Bitfield storage in memory

typedef struct {

    unsigned int    b1:1;
    unsigned int    b2:8;
    unsigned int    b3:7;
    unsigned int    b4:8;
    unsigned int    b5:7;
    unsigned int    b6:1;} BIT_STRUCT;

    Suppose you wanted to use the same structure definition to read in bit field data on an
big endian and little endian machine and you were willing to massage the data on input
before mapping the structure to it. Suppose we started with big endian data that was
32 bits long.

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Memory layout big/little endian:

              byte 0          byte 1       byte 2       byte 3

• |B31....B25     |B24....B16         |B15....B8   |B7....B0           |   Big endian
• |B7....B0         ||B15....B8          ||B24....B16||B31....B25      |   Little endian
• |b1[0],b2[7:1] | b2[0], b3[6:0]   | b4[7:0]       | b5[6:0] , b6[0]| physical layout when bigend generates.

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Mapping the bit fields to memory:

• Big endian.. .b1 is the most significant bit in memory
• For little endian i think that b1 is interpretted as the least significant bit in memory. This would be byte 0 B0.
• So things should almost work except that for the little endian:
• b1 is mapped to B7, not B0. flipping the
• b2[0] gets mapped to B15 not B8
The easiest thing would be to byte swap the data bytes. This won't work because the little endian bit field wants the first bit field close to the front of the first byte.

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Convert big endian to little endian.

1. bit flip each 8 bit byte of the structure.
2. The bit field elements will now extract the correct bits of data, but they are in the reverse order.
3. bit flip each of the structure elements. This needs to be done knowing how many bits each element uses. You can't transfer to a long , flip and then reassign. So you are stuck needing to know the length of each field (unfortunately only the compiler knows that).

1. byte swap the 4 bytes.
2. use the shift and mask operations to peel away Nbits at a time starting at the right side. This also needs to know the length of each structure element.

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