从传递给符合MISRA规则17.4的函数的void指针参数访问所有数组元素

Question

Only embedded C.

I need a function to copy unsigned data from a 4-bytes array byte per byte to an output parameter (both passed as reference). Function should be MISRA 17.4 compliant and should support different unsigned integer datatype for output parameter (considering input will always have the exact number of unsigned bytes to fill the output)

So my code is:

static void copy_array(const void * src, void * dest, const uint8_t lenght_bytes)
{
    uint8_t i;
    const uint8_t * src_8 = (const uint8_t*)src;
    uint8_t * dest_8 = (uint8_t*)dest;
    for (i = 0u; i < lenght_bytes; i++)
    {
        *(dest_8 + i) = *(src_8 + i);
    }
}

static void func(void)
{
    uint8_t data[] = {0xEFu, 0xCDu, 0x0u, 0x0u};
    uint16_t dest_16;
    uint32_t dest_32;

    copy_array(data, &dest_16, sizeof(dest_16));

    data[0] = 0xEFu;
    data[1] = 0xCDu;
    data[2] = 0xABu;
    data[3] = 0x89u;

    copy_array(data, &dest_32, sizeof(dest_32));
}

So, MISRA limits pointer arithmetic operations only to array indexing, therefore, my function is not compliant. Any smart way to avoid the rule or to perform same operation but MISRA compliant?

Answer 1

First of all, this is not valid C:

uint8_t data[4] = {0xEFu, 0xCDu, NULL, NULL};

Since NULL might be a null pointer constant of the form (void*)0 . Replace NULL with 0 here.

---

As for the old MISRA-C:2004 requirement about array indexing being the only allowed form, it was mostly nonsense and has been fixed in the current MISRA-C:2012. That being said, there is no need for explicit pointer arithmetic in your code, so that rule makes sense here.

Simply fix the function like this:

static void copy_array(const void* src, void* dest, const uint8_t lenght_bytes)
{
    uint8_t i;
    const uint8_t* src_8 = src;
    uint8_t* dest_8 = dest;

    for (i = 0u; i < lenght_bytes; i++)
    {
        dest_8[i] = src_8[i];
    }
}