简体   繁体   中英

C struct padding

I can use pragma pack on various compilers to force structs to have fields which are not on their natural alignments.

Is this recursive - so suppose struct typedef A contains a field of typedef struct B. If use a pragma to pack A will this force struct B to be packed?

Just don't. The exact behavior of an ugly nonstandard extension is not something you can even ask questions about without specifying an exact platform/compiler you're working with. If you find yourself needing packed structs, you'd be better to figure out why your code is broken and fix it. Packed structs are a band-aid for broken code that do nothing to fix the root cause of the brokenness.

You have to hope not! Lets say you have a function that takes a parameter struct A:

void fn( struct A x ) ;

and then a packed struct B that has struct A as a member:

#pragma pack(1)
struct B
{
    struct A a ; 
}

If you pass the member a of B to fn(), the function has no knowledge of the packing in this instance, and would fail.

Other answerer's results notwithstanding, I performed the following test on VC++ 2010:

struct A
{
    int a;
    char b;
    int c ;
} ;

struct B
{
    struct A  d ;
} ;

#pragma pack(1)
struct C
{
    struct A  d ;
} ;


#pragma pack(1)
struct D
{
    int a;
    char b;
    int c ;
} ;

#pragma pack(1)
struct E
{
    struct D ;
} ;

int main()
{
    int a = sizeof(struct A) ;
    int b = sizeof(struct B) ;
    int c = sizeof(struct C) ;
    int d = sizeof(struct D) ;
    int e = sizeof(struct E) ;
}

Inspection of a, b, c, d and e in main() in the debugger yield:

  • a = 12
  • b = 12
  • c = 12
  • d = 9
  • e = 9

The packing of struct C has no effect on the size of its struct A member.

Folowing Nathon's post, I tried about the same thing on my computer:

#include <stdio.h>

#if defined PACK_FIRST || defined PACK_BOTH
#pragma pack(1)
#endif
struct inner {
  char a;
  double b;
};

#if defined PACK_SECOND || defined PACK_BOTH
#pragma pack(1)
#endif
struct outer {
  char a;
  struct inner b;
  char c;
  double d;
};

int main(void) {
  printf("sizeof (char): %d (1, of course)\n", (int)sizeof (char));
  printf("sizeof (double): %d\n", (int)sizeof (double));
  printf("sizeof (inner): %d\n", (int)sizeof (struct inner));
  printf("sizeof (outer): %d\n", (int)sizeof (struct outer));
  return 0;
}
$ gcc 4128061.c
$ ./a.out 
sizeof (char): 1 (1, of course)
sizeof (double): 8
sizeof (inner): 16
sizeof (outer): 40
$ gcc -DPACK_FIRST 4128061.c
$ ./a.out 
sizeof (char): 1 (1, of course)
sizeof (double): 8
sizeof (inner): 9
sizeof (outer): 19
$ gcc -DPACK_SECOND 4128061.c
$ ./a.out 
sizeof (char): 1 (1, of course)
sizeof (double): 8
sizeof (inner): 16
sizeof (outer): 26
$ gcc -DPACK_BOTH 4128061.c
$ ./a.out 
sizeof (char): 1 (1, of course)
sizeof (double): 8
sizeof (inner): 9
sizeof (outer): 19

Apparently my gcc packs from the point where the #pragma line appears.

For the version of GCC I have handy, it looks like yes:

#include <stdio.h>
typedef struct
{
    char a;
    int  b;
}inner_t;
#pragma pack(1)
typedef struct
{
    char a;
    inner_t inner;
} outer_t;

int main()
{
    outer_t outer;

    outer.inner.a = 'a';
    outer.inner.b = 0xABCDEF01;
    printf ("outer.inner.a: %c\n", outer.inner.a);
    return 0;
}

And gdb breaking on the printf gives me...

(gdb) x/5xw &outer
0x7fffffffe4b0: 0xffff61a0      0xcdef01ff      0x000000ab      0x00000000
0x7fffffffe4c0: 0x00000000

inner.b is not word aligned. So under GCC 4.4.5 on a little endian 64-bit machine, nested structures are packed if the outer structure is packed. Pardon my typedefs, those of you who don't like them.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM