Structure padding and packing

Question

Consider   struct mystruct A      char a     int b     char c    x   struct mystruct B      int b     char a    y    The sizes of the structures are 12 and 8 respectively   Are these structures padded or packed   When does padding or packing take place

User · Answer

Structure packing suppresses structure padding, padding used when alignment matters most, packing used when space matters most.

Some compilers provide #pragma to suppress padding or to make it packed to n number of bytes. Some provide keywords to do this. Generally pragma which is used for modifying structure padding will be in the below format (depends on compiler):

#pragma pack(n)

For example ARM provides the __packed keyword to suppress structure padding. Go through your compiler manual to learn more about this.

So a packed structure is a structure without padding.

Generally packed structures will be used

to save space
to format a data structure to transmit over network using some protocol (this is not a good practice of course because you need to
deal with endianness)

User · Answer

There are no buts about it  Who want to grasp the subject must do the following ones         Peruse The Lost Art of Structure Packing written by Eric S     Raymond   Glance at Eric s code example   Last but not least  don t forget the following rule about padding that    a struct is aligned to the largest type   s alignment   requirements

User · Answer

The above answers explained the reason quite clearly  but seems not totally clear about the size of padding  so  I will add an answer according to what I learned from The Lost Art of Structure Packing  it has evolved to not limit to C  but also applicable to Go  Rust      Memory align  for struct   Rules    Before each individual member  there will be padding so that to make it start at an address that is divisible by its size  e g on 64 bit system int should start at address divisible by 4  and long by 8  short by 2  char and char   are special  could be any memory address  so they don t need padding before them  For struct  other than the alignment need for each individual member  the size of whole struct itself will be aligned to a size divisible by size of largest individual member  by padding at end  e g if struct s largest member is long then divisible by 8  int then by 4  short then by 2    Order of member    The order of member might affect actual size of struct  so take that in mind  e g the stu c and stu d from example below have the same members  but in different order  and result in different size for the 2 structs      Address in memory  for struct   Rules    64 bit system Struct address starts from  n   16  bytes   You can see in the example below  all printed hex addresses of structs end with 0   Reason  the possible largest individual struct member is 16 bytes  long double    Update  If a struct only contains a char as member  its address could start at any address    Empty space    Empty space between 2 structs could be used by non-struct variables that could fit in  e g in test struct address   below  the variable x resides between adjacent struct g and h  No matter whether x is declared  h s address won t change  x just reused the empty space that g wasted  Similar case for y      Example   for 64 bit system   memory align c          Memory align  amp  padding - for struct     compile  gcc memory align c    execute    a out       include  lt stdio h gt      size is 8  4   1  then round to multiple of 4  int s size   struct stu a       int i      char c         size is 16  8   1  then round to multiple of 8  long s size   struct stu b       long l      char c         size is 24  l need padding by 4 before it  then round to multiple of 8  long s size   struct stu c       int i      long l      char c         size is 16  8   4   1  then round to multiple of 8  long s size   struct stu d       long l      int i      char c         size is 16  8   4   1  then round to multiple of 8  double s size   struct stu e       double d      int i      char c         size is 24  d need align to 8  then round to multiple of 8  double s size   struct stu f       int i      double d      char c         size is 4  struct stu g       int i         size is 8  struct stu h       long l         test - padding within a single struct  int test struct padding         printf   s   ld n    stu a   sizeof struct stu a        printf   s   ld n    stu b   sizeof struct stu b        printf   s   ld n    stu c   sizeof struct stu c        printf   s   ld n    stu d   sizeof struct stu d        printf   s   ld n    stu e   sizeof struct stu e        printf   s   ld n    stu f   sizeof struct stu f         printf   s   ld n    stu g   sizeof struct stu g        printf   s   ld n    stu h   sizeof struct stu h         return 0        test - address of struct  int test struct address         printf   s   ld n    stu g   sizeof struct stu g        printf   s   ld n    stu h   sizeof struct stu h        printf   s   ld n    stu f   sizeof struct stu f         struct stu g g      struct stu h h      struct stu f f1      struct stu f f2      int x   1      long y   1       printf  address of  s   p n    g    amp g       printf  address of  s   p n    h    amp h       printf  address of  s   p n    f1    amp f1       printf  address of  s   p n    f2    amp f2       printf  address of  s   p n    x    amp x       printf  address of  s   p n    y    amp y           g is only 4 bytes itself  but distance to next struct is 16 bytes on 64 bit system  or 8 bytes on 32 bit system       printf  space between  s and  s   ld n    g    h    long   amp h  -  long   amp g            h is only 8 bytes itself  but distance to next struct is 16 bytes on 64 bit system  or 8 bytes on 32 bit system       printf  space between  s and  s   ld n    h    f1    long   amp f1  -  long   amp h            f1 is only 24 bytes itself  but distance to next struct is 32 bytes on 64 bit system  or 24 bytes on 32 bit system       printf  space between  s and  s   ld n    f1    f2    long   amp f2  -  long   amp f1            x is not a struct  and it reuse those empty space between struts  which exists due to padding  e g between g  amp  h      printf  space between  s and  s   ld n    x    f2    long   amp x  -  long   amp f2        printf  space between  s and  s   ld n    g    x    long   amp x  -  long   amp g            y is not a struct  and it reuse those empty space between struts  which exists due to padding  e g between h  amp  f1      printf  space between  s and  s   ld n    x    y    long   amp y  -  long   amp x        printf  space between  s and  s   ld n    h    y    long   amp y  -  long   amp h         return 0     int main int argc  char   argv          test struct padding           test struct address         return 0      Execution result - test struct padding     stu a  8 stu b  16 stu c  24 stu d  16 stu e  16 stu f  24 stu g  4 stu h  8   Execution result - test struct address     stu g  4 stu h  8 stu f  24 address of g  0x7fffd63a95d0     struct variable - address dividable by 16  address of h  0x7fffd63a95e0     struct variable - address dividable by 16  address of f1  0x7fffd63a95f0    struct variable - address dividable by 16  address of f2  0x7fffd63a9610    struct variable - address dividable by 16  address of x  0x7fffd63a95dc     non-struct variable - resides within the empty space between struct variable g  amp  h  address of y  0x7fffd63a95e8     non-struct variable - resides within the empty space between struct variable h  amp  f1  space between g and h  16 space between h and f1  16 space between f1 and f2  32 space between x and f2  -52 space between g and x  12 space between x and y  12 space between h and y  8   Thus address start for each variable is g d0 x dc h e0 y e8

