将 void* 转换为 char*

Question

I have a char * who points to the structure. Here is my structure:

struct prot
{
    int size;
    unsigned short codeAction;
    void *data;
};

I recovered size and codeAction , but now I want to recover data . And when I cast my last 8 bytes I have nothing in it.

The following code is just a test, it's a bad code:

char lol[4];
for (int i = 0; i < 4; i++)
    lol[i] = test[i];

int size = *(int*)lol;

char loli[2];
int index = 0;   
for (int i = 4; i < 6; i++)
{
    loli[index] = test[i];
    index++;
}

int code = *(short*)loli;

char lolo[8];
index = 0;
for (int i = 6; i < size; ++i)
{
    lolo[index] = test[i];
    index++;
}

void *newData = (char *)lolo; // how can I cast it?

How I can display the content of newData ?

Answer 1

Your problem is that when casting lolo you actually cast a pointer to the char array you defined. So the result of the cast would be a char pointer to the first cell of the array.

Why don't you just use this as a struct and access the fields regularly?

Anyway, you want to use lolo as a 64 bit type pointer and the access what's in it.

void* newData = *((uint64_t*)lolo)

Besides, don't loop until size in the last for loop, loop only 8 times, until lolo is full. The number of bytes in newData itself ( not what it points to) is constant, and is 4 bytes on 32bit machines, 8 bytes on 64bit ones.

Last thing - index++ , not o++ . o isn't defined, as much as I can see.