Question

following excerpted from here

pw = (widget *)malloc(sizeof(widget));

allocates raw storage. Indeed, the malloc call allocates storage
that’s big enough and suitably aligned to hold an object of type
widget

also see fast pImpl from herb sutter, he said:

Alignment. Any memory Alignment. Any memory that’s allocated
dynamically via new or malloc is guaranteed to be properly aligned for
objects of any type, but buffers that are not allocated dynamically
have no such guarantee

I am curious about this, how does malloc know alignment of the custom type?

Answer 1

Alignment requirements are recursive: The alignment of any struct is simply the largest alignment of any of its members, and this is understood recursively.

For example, and assuming that each fundamental type’s alignment equals its size (this is not always true in general), the struct X { int; char; double; } has the alignment of double, and it will be padded to be a multiple of the size of double (e.g. 4 (int), 1 (char), 3 (padding), 8 (double)). The struct Y { int; X; float; } has the alignment of X, which is the largest and equal to the alignment of double, and Y is laid out accordingly: 4 (int), 4 (padding), 16 (X), 4 (float), 4 (padding).

(All numbers are just examples and could differ on your machine.)

Therefore, by breaking it down to the fundamental types, we only need to know a handful of fundamental alignments, and among those there is a well-known largest. C++ even defines a type max_align_t whose alignment is that largest alignment.

All malloc() needs to do is to pick an address that’s a multiple of that value.