Why does 'sizeof' give wrong measurement?

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struct sizeof result not expected

I have this C++ struct:

struct bmp_header {

  //bitmap file header (14 bytes)
  char Sign1,Sign2; //2
  unsigned int File_Size; //4
  unsigned int Reserved_Dword; //4
  unsigned int Data_Offset; //4

  //bitmap info header (16 bytes)
  unsigned int Dib_Info_Size; //4
  unsigned int Image_Width; //4
  unsigned int Image_Height; //4

  unsigned short Planes; //2
  unsigned short Bits; //2  
};

It is supposed to be 30 bytes, but 'sizeof(bmp_header)' gives me value 32. What's wrong?

Struct data alignment: http://en.wikipedia.org/wiki/Data_structure_alignment — wkl, Sep 08 '11 at 16:51
`s/It is supposed to be 30 bytes/The sum total of each member is 30 bytes/` <-- BIG difference. Break down assumptions when breaking down problems. — Lightness Races in Orbit, Sep 08 '11 at 16:51
In the end of the day, though, it doesn't really make much difference because, due to padding, if you have an array of `bmp_header` structs or something like that, padding will be included between the elements of the array anyway. — Daniel, Sep 08 '11 at 18:50
that's why when i read the bmp header directly from the file to the structure, everything cockups coz of the alignment shift :), but fine now, i read to an byte array then convert to the struct — jondinham, Sep 08 '11 at 19:00

score 9 · Answer 1 · answered Sep 08 '11 at 16:50

It doesn't give a wrong measurement. You need to learn about alignment and padding.

The compiler can add padding in between structure members to respect alignment constraints. That said, it's possible to control padding with compiler specific directives (see GCC variable attributes or MSVC++ pragmas).

Daniel · Accepted Answer · 2011-09-08T16:57:33.793

The reason is because of padding. If you put the chars at the end of the struct, sizeof will probably give you 30 bytes. Integers are generally stored on memory addresses that are multiples of 4. Therefore, since the chars take up 2 bytes, there are two unused bytes between it and the first unsigned int. char, unlike int, is not usually padded.

In general, if space is a big concern, always order elements of structs from largest in size to smallest.

Note that padding is NOT always (or usually) the sizeof(element). It is a coincidence that int is aligned on 4 bytes and char is aligned on 1 byte.

score 2 · Answer 3 · answered Sep 08 '11 at 16:54

struct bmp_header {

  char Sign1,Sign2; //2
  // padding for 4 byte alignment of int: // 2
  unsigned int File_Size; //4
  unsigned int Reserved_Dword; //4
  unsigned int Data_Offset; //4

  unsigned int Dib_Info_Size; //4
  unsigned int Image_Width; //4
  unsigned int Image_Height; //4

  unsigned short Planes; //2
  unsigned short Bits; //2  
};

2+2+4+4+4+4+4+4+2+2 = 32. Looks correct to me. If you expect 30 it means you expect 1 byte padding, as in :

#pragma pack(push)
#pragma pack(1)

struct bmp_header {

  char Sign1,Sign2; //2
  unsigned int File_Size; //4
  unsigned int Reserved_Dword; //4
  unsigned int Data_Offset; //4

  unsigned int Dib_Info_Size; //4
  unsigned int Image_Width; //4
  unsigned int Image_Height; //4

  unsigned short Planes; //2
  unsigned short Bits; //2  
};

#pragma pack(pop)

Why does 'sizeof' give wrong measurement?

3 Answers3

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