How to set volatile array to zero using memset?

Question

volatile uint8_t reset_mask[768] = {0}

Now I am setting the values of this array elements to 1 during one of internal operations.

In another functional call, I need to set all the elements of this array to 0. One way is by using for loop but I believe better way to assign all the elements of array is by using memset

memset(reset_mask, 0, sizeof(reset_mask));

but I am getting this error :- "cast from type 'volatile uint8_t* {aka volatile unsigned char*}' to type 'void*' casts away qualifiers"

In case we cannot use memset here, is there a better way to set all elements of this volatile array in one go?

Answer 1

You "can" use memset casting away volatileness with const_cast, but it will remove the volatile semantics. So unless that is okay for you, you're stuck with the loop version. Unless your implementation gives you a version of memset that takes volatile* and is documented to do your job. (I'd think unlikely outside embedded systems, would surprise me even there.)

Answer 2

As Balog Pal said, casting away volatile to call memset breaks the volatile semantics. The whole memset may get optimized away, or undefined behaviour badness happens.

Either write your own loop to zero out the array, or use this memset_volatile:

void memset_volatile(volatile void *s, char c, size_t n)
{
    volatile char *p = s;
    while (n-- > 0) {
        *p++ = c;
    }
}

The real memset also returns s and c is an int, but this version is cleaner to me.

It is quite unclear if the compiler is allowed to optimize this function to write more than one byte at a time, so if the array is very large it could be a bit slow.

Edit: Since C11 there is memset_s. It doesn't take a pointer to volatile for some reason, so you have to cast away the volatile. It is however guaranteed to actually overwrite the memory.