math - high precision division in C -

i want following calculation without losing precision in c.

uint64_t ts_1 = 0x5212cb03ca115ac0;  uint64_t ts_2 = 0x5212cb03ca115cc0;  uint64_t ts_delta = (ts_2 - ts_1)  double scale_factor = ts_delta/(2^32) 

i getting value of ts_delta 0x200. value of scale_factor 15.000000.basically losing precision during calculation.

how do without losing precision.?

here short self contained example on how trying print.

#include <stdio.h> #include <stdint.h> #include <inttypes.h>  int main() {         uint64_t ts_1 = 0x5212cb03ca115ac0;         uint64_t ts_2 = 0x5212cb03ca115cc0;         uint64_t ts_delta = (ts_2 - ts_1);         double scale_factor = ((double)ts_delta) / (((uint64_t)1) << 32);          printf("ts_delta %"prix64" scale factor %f \n",ts_delta,scale_factor);          return 0; } 

you're not doing calculation think you're doing. ^ operator in c not perform exponentiation, performs bitwise exclusive or. dividing ts_delta 2 xor 32 == 34. i'm not sure how got 15.000000 that, should have come out 15.058823529411764.

the calculation wanted expressed in c this:

double scale_factor = ((double)ts_delta) / (((uint64_t)1) << 32); 

edit: other answer: if compiler supports c99's hexadecimal floating-point literals, can write

double scale_factor = ts_delta * 0x1p-32; 

written way, no casts necessary, because 1 side of multiplication double, integer on other side converted match. unfortunately, several compilers have uint64_t no other features of c99, e.g. recent msvc.