User · Answer

Are these structures padded or packed   They re padded  The only possibility that initially springs to mind  where they could be packed  is if char and int were the same size  so that the minimum size of the char int char structure would allow for no padding  ditto for the int char structure  However  that would require both sizeof int  and sizeof char  to be four  to get the twelve and eight sizes   The whole theory falls apart since it s guaranteed by the standard that sizeof char  is always one  Were char and int the same width  the sizes would be one and one  not four and four  So  in order to then get a size of twelve  there would have to be padding after the final field    When does padding or packing take place   Whenever the compiler implementation wants it to  Compilers are free to insert padding between fields  and following the final field  but not before the first field   This is usually done for performance as some types perform better when they re aligned on specific boundaries  There are even some architectures that will refuse to function  i e  crash  is you try to access unaligned data  yes  I m looking at you  ARM   You can generally control packing padding  which is really opposite ends of the same spectrum  with implementation-specific features such as  pragma pack  Even if you cannot do that in your specific implementation  you can check your code at compile time to ensure it meets your requirement  using standard C features  not implementation-specific stuff   For example     C11 or better      include  lt assert h gt  struct strA   char a  int  b  char c    x  struct strB   int  b  char a            y  static assert sizeof struct strA     sizeof char  2   sizeof int    quot No padding allowed quot    static assert sizeof struct strB     sizeof char      sizeof int    quot No padding allowed quot     Something like this will refuse to compile if there is any padding in those structures

User · Answer

Rules for padding    Every member of the struct should be at an address divisible by its size  Padding is inserted between elements or at the end of the struct to make sure this rule is met  This is done for easier and more efficient Bus access by the hardware  Padding at the end of the struct is decided based on the size of the largest member of the struct    Why Rule 2  Consider the following struct     If we were to create an array of 2 structs  of this struct  No padding will be  required at the end     Therefore  size of struct   8 bytes  Assume we were to create another struct as below     If we were to create an array of this struct  there are 2 possibilities  of the number of bytes of padding required at the end   A  If we add 3 bytes at the end and align it for int and not Long     B  If we add 7 bytes at the end and align it for Long     The start address of the second array is a multiple of 8 i e 24   The size of the struct   24 bytes  Therefore  by aligning the start address of the next array of the struct to a multiple of the largest member i e if we were to create an array of this struct  the first address of the second array must start at an address which is a multiple of the largest member of the struct  Here it is  24 3   8    we can calculate the number of padding bytes required at the end

