Converting a C byte array to a long long

Question

I have an 8-byte array in my application with this data:

00000133339e36a2

This data represents a long (on the platform the data was written in, on a Mac this would be a long long) with the value of

1319420966562

In the actual application this is a semi-randomized set of data, so the number will always be different. Therefore, I need to convert the byte array into a printable long long.

I've tried casting the data directly into a long long, but I came up with

1305392

where I should have been seeing the above number.

For those of you with more experience in C byte manipulation than I do, how would I correctly convert a byte array to a long long?

EDIT: Strangely, all of your solutions keep outputting the same number: 866006690. That is the decimal equivalent of the last four bytes of the data.

Answer 1

This seems one of the (rare) situations where an union is useful:

union number
{
    char charNum[16];
    long long longNum;
};

Answer 2

union le_long_long_array_u
{
  uint8_t byte[8];
  uint64_t longlong;
} le_long_long_array;

Copy (or just use) le_long_long_array.byte[] for the array; read back le_long_long_array.longlong

For example:

fill_array(&le_long_long_array.byte[0]);
return le_long_long_array.longlong;

Fuller example:

#include <stdint.h>
#include <stdio.h>

union le_long_long_array_u
{
  uint8_t byte[8];
  uint64_t longlong;
} le_long_long_array;

static uint8_t hex2int (char c)
{
    return (c >= '0' && c <= '9') 
            ? (uint8_t )(c  - '0')
            : ((c >= 'a' && c <= 'f') 
                ? (uint8_t )(c  - 'a')
                : ((c >= 'A' && c <= 'F') 
                    ? (uint8_t )(c  - 'A')
                    : 0));
}

int main (int argc, char **argv)
{
    if (argc == 2)
    {
        int i;
        for (i = 0; i <= 7; i++)
        {
            char *str = argv[1];

            if (str[2*i] == '\0' || str[2*i+1] == '\0')
            {
                printf("Got short string.\n");
                return 1;
            }

            le_long_long_array.byte[i] = (hex2int(str[2*i]) << 4) + hex2int(str[2*i+1]);
        }
        printf("Got %lld\n", le_long_long_array.longlong);
    }
    else
    {
        printf("Got %d args wanted 1.\n", argc - 1);
        return 1;
    }
    return 0;
}

Produces:

e e$ gcc -c un.c
e e$ gcc -o un un.o
e e$ ./un 0100000000000000
Got 1
e e$ ./un 0101000000000000
Got 257
e e$ ./un a23639333301000000
Got 1319414347266
e e$ 

as you would expect for little endian data.

Answer 3

Depending on your desired endianness, something like this would work:

int i;
char inArray[8] = { 0x00,0x00,0x01,0x33,0x33,0x9e,0x36,0xa2 };
long long outLong = 0;

for(i=0; i<8; i++)
    outLong |= ( (long long)inArray[i]) << (i*8) );

Answer 4

long long num = 0;
for(int i = 0; i < 8; i++) {
    num = (num << 8) | byte_array[i];
}

Assuming they're stored big-endian.

Answer 5

Great question! And great answers! To throw in a cent:

// Getting stuff "in"
unsigned long long ll1 = 1319420966562;
unsigned char * c = reinterpret_cast<unsigned char *>( &ll1 );

// Getting stuff "out"
unsigned long long ll2;
memcpy( &ll2, c, 8 );

ll2 is now equal to ll1.

I must admin, however, I like the union examples better. :)

Answer 6

Perhaps a pointer notation?

uint8_t Array[8] = { 0x00,0x00,0x01,0x33,0x33,0x9e,0x36,0xa2 };
uint32_t *Long = Array;

Or if possible

*((uint32_t*)Array)