Question

My data has structure like the following:

typedef struct struct2 {
  uint8_t  num;
  uint8_t  action;
  uint8_t  id;
} struct2_t;

typedef struct struct1 {
  uint8_t  title;
  struct2_t* content;
} struct1_t;

I assigned a memory space to a structure pointer and printed out the value.

void main(){

    const uint8_t msg[5] = {
            5, 4, 3, 2, 1
    };

    struct1_t *req = NULL;
    req = (struct1_t *)msg;

    printf("%d ", req->title);
    printf("%d ", req->content->num);
    printf("%d ", req->content->action);
    printf("%d ", req->content->id);

}

The result is: 5 5 4 3 while I expected it to be 5 4 3 2?

Besides, in another place I copy the content in my buffer and do the similar thing.

void main(){

    const uint8_t msg[5] = {
            5, 4, 3, 2, 1
    };

    uint8_t *test = NULL;
    test = (uint8_t*)malloc(5 + 1);
    memset((test), 0x00, 5);
    memcpy(test, msg , 5);

    struct1_t *req = NULL;
    req = (struct1_t *)test;

    printf("%d ", req->title);
    printf("%d ", req->content->num);
    printf("%d ", req->content->action);
    printf("%d \n", req->content->id);
}

But this time I can not access the struct2_t pointer

5 Segmentation fault (core dumped)

Can anyone tell me what I'm missing? Thanks for you help

Answer 1

您似乎缺少的是明确了解何时将dot（.）运算符与结构一起使用以及何时使用arrow（->）运营商。

规则很简单。如果您有struct，则可以使用.运算符访问其成员。如果您有pointer to a struct，则可以使用->箭头运算符访问其成员。当您的struct包含pointer to a struct时，请使用其中一个。

以下示例应该明确说明并提供使用content指针指向具有自动存储s2的结构和存储在动态分配的内存中的结构的示例＆＃39; s3＆＃39;

#include <stdio.h>
#include <stdlib.h>
#include <stdint.h>
#include <inttypes.h>

enum { MAXL = 128, MAXC = 512 };

typedef struct struct2 {
uint8_t  num;
uint8_t  action;
uint8_t  id;
} struct2_t;

typedef struct struct1 {
uint8_t  title;
struct2_t *content;
} struct1_t;

int main (void) {

    struct2_t s2 = { 5, 4, 3 }, *s3 = NULL;
    struct1_t s1 = { 2, &s2 };  /* content initialized to s2 */

    /* using address of s2 as content */
    printf ("\n title  : %" PRIu8 "\n"
            " num    : %"   PRIu8 "\n"
            " action : %"   PRIu8 "\n"
            " id     : %"   PRIu8 "\n", s1.title, 
            s1.content->num, s1.content->action, 
            s1.content->id);

    /* dynamic allocation of s3 */
    if (!(s3 = calloc (1, sizeof *s3))) {
        fprintf (stderr, "error: virtual memory exhausted.\n");
        return 0;
    }

    s3->num    = 8;  /* assignment of values to s3 */
    s3->action = 7;
    s3->id     = 6;
    s1.content = s3; /* update content pointer to s3 */

    /* using address of s3 as content */
    printf ("\n title  : %" PRIu8 "\n"
            " num    : %"   PRIu8 "\n"
            " action : %"   PRIu8 "\n"
            " id     : %"   PRIu8 "\n", s1.title, 
            s1.content->num, s1.content->action, 
            s1.content->id);

    free (s3);  /* don't forget to free all memory you allocate */

    return 0;
}

示例输出

$ ./bin/structnested

 title  : 2
 num    : 5
 action : 4
 id     : 3

 title  : 2
 num    : 8
 action : 7
 id     : 6

查看示例并告诉我您是否有任何其他问题，或者如果我错过了您的部分问题，请告诉我。

Answer 2

您永远不会初始化content，因此您会获得未定义的行为。你需要这样做：

/* ... */

    req = (struct1_t *)msg;
    req->content = (struct2_t *)&msg[1];

    printf("%d ", req->title);

/* ... */

Typepunning不是递归的。

Answer 3

你的代码中有两个faws，首先在struct1中使用指向struct2的指针，但是你正在用数据填充struct1实例的内存，因此你的一个值被认为是一个指针。其次，您必须考虑结构内数据的对齐。 Teh编译器根据数据类型和目标体系结构放置数据，并使用填充字节填充空格。由于32位架构通常在4字节对齐时更好地访问数据，因此编译器会将struct1（或struct1指针）移动到下一个4字节对齐的地址。所以你有代码

0xYYYYYYYY+00 struct1.title (byte0)
0xYYYYYYYY+01 padding byte0
0xYYYYYYYY+02 padding byte1
0xYYYYYYYY+03 padding byte2
0xYYYYYYYY+04 struct1.struct2*

如果你纠正指针问题：那么你有

0xYYYYYYYY+00 struct1.title (byte0)
0xYYYYYYYY+01 padding byte0
0xYYYYYYYY+02 padding byte1
0xYYYYYYYY+03 padding byte2
0xYYYYYYYY+04 struct1.struct2.num
0xYYYYYYYY+05 struct1.struct2.action ( here is no padding,as the data iself is u8)
0xYYYYYYYY+06 struct1.struct2.id( here is one paddingbyte,as the data iself is u8, but at the end of the structure).

一般来说，我建议不要以这种方式初始化srtuct，因为你必须考虑编译器的对齐逻辑，这可能有点棘手。

Type conversion to a structure which has another structure ptr in it

3 个答案: