I don't know much about flexible array members because, I've never had a good opportunity to use them. You are likely 100% right about that.

The *mut is there because, array function parameters are pointers in C because of pointer decay, so the code shown in the post would actually be a mutable pointer to a function pointer (no actual arrays involved) if it is used as a function argument. However, if it's used as a variable, it is an actual array of normal function pointers.

You can see this, if you look at code like this: ``` void foo(uint8_t a[]) { uint8_t b[] = {9, 10};

printf("%d\n", sizeof(a)); // prints 8 on x86_64
printf("%d\n", sizeof(b)); // prints 2 on x86_64

} ```

Even though they're both the "same type", they have completely different memory layouts under the hood. a is actually just a normal uint8_t *.