Question

Possible Duplicate:
Size of struct with a single element

Given any type A and the following struct:

struct S
{
    A a;
};

Are there any cases where sizeof(S) is greater than sizeof(A)?

For example, can sizeof(std::array<T, n>) be greater than sizeof(T[n])?

Answer 1

While the struct can be padded, on all systems I know, the compiler will pad so that the alignment of the structure is the same as the largest alignment of its members. It does this so that an array of the structure will always be correctly aligned.

So:

struct S 
{
   char a;
} // Size 1, no padding

struct S2 
{
   unsigned int a;
   char b;
} // Size 8, 3 bytes padding (assuming 32 bit integer)

Edit: Note, that compilers can also add internal padding, to keep the alignment of the data correct.

The C/C++ standard doesn’t specify any of these detail. What you want is the C ABI (application binary interface) for the system you’re running on, which should specify default layout for structs (compilers can choose to override this if they see fit, see also #pragma pack). For an example, look at the X86_64 ABI page 13, which states:

Aggregates and Unions Structures and unions assume the alignment of
their most strictly aligned compo- nent. Each member is assigned to
the lowest available offset with the appropriate alignment. The size
of any object is always a multiple of the object‘s alignment. An array
uses the same alignment as its elements, except that a local or global
array variable of length at least 16 bytes or a C99 variable-length
array variable always has alignment of at least 16 bytes. Structure
and union objects can require padding to meet size and alignment
constraints. The contents of any padding is undefined.