c++ - Size of a structure having unsigned short ints -

i surfing in 1 of our organisational data documents , came across following piece of code.

struct {  unsigned short int i:1;  unsigned short int j:1;  unsigned short int k:14; };   int main(){  aa;  int n = sizeof(aa);  cout << n; } 

initially thought size 6 bytes size of unsigned short int 2 bytes. output of above code 2 bytes(on visual studio 2008).

is there slight possibility i:1, j:1 , k:14 makes bit field or something? guess , not sure it. can please me in this?

yes, bitfield, indeed.

well, i'm not sure c++, but in c99 standard, per chapter 6.7.2.1 (10):

an implementation may allocate addressable storage unit large enough hold bit-field. if enough space remains, bit-field follows bit-field in structure shall packed adjacent bits of same unit. if insufficient space remains, whether bit-field not fit put next bits or overlaps adjacent units implementation-defined. order of allocation of bit-fields within unit (high-order low-order or low-order high-order) implementation-defined. alignment of addressable storage unit unspecified.

that makes structure size (1 bit + 1 bit + 14 bits) = 16 bits = 2 bytes.

note: no structure padding considered here.

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edit:

as per c++14 standard, chapter §9.7,

a member-declarator of form

identifier_opt attribute-specifier-seq_opt: constant-expression

specifies bit-field; length set off bit-field name colon. [...] allocation of bit-fields within class object implementation-defined. alignment of bit-fields implementation-defined. bit-fields packed addressable allocation unit.