Size of structure with a char, a double, an int and a t [duplicate]

Answer 1

It's 24 bytes due to padding. Most compilers pad data to a multiple of its size. So, a 4-byte int is padded to a multiple of 4 bytes. A 8-byte double is padded to a multiple of 8 bytes. For your structure, this means:

struct struct_type{
  int i; // offset 0 (0*4)
  char ch; // offset 4 (4*1)
  char padding1[3];
  int *p; // offset 8 (2*4)
  char padding1[4];
  double d; // offset 16 (2*8)
}s;

You can optimize your struct like that:

struct struct_type{
  double d;
  int i;
  int *p;
  char ch;
}s;

sizeof(s)==17 on most compilers (20 on some others)

Answer 2

$9.2/12 states - "Nonstatic data members of a (non-union) class declared without an intervening access-specifier are allocated so that later members have higher addresses within a class object. The order of allocation of nonstatic data members separated by an access-specifier is unspecified (11.1). Implementation alignment requirements might cause two adjacent members not to be allocated immediately after each other; so might requirements for space for managing virtual functions (10.3) and virtual base classes (10.1)."

So just like the sizeof(double) and sizeof(int), the offsets at which structure members would be aligned is unspecified, except that members that are declared later are at higher addresses.

Answer 3

The additional size comes from data alignment, i.e. the members are aligned to multiples 4 or 8 bytes.

Your compiler probably aligns int and pointers to multiples for 4 bytes and the double to multiples for 8 bytes.

If you move the double to a different position within the struct, you might be able to reduce the size of the struct from 24 to 20 bytes. But it depends on the compiler.