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cbit-shiftuint64uint8tuint32

convert uint8 and uint32 variables to one uint64

I have the following variables.

uint8_t flags;
uint32_t token;

I need to write function that cobine them into one uint64_t, and than parse them back into two variables, one of uint8_t and one of uint32_t.

uint64 convert(uint8_t flags, uint32_t token);
void convertBack(uint64 allTogether, uint8_t* flags, uint32_t* token);

I tried to found something that doing the following, but most of what I found is convert two of the same to one bigger, like two uint32_t to one uint64_t

Thank You

Solution

  • How about:

    uint64_t convert(uint8_t flags, uint32_t token)
{
  return ((uint64_t) flags << 32) | token;
}

    This puts the 8-bit field "above" the 32-bit one, i.e. the returned uint64_t has its bits arranged like this:

     +---------------------------------+--------------------------------+
 |666655555555554444444444|33333333|3322222222221111111111          |
 |321098765432109876543210|98765432|10987654321098765432109876543210|
 +------------------------+--------+--------------------------------+
 |        unused(24)      |flags(8)|          token(32)             |
 +------------------------+-----------------------------------------+

    Bit numbers in decimal should be read downwards, bit 0 is on the far right (the LSB).

    I'll leave convertBack() (which is a pretty bad name, both of these names are sub-optimal) to you. :)