CODEWITHSUNDEEP
c++structendiannessmemory-alignment
Petruza
Petruza

Reputation: 12276

Casting a byte array to a struct pointer depends on endianness or memory alignment?

Suppose this code:

unsigned char list[3] = { 1, 2, 3 };
struct _struct{
  unsigned char a;
  unsigned char b;
  unsigned char c;
} *s;
s = ( _struct * ) list;

Can I assume that always s->a == 1, s->b == 2, s->c == 3 ?
Or it will depend on the system's endianness or memory alignment?

Upvotes: 7

Views: 2903

Answers (2)

David Thornley
David Thornley

Reputation: 57036

Let's dissect this.

In all cases, sizeof(char) == 1, and the list array will have its three members at memory locations list, list + 1, and list + 2.

The situation with the struct is not quite as clear. The Standard guarantees that the members will be allocated in increasing memory locations, but not that they will be contiguous. The compiler is free to introduce padding between members, and padding at the end.

Therefore, s->a == 1 will always be true. If the implementation puts the unsigned chars in the struct adjacent (and most will), then the other equalities will necessarily be true.

By the way, calling a struct _struct may cause problems. A name beginning with an underscore in the global namespace is reserved for the implementation.

Upvotes: 7

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272487

Yes, it will depend on the system's and compiler's memory alignment and packing rules.

Upvotes: 4

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