Array doubles in size if a struct defines both its uint16_t words and uint8_t bytes

问题

I have an array each of whose elements could be either uint16_t or a pair of uint8_t.

Its elements are defined as a union of a uint16_t and a sub-array of 2 uint8_t.

Unfortunately, the compiler (MicroChip XC16) allocates twice as much memory as it should for the array.

typedef union {
   uint16_t u16;   // As uint16_t
   uint8_t  u8[2]; // As uint8_t
} my_array_t;

my_array_t my_array[1]; // 1 word array, for testing

my_array[0].u8[0] = 1;
my_array[0].u8[1] = 2;
uint8_t byte_0 = my_array[0].u8[0]; // Gets 0x01
uint8_t byte_1 = my_array[0].u8[1]; // Gets 0x02
uint16_t byte_0 = my_array[0].u16; // Gets 0x0201

The compiler allocates 4 bytes instead of 2 bytes as it should.

Workaround: if I change the struct to:

typedef union {
   uint16_t u16;   // As uint16_t
   uint8_t  u8[1];   // As uint8_t
} my_array_t;

The compiler allocates 2 bytes as it should, but then this is incorrect:

my_array[0].u8[1] = 2;

though it still works:

uint8_t byte_1 = my_array[0].u8[1]; // Gets 0x02

(except for the inconvenience that the debugger doesn't show its value).

Question: should I live with the workaround, or should I use a better solution?

Please refer to a previous discussion on this, where the above solution was suggested.

EDIT.

Per EOF's suggestion (below), I checked sizeof.

Before the workaround:

sizeof(my_array_t) // Is 4
sizeof(my_array[0]) // Is 4
sizeof(my_array[0].u8) // Is 2

After the workaround:

sizeof(my_array_t) // Is 2
sizeof(my_array[0]) // Is 2
sizeof(my_array[0].u8) // Is 2

That would indicate that it's a compiler bug.

回答1:

Instead of an array of 2 bytes, use a structure of 2 bytes:

// Two bytes in a 16-bit word
typedef struct{
    uint8_t     lsb;    // As uint8_t, LSB
    uint8_t     msb;    // As uint8_t. MSB
} two_bytes_t;

typedef union {
   uint16_t u16;   // As uint16_t
   two_bytes_t  u8x2; // As 2 each of uint8_t
} my_array_t;


my_array_t my_array[1]; // 1 word array, for testing

my_array[0].u8x2.msb = 1;
my_array[0].u8x2.lsb = 2;

The XC16 compiler correctly allocates only 2 bytes for each element, and the debugger correctly shows the individual bytes.

回答2:

It looks like this issue was fixed in the compiler. I checked it in XC16 1.26 and got these results from my code (optimization 0):

#include "mcc_generated_files/mcc.h"
#include "stddef.h"

typedef union
{
    uint16_t u16;
    uint8_t u8[2];
} example_1_t;

typedef union
{
    uint16_t u16;

    struct
    {
        uint8_t lsb;
        uint8_t msb;
    };
} example_2_t;

int main(void)
{
    SYSTEM_Initialize();

    size_t typeSize1 = sizeof (example_1_t); // debugger shows 2
    size_t typeSize2 = sizeof (example_2_t); // debugger shows 2

    example_1_t ex1; // Can see all values in debugger
    ex1.u16 = 0x4321; // u8[0] = 0x21, u8[1] = 0x43
    example_2_t ex2; // Can see all values in debugger
    ex2.u16 = 0x4321; // lsb = 0x21, msb = 0x43

    size_t objSize1 = sizeof (ex1); // debugger shows 2
    size_t objSize2 = sizeof (ex2); // debugger shows 2

    while (1)
    {
    }

    return -1;
}

来源：https://stackoverflow.com/questions/26999975/array-doubles-in-size-if-a-struct-defines-both-its-uint16-t-words-and-uint8-t-by

标签

arrays

struct

microchip

xc16