How do I find the size of a struct

Question

struct a       char  c      char b       What is sizeof a

User · Answer

If you want to manually count it  the size of a struct is just the size of each of its data members after accounting for alignment  There s no magic overhead bytes for a struct

User · Answer

The exact value is sizeof a   You might also take a risk and assume that it is in this case no less than 2 and no greater than 16

User · Answer

I assume you mean struct and not strict  but on a 32-bit system it ll be either 5 or 8 bytes  depending on if the compiler is padding the struct

User · Answer

This will vary depending on your architecture and how it treats basic data types  It will also depend on whether the system requires natural alignment

User · Answer

I suspect you mean  struct   not  strict   and  char  instead of  Char    The size will be implementation dependent   On most 32-bit systems  it will probably be 5 -- 4 bytes for the pointer  one for the char   I don t believe alignment will come into play here   If you swapped  c  and  b   however  the size may grow to 8 bytes   Ok  I tried it out  g   4 2 3  with -g option  and I get 8

User · Answer

include  lt stdio h gt   typedef struct   char  c  char b    a   int main         printf  sizeof a      d   sizeof a        I get  sizeof a     8   on a 32-bit machine  The total size of the structure will depend on the packing  In my case  the default packing is 4  so  c  takes 4 bytes   b  takes one byte  leaving 3 padding bytes to bring it to the next multiple of 4  8  If you want to alter this packing  most compilers have a way to alter it  for example  on MSVC    pragma pack 1  typedef struct   char  c  char b    a    gives sizeof a     5  If you do this  be careful to reset the packing before any library headers

User · Answer

Contrary to what some of the other answers have said  on most systems  in the absence of a pragma or compiler option  the size of the structure will be at least 6 bytes and  on most 32-bit systems  8 bytes   For 64-bit systems  the size could easily be 16 bytes   Alignment does come into play  always   The sizeof a single struct has to be such that an array of those sizes can be allocated and the individual members of the array are sufficiently aligned for the processor in question   Consequently  if the size of the struct was 5 as others have hypothesized  then an array of two such structures would be 10 bytes long  and the char pointer in the second array member would be aligned on an odd byte  which would  on most processors  cause a major bottleneck in the performance

User · Answer

The sizeof the structure should be 8 bytes on a 32 bit system  so that the size of the structure becomes multiple of 2  This makes individual structures available at the correct byte boundaries when an array of structures is declared  This is achieved by padding the structure with 3 bytes at the end   If the structure had the pointer declared after the char  it would still be 8 bytes in size  but the 3 byte padding would have been added to keep the pointer  which is a 4 byte element  aligned at a 4 byte address boundary     The rule of thumb is that elements should be at an offset which is the multiple of their byte size and the structure itself should be of a size which is a multiple of 2

[c] How do I find the size of a struct?

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