c - Efficiency of structs of length 8, and uint64_t -

tldr: 8-byte structures handled efficiently 8-byte uint64_t?

i have 3 data structures similar. 6, 7 , 8 bytes long. want put them in uint64_t variables. goal comparisons , assignments efficient. (these values used key in several (large) trees).

example: have defined following data-structure, 1 7 bytes long.

typedef struct {   union {     uint64_t raw;     struct {       uint8_t unused;       uint8_t node_number;       uint8_t systemid[systemid_length]; /* systemid_length 6 bytes. */     } nodeid;   }; } nodeid_t; 

i can quick assignments , copies via raw union member.

nodeid_t x, y;  x.raw = y.raw if (x.raw > y.raw) { 

etc, etc.

my question use in functions , in assignments. when pass struct value, compiler (gcc) recognize these structures 8 bytes long. , therefore treat if int64_t?

example: there efficiency/performance differences between:

int64_t my_function(); nodeid_t my_function(); 

in other words, gcc use 1 instruction put nodeid_t on stack, if integer? or create loop, , copy 8 bytes 1 one? depend on -o optimization?

same question assignment.

int64_t a, b; nodeid_t x, y;  = b; /* 1 machine instruction, hope. */ x = y; /* 1 instruction, or loop ? */ 

you cannot union has same size uint64_t.

this due packing in nodeid struct: compilers insert gaps between struct members. compilers allow change packing arrangements code not portable.

it safer have array of uint8_t: memory contiguous.

a compiler copy memory on assignment, may use nodeid_t function return types.

your second job rename nodeid_t: _t suffixes reserved in posix c.