为什么在调用memcpy bool值后memcpy to int不起作用

Question

// Simple struct containing 3 floats
struct vector
{
    float x;
    float y;
    float z;
};

// My test struct
struct test2
{
    float a;
    vector b;
    bool c;
    int d;
};

int main()
{
    // I create my structure on the heap here and assign values
    test2* test2ptr = new test2();
    test2ptr->a = 50;
    test2ptr->b.x = 100;
    test2ptr->b.y = 101;
    test2ptr->b.z = 102;
    test2ptr->c = true;
    test2ptr->d = 5;

    // Then turn the struct into an array of single bytes
    unsigned char* data = (unsigned char*)test2ptr;
    // Variable for keeping track of the offset
    unsigned int offset = 0;

    // Variables that I want the memory copied into they
    float a;
    vector b;
    bool c;
    int d;

    // I copy the memory here in the same order as it is defined in the struct
    std::memcpy(&a, data, sizeof(float));
    // Add the copied data size in bytes to the offset
    offset += sizeof(float);

    std::memcpy(&b, data + offset, sizeof(vector));
    offset += sizeof(vector);

    std::memcpy(&c, data + offset, sizeof(bool));
    offset += sizeof(bool);

    // It all works until here the results are the same as the ones I assigned
    // however the int value becomes 83886080 instead of 5
    // moving this above the bool memcpy (and moving the variable in the struct too) fixes the problem
    std::memcpy(&d, data + offset, sizeof(int));
    offset += sizeof(int);

    return 0;
}

Answer 1

忽略数据在结构中的填充。

看看下面的简化示例：

struct X
{
    bool b;
    int i;
}; 

int main()
{
    X x;
    std::cout << "Address of b " << (void*)(&x.b) << std::endl;
    std::cout << "Address of i " << (void*)(&x.i) << std::endl;
}

这将在我的PC上产生：

如果在程序之前添加pragma pack(1)，则输出为：

b 0x7FFD16C79CFB的地址

I0x7FFD16C79CFC的地址