linux - How to print hexadecimal double in C? -

i have number in hexadecimal:

ffffffffffff

and need save it, used double

double a=0xffffffffffff; 

but need print , don't know how to. each time use %f, %d, %x, doesn't print value of it; need print ffffffffffff. code:

int main() {     double a=0xffffffffffff;     printf("%x\n",a);     printf("%d\n",a);     printf("%x\n",a);     printf("%f\n",a);     return 0; } 

the true value %f; returns decimal value of hexadecimal — returns this:

 ffffffe0  -32   ffffffe0  281474976710655.000000 

with need change hexadecimal string, compare it, because have ffffffffffff in string , need compare both. if can't printf it, neither sprintf work.

that's integer, , long one. don't use double store such value, that's floating-point.

just use:

unsigned long long temp = 0xffffffffffffull; 

you have 12 hexadecimal digits, number needs @ least 12 * 4 = 48 bits. platforms should have unsigned long long of 64 bits, should fine.

if compiler supported enough support c99, can do:

#include <stdint.h> 

and use uint_least64_t type suggested in comment. in linux guess you're using gcc should fine, might need specify intend code compiled c99 (with -std=c99).

to print it, use:

printf("temp=%llx\n", temp); 

also note value not hexadecimal, can print as hexadecimal. value value, base matters when converting to/from text, i.e. external representation of number. internally on binary computer, number stored in binary.