
struct pppoe_tag
{2
__u16 tag_type;3
__u16 tag_len;4
char tag_data[0];5
} __attribute ((packed));
struct pppoe_tag *sample_tag;2
__u16 sample_tag_len = 10;3
sample_tag = (struct pppoe_tag *)malloc(sizeof(struct pppoe_tag)+sizeof(char)*sample_tag_len);4
sample_tag->tag_type = 0xffff;5
sample_tag->tag_len = sample_tag_len;6
sample_tag->tag_data[0]=
.7


free(sample_tag)例1:test_size.c
struct tag12


{3
int a;4
int b;5
}__attribute ((packed));6

7
struct tag28


{9
int a;10
int b;11
char *c;12
}__attribute ((packed));13

14
struct tag315


{16
int a;17
int b;18
char c[0];19
}__attribute ((packed));20

21
struct tag422


{23
int a;24
int b;25
char c[1];26
}__attribute ((packed));27

28
int main()29


{30
struct tag2 l_tag2;31
struct tag3 l_tag3;32
struct tag4 l_tag4;33
34
memset(&l_tag2,0,sizeof(struct tag2));35
memset(&l_tag3,0,sizeof(struct tag3));36
memset(&l_tag4,0,sizeof(struct tag4));37

38
printf("size of tag1 = %d\n",sizeof(struct tag1));39
printf("size of tag2 = %d\n",sizeof(struct tag2));40
printf("size of tag3 = %d\n",sizeof(struct tag3));41
42
printf("l_tag2 = %p,&l_tag2.c = %p,l_tag2.c = %p\n",&l_tag2,&l_tag2.c,l_tag2.c);43
printf("l_tag3 = %p,l_tag3.c = %p\n",&l_tag3,l_tag3.c);44
printf("l_tag4 = %p,l_tag4.c = %p\n",&l_tag4,l_tag4.c);45
exit(0);46
}程序的运行结果如下:
size of tag1 = 8
size of tag2 = 12
size of tag3 = 8
size of tag4 = 9
l_tag2 = 0xbffffad0,&l_tag2.c = 0xbffffad8,l_tag2.c = (nil)
l_tag3 = 0xbffffac8,l_tag3.c = 0xbffffad0
l_tag4 = 0xbffffabc,l_tag4.c = 0xbffffac4
struct pppoe_tag
{2
__u16 tag_type;3
__u16 tag_len;4
char *tag_data;5
} __attribute ((packed));6

7
struct pppoe_tag *sample_tag;8
__u16 sample_tag_len = 10;
sample_tag = (struct pppoe_tag *)malloc(sizeof(struct pppoe_tag));2
sample_tag->tag_len = sample_tag_len;3
sample_tag->tag_data = malloc(sizeof(char)*sample_tag_len);4
sample_tag->tag_data[0]=
free(sample_tag->tag_data);2
free(sample_tag);
sample_tag = (struct pppoe_tag *)malloc(sizeof(struct pppoe_tag)+sizeof(char)*sample_tag_len);2
sample_tag->tag_len = sample_tag_len;3
sample_tag->tag_data = ((char *)sample_tag)+sizeof(struct pppoe_tag);4
sample_tag->tag_data[0]=
free(sample_tag);所以无论使用那种方法,都没有char tag_data[0]这样的定义来得方便。
讲了这么多,其实本质上涉及到的是一个C语言里面的数组和指针的区别问题。char a[1]里面的a和char *b的b相同吗?《Programming Abstractions in C》(Roberts, E. S.,机械工业出版社,2004.6)82页里面说:“arr is defined to be identical to &arr[0]”。也就是说,char a[1]里面的a实际是一个常量,等于&a[0]。而char *b是有一个实实在在的指针变量b存在。所以,a=b是不允许的,而b=a是允许的。两种变量都支持下标式的访问,那么对于a[0]和b[0]本质上是否有区别?我们可以通过一个例子来说明。
例二:
10 #include <stdio.h>
20 #include <stdlib.h>
30
40 int main()
50
{
60 char a[10];
70 char *b;
80
90 a[2]=0xfe;
100 b[2]=0xfe;
110 exit(0);
120 }
080483f0 <main>:
80483f0: 55 push %ebp
80483f1: 89 e5 mov %esp,%ebp
80483f3: 83 ec 18 sub $0x18,%esp
80483f6: c6 45 f6 fe movb $0xfe,0xfffffff6(%ebp)
80483fa: 8b 45 f0 mov 0xfffffff0(%ebp),%eax
80483fd: 83 c0 02 add $0x2,%eax
8048400: c6 00 fe movb $0xfe,(%eax)
8048403: 83 c4 f4 add $0xfffffff4,%esp
8048406: 6a 00 push $0x0
8048408: e8 f3 fe ff ff call 8048300 <_init+0x68>
804840d: 83 c4 10 add $0x10,%esp
8048410: c9 leave
8048411: c3 ret
8048412: 8d b4 26 00 00 00 00 lea 0x0(%esi,1),%esi
8048419: 8d bc 27 00 00 00 00 lea 0x0(%edi,1),%edi
int do1(char a[],int len);
int do2(char *a,int len);这两个函数中的a并无任何区别。都是实实在在存在的指针变量。
顺便再说一下,对于struct pppoe_tag的最后一个成员的定义是char tag_data[0],某些编译器不支持长度为0的数组的定义,在这种情况下,只能将它定义成char tag_data[1],使用方法相同
本文探讨了C语言中结构体使用变长数组的技巧,对比了与指针的不同,并通过实例展示了如何在Linux环境下高效利用变长数组。
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