标题：各位对C语言结构中成员的对齐(alignment)怎样理解？

本来觉得自已很清楚了，后来看到一个例子，又糊涂了。先看一个WATCOM给出的表：(ZP1表示按一个字符对齐) zp1 zp2 zp4 zp8 zp16 sizeof(member) \--------------------------------------- 1 | 0 0 0 0 0 2 | 0 2 2 2 2 4 | 0 2 4 4 4 8 | 0 2 4 8 8 16 | 0 2 4 8 16 x | aligned to largest member 看一段代码： #include #include typedef struct memo_el { char date[9]; struct memo_el *prev,*next; int ref_number; char sex; } memo; main( ) { printf( "Offset of %s is %d\n", "date", offsetof( memo, date ) ); printf( "Offset of %s is %d\n", "prev", offsetof( memo, prev ) ); printf( "Offset of %s is %d\n", "next", offsetof( memo, next ) ); printf( "Offset of %s is %d\n", "ref_number", offsetof( memo, ref_number ) ); printf( "Offset of %s is %d\n", "sex", offsetof( memo, sex ) ); printf( "Size of %s is %d\n", "memo", sizeof( memo ) ); printf( "Number of padding bytes is %d\n", sizeof( memo ) - (offsetof( memo, sex ) + sizeof( char )) ); return 0; } 输出： 16位zp1:（毫无疑问） Offset of date is 0 Offset of prev is 9 Offset of next is 11 Offset of ref_number is 13 Offset of sex is 15 Size of memo is 16 Number of padding bytes is 0 16位 zp2、4、8结果都一样: Offset of date is 0 Offset of prev is 10 Offset of next is 12 Offset of ref_number is 14 Offset of sex is 16 Size of memo is 18 Number of padding bytes is 1 如果再把成员改一个长整类型， typedef struct memo_el { char date[9]; struct memo_el *prev,*next; long int ref_number; char sex; } memo; 结果在16位ZP2-8都是这样： Offset of date is 0 Offset of prev is 10 Offset of next is 12 Offset of ref_number is 16 Offset of sex is 20 Size of memo is 24 Number of padding bytes is 3 现在我无法解释了~