c - C/GCC 4.6.3 中的 union 填充值

Question

我的代码初始化了 6 个结构、2 个 union ，并使用“转储”函数显示为它们分配的字节地址，以及这些字节中的值。

短代码:

#include <stdio.h>
// #DEFINE PD = padding

void dump (void *p, int n);

/* (...) */

union U1
{
    int i;
    char c[5];
};

/*  has 8 bytes of information, organized as follows:
    | i/c[0] | i/c[1] | i/c[2] | i/c[3] | PD/c[4] | PD | PD | PD |                                      */

union U2
{
    short s;
    char c[5];
};

/*  has 6 bytes of information, organized as follows:
    | s/c[0] | s/c[1] | PD/c[2] | PD/c[3] | PD/c[4] | PD |                                          */



int main (void)
{
    int i;

    union U1 u1;
    union U2 u2;

    /* (...) */

         u1.i = 0x01020304;      // initializes int
    printf("\nu1 (int)\n");
        dump(&u1, sizeof(u1));
        for (i=0;i<5;i++)        // initializes char
            u1.c[i] = 0xcc;
    printf("u1 (char)\n");
        dump(&u1, sizeof(u1));

        u2.s = 0x0102;           // initializes short
    printf("\nu2 (short)\n");
        dump(&u2, sizeof(u2));   
        for (i=0;i<5;i++)        // initializes char
            u2.c[i] = 0xcc;
    printf("u2 (char)\n");
        dump(&u2, sizeof(u2));

    return 0;
} 

从所有结构和第一个 union 中，我得到了预期的字节数和值 - 所有填充字节都是 00 - 但从最后一个 union 中，我得到了这个:

u2 (short)
0x7fff825a05a0 - 02
0x7fff825a05a1 - 01
0x7fff825a05a2 - 5A
0x7fff825a05a3 - 82
0x7fff825a05a4 - FF
0x7fff825a05a5 - 7F
u2 (char)
0x7fff825a05a0 - CC
0x7fff825a05a1 - CC
0x7fff825a05a2 - CC
0x7fff825a05a3 - CC
0x7fff825a05a4 - CC
0x7fff825a05a5 - 7F

short 的中间 2 个字节的值是随机的 - 可能是随机内存，每次运行都会改变 - short 的最后 2 个和 char 数组的最后一个是固定的。

为什么我得到这个值？不应该所有的填充字节都是 0x00 吗？即使第一个 union 进展顺利，从 int 到 short 的变化也会改变什么？为什么用 short 变量初始化 union 上的部分内存垃圾和部分固定值？

我想这是一个理论上的答案，如果是这样，你能引用引用文献吗？那太好了。

提前打扰一下可能会发这么大的帖子，这是我的第一个帖子。 :)

完整代码:

#include <stdio.h>
// #DEFINE PD = padding

void dump (void *p, int n)
{
     unsigned char *p1 = p;

     while (n--)
     {
          printf("%p - %02X\n",p1, *p1);
          p1++;
     }
}

struct X1
{
    char c1;
    int i;
    char c2;
} x1 = {0xc1, 0x01020304, 0xc2};

/*  possui 12 bytes de informação, organizados em:
    | c1 | PD | PD | PD | i | i | i | i | c2 | PD | PD | PD |                                       */

struct X2
{
    int i;
    char c;
} x2 = {0x01020304, 0xc2};

/*  possui 8 bytes de informação, organizados em:
    | i | i | i | i | c | PD | PD | PD |                                                            */

struct X3
{
    int i;
    char c1;
    char c2;
} x3 = {0x01020304, 0xc1, 0xc2};

/*  possui 8 bytes de informação, organizados em:
    | i | i | i | i | c1 | c2 | PD | PD |                                                           */

struct X4
{
    struct X2 x;
    char c;
} x4 = {{0x01020304, 0xc1}, 0xc2};

/*  possui 8 bytes de informação, organizados em:
    | X2.i | X2.i | X2.i | X2.i | X2.c | PD | PD | PD | c | PD | PD | PD |                          */

struct X5
{
    char c1;
    char c2;
    char c3;
} x5 = {0xc1, 0xc2, 0xc3};

/*  possui 3 bytes de informação, organizados em:
    | c1 | c2 | c3 |                                                                            */

struct X6
{
    short s1;
    int i;
    char c[3];
    short s2;
} x6 = {0x0102, 0x01020304, {0xc1, 0xc2, 0xc3}, 0x0102};