User · Answer

Data structure alignment is the way data is arranged and accessed in computer memory  It consists of two separate but related issues  data alignment and data structure padding  When a modern computer reads from or writes to a memory address  it will do this in word sized chunks  e g  4 byte chunks on a 32-bit system  or larger  Data alignment means putting the data at a memory address equal to some multiple of the word size  which increases the system   s performance due to the way the CPU handles memory  To align the data  it may be necessary to insert some meaningless bytes between the end of the last data structure and the start of the next  which is data structure padding    In order to align the data in memory   one or more empty bytes  addresses  are inserted  or left empty  between memory addresses which are allocated for other structure members while memory allocation  This concept is called structure padding  Architecture of a computer processor is such a way that it can read 1 word  4 byte in 32 bit processor  from memory at a time  To make use of this advantage of processor  data are always aligned as 4 bytes package which leads to insert empty addresses between other member   s address  Because of this structure padding concept in C  size of the structure is always not same as what we think

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I know this question is old and most answers here explains padding really well  but while trying to understand it myself I figured having a  visual  image of what is happening helped   The processor reads the memory in  chunks  of a definite size  word   Say the processor word is 8 bytes long  It will look at the memory as a big row of 8 bytes building blocks  Every time it needs to get some information from the memory  it will reach one of those blocks and get it     As seem in the image above  doesn t matter where a Char  1 byte long  is  since it will be inside one of those blocks  requiring the CPU to process only 1 word   When we deal with data larger than one byte  like a 4 byte int or a 8 byte double  the way they are aligned in the memory makes a difference on how many words will have to be processed by the CPU  If 4-byte chunks are aligned in a way they always fit the inside of a block  memory address being a multiple of 4  only one word will have to be processed  Otherwise a chunk of 4-bytes could have part of itself on one block and part on another  requiring the processor to process 2 words to read this data   The same applies to a 8-byte double  except now it must be in a memory address multiple of 8 to guarantee it will always be inside a block   This considers a 8-byte word processor  but the concept applies to other sizes of words   The padding works by filling the gaps between those data to make sure they are aligned with those blocks  thus improving the performance while reading the memory   However  as stated on others answers  sometimes the space matters more then performance itself  Maybe you are processing lots of data on a computer that doesn t have much RAM  swap space could be used but it is MUCH slower   You could arrange the variables in the program until the least padding is done  as it was greatly exemplified in some other answers  but if that s not enough you could explicitly disable padding  which is what packing is

User · Answer

Padding and packing are just two aspects of the same thing    packing or alignment is the size to which each member is rounded off padding is the extra space added to match the alignment   In mystruct A  assuming a default alignment of 4  each member is aligned on a multiple of 4 bytes  Since the size of char is 1  the padding for a and c is 4 - 1   3 bytes while no padding is required for int b which is already 4 bytes  It works the same way for mystruct B

User · Answer

Structure packing is only done when you tell your compiler explicitly to pack the structure  Padding is what you re seeing  Your 32-bit system is padding each field to word alignment  If you had told your compiler to pack the structures  they d be 6 and 5 bytes  respectively  Don t do that though  It s not portable and makes compilers generate much slower  and sometimes even buggy  code

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Padding aligns structure members to  natural  address boundaries - say  int members would have offsets  which are mod 4     0 on 32-bit platform  Padding is on by default  It inserts the following  gaps  into your first structure   struct mystruct A       char a      char gap 0 3      inserted by compiler  for alignment of b        int b      char c      char gap 1 3      - -  for alignment of the whole struct in an array      x    Packing  on the other hand prevents compiler from doing padding - this has to be explicitly requested - under GCC it s   attribute      packed      so the following   struct   attribute      packed     mystruct A       char a      int b      char c       would produce structure of size 6 on a 32-bit architecture   A note though - unaligned memory access is slower on architectures that allow it  like x86 and amd64   and is explicitly prohibited on strict alignment architectures like SPARC

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