How do I convert a 64bit integer to a char array and back?

Question

I am having trouble converting a int64_t to a char array and back. I don\'t know what is wrong with the code below, it makes complete logical sense to me. The code works for

Answer 1

This is rather simple, the problem is that you are shifting bits in the char array but size of a[i] is 4 byes (upcast to int), so your shift just goes over range. Try replacing this in your code:

int64_t charTo64bitNum(char a[]) {
  int64_t n = 0;
  n = (((int64_t)a[0] << 56) & 0xFF00000000000000U)
    | (((int64_t)a[1] << 48) & 0x00FF000000000000U)
    | (((int64_t)a[2] << 40) & 0x0000FF0000000000U)
    | (((int64_t)a[3] << 32) & 0x000000FF00000000U)
    | ((a[4] << 24) & 0x00000000FF000000U)
    | ((a[5] << 16) & 0x0000000000FF0000U)
    | ((a[6] <<  8) & 0x000000000000FF00U)
    | (a[7]        & 0x00000000000000FFU);
  return n;
}

In this way you'll cast the char to a 64bit number before doing the shift and you won't go over range. You'll obtain correct results:

entity:Dev jack$ ./a.out 
aNum = 123456789
bNum = 51544720029426255

Just a side note, I think this would work fine too, assuming you don't need to peek inside the char array:

#include <string.h>

void int64ToChar(char a[], int64_t n) {
  memcpy(a, &n, 8);
}

int64_t charTo64bitNum(char a[]) {
  int64_t n = 0;
  memcpy(&n, a, 8);
  return n;
}

Answer 2

Are you on a 32bit machine? IIRC I think that the U suffix only means either 'unsigned int' or 'unsigned long', both of which are 32bits on a 32bit machine. I think you want a 'unsigned long long' (64bit) literal in your bit shifts, or 0xFF00000000000000ULL

http://en.kioskea.net/faq/978-c-language-handling-64-bit-integers#unsigned-64-bit-integer

Answer 3

void int64ToChar(char mesg[], int64_t num) {
    *(int64_t *)mesg = num; //or *(int64_t *)mesg = htonl(num);

}

Answer 4

In charTo64bitNum you need to cast the char to 64-bit before you shift it:

(((int64_t)a[0] << 56) & 0xFF00000000000000U)