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Why does this implementation of offsetof() work?

In ANSI C, offsetof is defined as below.

#define offsetof(st, m) \     ((size_t) ( (char *)&((st *)(0))->m - (char *)0 )) 

Why won't this throw a segmentation fault since we are dereferencing a NULL pointer? Or is this some sort of compiler hack where it sees that only address of the offset is taken out, so it statically calculates the address without actually dereferencing it? Also is this code portable?

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chappar Avatar asked Apr 03 '09 13:04

chappar


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How offsetof works?

It evaluates to the offset (in bytes) of a given member within a struct or union type, an expression of type size_t. The offsetof() macro takes two parameters, the first being a structure name, and the second being the name of a member within the structure. It cannot be described as a C prototype.

What is offsetof function in C?

Description. The C library macro offsetof(type, member-designator) results in a constant integer of type size_t which is the offset in bytes of a structure member from the beginning of the structure. The member is given by member-designator, and the name of the structure is given in type.

How do you find the offset of a structure member?

Use offsetof() to find the offset from the start of z or from the start of x . #include <stddef. h> size_t offsetof(type, member); offsetof() returns the offset of the field member from the start of the structure type.


2 Answers

At no point in the above code is anything dereferenced. A dereference occurs when the * or -> is used on an address value to find referenced value. The only use of * above is in a type declaration for the purpose of casting.

The -> operator is used above but it's not used to access the value. Instead it's used to grab the address of the value. Here is a non-macro code sample that should make it a bit clearer

SomeType *pSomeType = GetTheValue(); int* pMember = &(pSomeType->SomeIntMember); 

The second line does not actually cause a dereference (implementation dependent). It simply returns the address of SomeIntMember within the pSomeType value.

What you see is a lot of casting between arbitrary types and char pointers. The reason for char is that it's one of the only type (perhaps the only) type in the C89 standard which has an explicit size. The size is 1. By ensuring the size is one, the above code can do the evil magic of calculating the true offset of the value.

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JaredPar Avatar answered Sep 27 '22 17:09

JaredPar


Although that is a typical implementation of offsetof, it is not mandated by the standard, which just says:

The following types and macros are defined in the standard header <stddef.h> [...]

offsetof(type,member-designator)

which expands to an integer constant expression that has type size_t, the value of which is the offset in bytes, to the structure member (designated by member-designator), from the beginning of its structure (designated by type). The type and member designator shall be such that given

statictypet;

then the expression &(t.member-designator) evaluates to an address constant. (If the specified member is a bit-field, the behavior is undefined.)

Read P J Plauger's "The Standard C Library" for a discussion of it and the other items in <stddef.h> which are all border-line features that could (should?) be in the language proper, and which might require special compiler support.

It's of historic interest only, but I used an early ANSI C compiler on 386/IX (see, I told you of historic interest, circa 1990) that crashed on that version of offsetof but worked when I revised it to:

#define offsetof(st, m) ((size_t)((char *)&((st *)(1024))->m - (char *)1024)) 

That was a compiler bug of sorts, not least because the header was distributed with the compiler and didn't work.

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Jonathan Leffler Avatar answered Sep 27 '22 15:09

Jonathan Leffler