shifting in registers -

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i have update 32 bit register using these data includes bit shifting, confused 2 things:

  1. which lsb , msb,
  2. what operator |

given expression:

3 << 0 |  7 << 3 |  1 << 6 |  0 << 7 |  1 << 7 |  0 << 8 |  0 << 10 | 0 << 11 |  0 << 12 |  0 << 13 |  0 << 14

and remaining 15 bits 0.

how data shifted, assuming initial bits in register 0's?

011 111 1 0 1 0 0 0 0 0 0 x.......x

or

x .....x 0 0 0 0 0 0 1 0 1 111 011

the lsb (least-significant bit) bit value represents 1 (2^0) , msb bit value represents 2^(n-1), n number of bits in register. in general, when written out in binary, msb left-most bit , lsb right-most bit. more not, lsb shown bit 0 in hardware documentation, though know 1 company reverses bit numbering msb numbered 0.

<< c shift-left operator, shifting bit away lsb , toward msb. therefore 7<<3 represents 111000 in binary.

| c bit-wise or operator. used combine values resulting bits 1 if either of corresponding input bits one.

looking @ original value 3 << 0 | 7 << 3 | 1 << 6 | 0 << 7 | 1 << 7 | 0 << 8 | 0 << 10 | 0 << 11| 0 << 12 | 0 << 13 | 0 << 14

0000 0000 0000 0011 3<<0

0000 0000 0011 1000 7<<3

0000 0000 0100 0000 1<<6

0000 0000 0000 0000 0<<7

etc.

this type of construct typically used describe value going register noting individual fields of register.


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