Thread: typecast array of bytes to larger data types

You want something like this.

Code:

#include <stdio.h>
#include <inttypes.h>

int main(void) {
    unsigned char array[10] = { 0 };
    uint16_t i = *(uint16_t*)&array[2];
    printf("%d\n", i);
    return 0;
}

But beware of issues of endianness.

You could create a pointer to a UINT16 and set the pointer to the address of array[2], thusly:

Code:

UINT16 *derp, value;

derp = (UINT16 *) &array[2];
value = *derp;

Originally Posted by codegrush

Hello all, I'm working with a serial protocol where I get an array of bytes. In the array there is are different data types serialized. Is there a way of type casting the array to extract the data directly like perhaps: (UINT16)*(&array[2]) ? I want to treat array[2] and array[3] as a 16bit data type but that code didn't do the job. I worked around it by using memcpy to copy those 2 memory locations into a 16bit data type but I bet one of you geniuses knows what's the proper way to typecast the array location directly.

Yes I know the way of casting it and accesssing directly, but that is the worst way to do it. If you typecast a misaligned address then either your program will crash, or it will take at least twice as long as a normal read, depending on the architecture.

Using memcpy on the other hand wont crash, but will give the wrong result if the data is not in the same endian format as your machine.

The correct way to do it is by reconstructing the value using bitshifts. You can get away with doing it other ways in several cases, but it's not kosher.

As iMalc said, you actually need to do this by shifting a byte at a time into the larger variable in order to avoid alignment problems, and also to adjust the endianness. So something like this:

Code:

#include <inttypes.h> // for uint16_t, etc.

unsigned char a[100];
uint16_t u;
int i = 0;

// TODO: read data into a

u = (uint16_t)a[i++] << 8;
u |= a[i++];

// or for the other endianness
u = a[i++];
u |= (uint16_t)a[i++] << 8;