/*  possui 16 bytes de informação, organizados em:
    | s1 | s1 | PD | PD | i | i | i | i | c[0] | c[1] | c[2] | PD | s2 | s2| PD | PD |              */


union U1
{
    int i;
    char c[5];
};

/*  possui 8 bytes de informação, organizados em:
    | i/c[0] | i/c[1] | i/c[2] | i/c[3] | PD/c[4] | PD | PD | PD |                                      */

union U2
{
    short s;
    char c[5];
};

/*  possui 8 bytes de informação, organizados em:
    | s/c[0] | s/c[1] | PD/c[2] | PD/c[3] | PD/c[4] | PD | PD | PD |                                            */



int main (void)
{
    int i;

    union U1 u1;
    union U2 u2;

    printf("\nx1: \n");
        dump(&x1, sizeof(x1));
    printf("\nx2: \n");
        dump(&x2, sizeof(x2));
    printf("\nx3: \n");
        dump(&x3, sizeof(x3));
    printf("\nx4: \n");
        dump(&x4, sizeof(x4));
    printf("\nx5: \n");
        dump(&x5, sizeof(x5));
    printf("\nx6: \n");
        dump(&x6, sizeof(x6));

         u1.i = 0x01020304;
    printf("\nu1 (int)\n");
        dump(&u1, sizeof(u1));
        for (i=0;i<5;i++)
            u1.c[i] = 0xcc;
    printf("u1 (char)\n");
        dump(&u1, sizeof(u1));

        u2.s = 0x0102;
    printf("\nu2 (short)\n");
        dump(&u2, sizeof(u2));
        for (i=0;i<5;i++)
            u2.c[i] = 0xcc;
    printf("u2 (char)\n");
        dump(&u2, sizeof(u2));

    return 0;
    }

完整输出:

x1: 
0x601030 - C1
0x601031 - 00
0x601032 - 00
0x601033 - 00
0x601034 - 04
0x601035 - 03
0x601036 - 02
0x601037 - 01
0x601038 - C2
0x601039 - 00
0x60103a - 00
0x60103b - 00

x2: 
0x60103c - 04
0x60103d - 03
0x60103e - 02
0x60103f - 01
0x601040 - C2
0x601041 - 00
0x601042 - 00
0x601043 - 00

x3: 
0x601044 - 04
0x601045 - 03
0x601046 - 02
0x601047 - 01
0x601048 - C1
0x601049 - C2
0x60104a - 00
0x60104b - 00

x4: 
0x60104c - 04
0x60104d - 03
0x60104e - 02
0x60104f - 01
0x601050 - C1
0x601051 - 00
0x601052 - 00
0x601053 - 00
0x601054 - C2
0x601055 - 00
0x601056 - 00
0x601057 - 00

x5: 
0x601058 - C1
0x601059 - C2
0x60105a - C3

x6: 
0x601060 - 02
0x601061 - 01
0x601062 - 00
0x601063 - 00
0x601064 - 04
0x601065 - 03
0x601066 - 02
0x601067 - 01
0x601068 - C1
0x601069 - C2
0x60106a - C3
0x60106b - 00
0x60106c - 02
0x60106d - 01
0x60106e - 00
0x60106f - 00

u1 (int)
0x7fff825a0590 - 04
0x7fff825a0591 - 03
0x7fff825a0592 - 02
0x7fff825a0593 - 01
0x7fff825a0594 - 00
0x7fff825a0595 - 00
0x7fff825a0596 - 00
0x7fff825a0597 - 00
u1 (char)
0x7fff825a0590 - CC
0x7fff825a0591 - CC
0x7fff825a0592 - CC
0x7fff825a0593 - CC
0x7fff825a0594 - CC
0x7fff825a0595 - 00
0x7fff825a0596 - 00
0x7fff825a0597 - 00

u2 (short)
0x7fff825a05a0 - 02
0x7fff825a05a1 - 01
0x7fff825a05a2 - 5A
0x7fff825a05a3 - 82
0x7fff825a05a4 - FF
0x7fff825a05a5 - 7F
u2 (char)
0x7fff825a05a0 - CC
0x7fff825a05a1 - CC
0x7fff825a05a2 - CC
0x7fff825a05a3 - CC
0x7fff825a05a4 - CC
0x7fff825a05a5 - 7F

Answer 1

来自 C99 标准，6.2.6.1p7:

When a value is stored in an object of structure or union type, including in a member object, the bytes of the object representation that correspond to any padding bytes take unspecified values.