> struct foo *fee;
> It's possible that:
> sizeof(struct foo) != (((char *)&fee) - ((char *)&fee))
> because of end-padding, which is not accounted for in arrays,
Er, no, that's not right. Otherwise
fee = malloc(n * sizeof(struct foo))
There may also be unnamed padding at the end of a structure or union,
as necessary to achieve the proper alignment were the structure or
union to be an element of an array.
the result [of sizeof] is the total number of bytes in such an object,
including internal and trailing padding.
So if a struct needs padding in an array, it has it even when it isn't
in an array.
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