How is the size of a struct with Bit Fields determined/measured?

Question

#include 

typedef struct size
{
        unsigned int a:1;
        unsigned int b:31;
        unsigned int c:1;
} mystruct;

int main()
{
        myst         


        

Answer 1

According to Steve jessop's answer just to fulfill his answer by adding some documents which may help.

A member of a structure or union may have any complete object type other than a variably modified type.In addition, a member may be declared to consist of a specified number of bits (including a sign bit, if any). Such a member is called a bit-field its width is preceded by a colon

An implementation may allocate any addressable storage unit large enough to hold a bit- field. If enough space remains, a bit-field that immediately follows another bit-field in a structure shall be packed into adjacent bits of the same unit. If insufficient space remains, whether a bit-field that does not fit is put into the next unit or overlaps adjacent units is implementation-defined. The order of allocation of bit-fields within a unit (high-order to low-order or low-order to high-order) is implementation-defined. The alignment of the addressable storage unit is unspecified.

Within a structure object, the non-bit-field members and the units in which bit-fields reside have addresses that increase in the order in which they are declared. A pointer to a structure object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides), and vice versa. There may be unnamed padding within a structure object, but not at its beginning.

——ISO/IEC 9899:201x 6.7.2.1