[SOLVED] Have fields in bit fields got contiguous memory location for its elements? Is this behavior compiler independent?

Have fields in bit fields got contiguous memory location for its elements? Is this behavior compiler independent?

In a bit field in C, are the memory locations of its elements contiguous? If yes, is this behavior equal for all compilers?

Example:

typedef struct
{
   uint8_t        in_alarm      :1;
   uint8_t        fault         :1;
   uint8_t        overridden    :1;
   uint8_t        out_of_service:1;
}StatusFlag_t;

Will the fields in_alarm, fault etc have contiguous memory locations?

Solution

Almost every aspect of bit-field handling is compiler dependent according to the C standard. The chances are that the 4 bits in your structure will be contiguous in a single byte, but it is not guaranteed whether they'll be the most-significant or least-significant 4 bits. If there were more than 8 bits, then the values would cross more than one storage unit (because the base type is uint8_t; note that the C standard does not require a compiler to support the use of uint8_t as the type of a bit-field). Note that bit-field members usually do not have different addresses; you can't take the address of a bit-field element.

There is no guarantee that different compilers will behave the same across different platforms. There is typically an ABI (application binary interface) which defines the behaviour for a particular O/S, and the compiler will adhere to the ABI for the platform on which it is running. But the C standard does not mandate this behaviour.

For relevant quotes from the standard, see (amongst other possibilities) How do bit-fields and their alignments work in C programming?