请问这个结构体在32位系统中占几个字节？

typedef   struct   _A  
  {  
  int   a;  
  char   b;  
  DWORD   c;  
  _A   *   next;  
  }abc;  
   
  看它占几个字节怎么看？  
  问题点数：50、回复次数：8Top

1 楼99263（）回复于 2002-10-12 10:59:09 得分 10

int   len=0;  
  len   =   sizeof(abc);  
   
  另外，你在编译时字节对齐方式不同，这个len的结果也会不同，有1字节对齐的，也有2、4、8字节对齐的。Top

2 楼igogo（gogo）回复于 2002-10-12 11:10:55 得分 0

谢谢99263，1、2、4、8字节对齐是什么意思？Top

3 楼sjsj（虚怀若谷）回复于 2002-10-12 11:45:43 得分 10

使用sizeof可以得到单个变量的存储大小  
   
  字节对齐简单的讲：  
  在内存中有四个字节，为[0],[1],[2],[3]  
  你定义了两个char类型的变量，c1，c2  
  常规下是c1,c2分别对应[0],[1]  
  如果是2字节对齐，那么c1,c2分别对应[0],[2]Top

4 楼everandforever（Forever）回复于 2002-10-12 13:12:16 得分 10

默认的情况：  
   
  typedef   struct   _A  
  {  
  int   a;           //4字节  
  char   b;         //4字节  
  DWORD   c;       //4字节  
  _A   *   next;   //4字节  
  }abc;Top

5 楼qing_li73（Vincent Lee）回复于 2002-10-12 13:18:26 得分 10

Just   test   via   ur   computer.....:)  
   
  Storage   and   Alignment   of   Structures  
  Microsoft   Specific   —>  
   
  Structure   members   are   stored   sequentially   in   the   order   in   which   they   are   declared:   the   first   member   has   the   lowest   memory   address   and   the   last   member   the   highest.    
   
  Every   data   object   has   an   alignment-requirement.   For   structures,   the   requirement   is   the   largest   of   its   members.   Every   object   is   allocated   an   offset   so   that  
   
  offset   %   alignment-requirement   ==   0  
   
  Adjacent   bit   fields   are   packed   into   the   same   1-,   2-,   or   4-byte   allocation   unit   if   the   integral   types   are   the   same   size   and   if   the   next   bit   field   fits   into   the   current   allocation   unit   without   crossing   the   boundary   imposed   by   the   common   alignment   requirements   of   the   bit   fields.    
   
  To   conserve   space   or   to   conform   to   existing   data   structures,   you   may   want   to   store   structures   more   or   less   compactly.   The/Zp[n]   compiler   option   and   the#pragma   pack   control   how   structure   data   is   “packed”   into   memory.   When   you   use   the   /Zp[n]   option,   where   n   is   1,   2,   4,   8,   or   16,   each   structure   member   after   the   first   is   stored   on   byte   boundaries   that   are   either   the   alignment   requirement   of   the   field   or   the   packing   size   (n),   whichever   is   smaller.   Expressed   as   a   formula,   the   byte   boundaries   are   the    
   
  min(   n,   sizeof(   item   )   )  
   
  where   n   is   the   packing   size   expressed   with   the   /Zp[n]   option   and   item   is   the   structure   member.   The   default   packing   size   is   /Zp8.  
   
  To   use   the   pack   pragma   to   specify   packing   other   than   the   packing   specified   on   the   command   line   for   a   particular   structure,   give   the   pack   pragma,   where   the   packing   size   is   1,   2,   4,   8,   or   16,   before   the   structure.   To   reinstate   the   packing   given   on   the   command   line,   specify   the   pack   pragma   with   no   arguments.    
   
  Bit   fields   default   to   size   long   for   the   Microsoft   C   compiler.   Structure   members   are   aligned   on   the   size   of   the   type   or   the   /Zp[n]   size,   whichever   is   smaller.   The   default   size   is   4.  
   
  END   Microsoft   Specific  
   
  Top

6 楼igogo（gogo）回复于 2002-10-12 21:21:47 得分 0

everandforever(Forever）   为什么char型和next指针也是4个字节？Top

7 楼banalman（IT解放者）回复于 2002-10-12 21:59:54 得分 0

gzTop

8 楼everandforever（Forever）回复于 2002-10-12 22:12:52 得分 10

WIN32下指针都是4字节啊？   有问题么？  
  char是4字节是因为数据对齐。Top

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