Misalignment of members in structures [duplicate] - c

Question

This question already has answers here:
Practical Use of Zero-Length Bitfields
(5 answers)
Closed 8 years ago.
In C, sometimes certain members of a structure tend to have misaligned offsets, as in case of this thread in HPUX community
In such a case, one is suggested to use zero-width bit field to align the(misaligned) next member.
Under what circumstance does misalignment of structure members happen? Is it not the job of the compiler to align offsets of members at word boundary?

Answer 1

"Misalignment" of a structure member can only occur if the alignment requirements of the structure member are deliberately hidden. (Or if some implementation-specific mechanism is used to suppress alignment, such as gcc's packed attribute`.)
For example, in the referenced problem, the issue is that there is a struct:
struct {
// ... stuff
int val;
unsigned char data[DATA_SIZE];
// ... more stuff
}
and the programmer attempts to use data as though it were a size_t:
*(size_t*)s->data
However, the programmer has declared data as unsigned char and the compiler therefore only guarantees that it is aligned for use as an unsigned char.
As it happens, data follows an int and is therefore also aligned for an int. On some architectures this would work, but on the target architecture a size_t is bigger than an int and requires a stricter alignment.
Obviously the compiler cannot know that you intend to use a structure member as though it were some other type. If you do that and compile for an architecture which requires proper alignment, you are likely to experience problems.
The referenced thread suggests inserting a zero-length size_t bit-field before the declaration of the unsigned char array in order to force the array to be aligned for size_t. While that solution may work on the target architecture, it is not portable and should not be used in portable code. There is no guarantee that a 0-length bit-field will occupy 0 bits, nor is there any guarantee that a bit-field based on size_t will actually be stored in a size_t or be appropriately aligned for any non bit-field use.
A better solution would be to use an anonymous union:
// ...
int val;
union {
size_t dummy;
unsigned char data[DATA_SIZE];
};
// ...
With C11, you can specify a minimum alignment explicitly:
// ...
int val;
_Alignas(size_t) unsigned char data[DATA_SIZE];
// ...
In this case, if you #include <stdalign.h>, you can spell _Alignas in a way which will also work with C++11:
int val;
alignas(size_t) unsigned char data[DATA_SIZE];

Answer 2

Q: Why does it misalignment happen? Is it not the job of the compiler to align offsets of members at word boundary?
You are probably aware that the reason that structure fields are aligned to specific boundaries is to improve performance. A properly aligned field may only require a single memory fetch operation by the CPU; where a mis-aligned field will require at least two memory fetch operations (twice the CPU time).
As you indicated, it is the compilers job to align structure fields for fastest CPU access; unless a programmer over-rides the compiler's default behavior.
Then the question might be; Why would the programmer over-ride the compiler's default alignment of structure fields?
One example of why a programmer would want to over-ride the default alignment is when sending a structure 'over the wire' to another computer. Generally, a programmer wants to pack as much data as possible, into fewest number of bytes.
Hence, the programmer will disable the default alignment when structure density is more important than CPU performance accessing structure fields.