在c中将char数组转换为structs数组

Question

我有一个 unsigned char*（从服务器发送的字节），我需要将它转换/转换为一个结构数组，其大小与 unsigned char* 的大小完全匹配。 不复制值，我该如何实现？在 uchar* 中，可以有任意数量的结构。

我如何做到这一点？

我尝试了以下（伪代码）：

// Define the struct:
struct MyStruct:
    unsigned char name[6]
    long time
    double speed

// ...some code...

// Get the uchar*.
unsigned char* chars = getUchars()

// The uchar*'s first byte is a special flag,
// so the structs start at index 1 not 0.
unsigned char* chars2 = &chars[1]

// Convert chars2 to an array of MyStructs.
MyStruct *structs = (MyStruct*) chars2

// Get the first struct and print values.
MyStruct s1 = structs[0]
_print(charArrayToString(s1.name), s1.time, s1.speed)
// When this _print() is called, the struct's
// name value is printed correctly, but its time
// and speed are not printed correctly.

// Get the second struct.
MyStruct s2 = structs[1]
_print(charArrayToString(s2.name), s2.time, s2.speed)
// None of this second struct's name, time and speed
// is printed correctly.

我在哪里犯错了？

Answer 1

使用编译指示包，它适用于大多数编译器

#include <stdio.h>

struct DefaultPack
{
    char str[6];
    long time;
    double speed;
};

#pragma pack(push,1)
struct Packed
{
    char str[6];
    long time;
    double speed;
};
#pragma pack(pop)


int main (int argc, char *argv[])
{
    struct DefaultPack n;
    struct Packed p;
    printf("Sizes of struct:\n"
           "Default packing: %lu\n"
           "Packed: %lu\n"
           "in packed struct:\n"
           "offset of time: %lu\n"
           "offset of speed: %lu\n"
           "in default packed struct:\n"
           "offset of time: %lu\n"
           "offset of speed: %lu\n",
           sizeof(struct DefaultPack),
           sizeof(struct Packed),
           (void*)&p.time  - (void*)&p.str[0],
           (void*)&p.speed - (void*)&p.str[0],
           (void*)&n.time  - (void*)&n.str[0],
           (void*)&n.speed - (void*)&n.str[0]
           );
    return(0);
}