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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 ...
- Casting the value zero to the struct pointer type
a* - Getting the struct field b of this (illegally placed) struct object
- Getting the address of this
bfield - 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.
It has no advantages and should not be used, since it invokes undefined behavior (and uses the wrong type - int instead of size_t).
The C standard defines an offsetof macro in stddef.h which actually works, for cases where you need the offset of an element in a structure, such as:
#include <stddef.h>
struct foo {
int a;
int b;
char *c;
};
struct struct_desc {
const char *name;
int type;
size_t off;
};
static const struct struct_desc foo_desc[] = {
{ "a", INT, offsetof(struct foo, a) },
{ "b", INT, offsetof(struct foo, b) },
{ "c", CHARPTR, offsetof(struct foo, c) },
};
which would let you programmatically fill the fields of a struct foo by name, e.g. when reading a JSON file.