如何通过指针读取 memcpy 结构结果

Question

I want to copy struct content in memory via char* pc the print it back but hear I have exception (reading violation)

struct af {
        bool a;
        uint8_t b;
        uint16_t c;
    };

int main() {

        af t;
        t.a = true;
        t.b = 3;
        t.c = 20;

        char* pc = nullptr;

        memcpy(&pc, &t, sizeof(t));

        std::cout << "msg is " << pc << std::endl; // here the exception

        return 0;
    }

then I want to recover data from memory to another structure of same type I did af* tt = (af*)(pc); then tried to access to tt->a but always exception

Answer 1

You need to allocate memory before you can copy something into it. Also, pc is already the pointer, you need not take the address of it again. Moreover, the byte representation is very likely to contain non-printable characters. To see the actual effect the following copies from the buffer back to an af and prints its members (note that a cast is needed to prevent std::cout to interpret the uint8_t as a character):

#include <iostream>
#include <cstring>

struct af {
        bool a;
        uint8_t b;
        uint16_t c;
    };

int main() {
    af t;
    t.a = true;
    t.b = 3;
    t.c = 20;

    char pc[sizeof(af)];

    std::memcpy(pc, &t, sizeof(t)); // array pc decays to pointer to first element

    for (int i=0;i<sizeof(af); ++i){
        std::cout << i << " " << pc[i] << "\n";
    }
    af t2;
    std::memcpy(&t2, pc,sizeof(t));
    std::cout << t2.a << " " << static_cast<unsigned>(t2.b) << " " << t2.c;

}

Output :

0 
1 
2 
3 
1 3 20

Note that I replaced the output of pc with a loop that prints individual characters, because the binary representation might contain null terminators and pc is not a null terminated string. If you want it to be a null-terminated string, it must be of size sizeof(af) +1 and have a terminating '\\0' .