How does the C offsetof macro work? [duplicate]

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R.. is correct in his answer to the second part of your question: this code is not advised when using a modern C compiler.

But to answer the first part of your question, what this is actually doing is:

(
  (int)(         // 4.
    &( (         // 3.
      (a*)(0)    // 1.
     )->b )      // 2.
  )
)

Working from the inside out, this is …

  1. Casting the value zero to the struct pointer type a*
  2. Getting the struct field b of this (illegally placed) struct object
  3. Getting the address of this b field
  4. Casting the address to an int

Conceptually this is placing a struct object at memory address zero and then finding out at what the address of a particular field is. This could allow you to figure out the offsets in memory of each field in a struct so you could write your own serializers and deserializers to convert structs to and from byte arrays.

Of course if you would actually dereference a zero pointer your program would crash, but actually everything happens in the compiler and no actual zero pointer is dereferenced at runtime.

In most of the original systems that C ran on the size of an int was 32 bits and was the same as a pointer, so this actually worked.

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