解引用指向结构变量的指针

Question

I try to understand what value we get when dereference the pointer to struct.

My struct is.

    struct car
{
    int a = 5;
    int b = 10;
    int c = 2;
    int d = 14;
    int e = 20;
};

My code in main function is

int main()
{
    car  k1,k2,k3,k4,k5; // declare struct variable
    car* ptr1,*ptr2,*ptr3,*ptr4,*ptr5;// declare pointer to struct variable
    ptr1 = &k1;
    ptr2 = &k2;
    ptr3 = &k3;
    ptr4 = &k4;
    ptr5 = &k5;

    printf("*ptr1=%d *ptr2=%d *ptr3=%d *ptr4=%d *ptr5=%d  \n", *ptr1,*ptr2,*ptr3,*ptr4,*ptr5);
 };

When I dereference ptr1 to ptr5.I get the value correspond with the value of member in struct(value a to e).Why I get this value.

Answer 1

To be able to print struct you need to overload operator << . Your struct then should look like this:

struct car
{
    int a = 5;
    int b = 10;
    int c = 2;
    int d = 14;
    int e = 20;

    friend std::ostream& operator<<(std::ostream& os, const car& c) {
      os << "Struct data:\n";
      os << "------------\n";
      os << "a var: " + std::to_string(c.a) + "\n";
      os << "b var: " + std::to_string(c.b) + "\n";
      os << "c var: " + std::to_string(c.b) + "\n";
      os << "d var: " + std::to_string(c.b) + "\n";
      os << "e var: " + std::to_string(c.b) + "\n";
      return os;
    }
};

Then if you would print dereferenced ptr like this:

std::cout << *ptr1 << std::endl;

You will see output like this:

Struct data:
------------
a var: 5
b var: 10
c var: 10
d var: 10
e var: 10

You can return what you want, depends on you.

Answer 2

Variadic functions do not have type checking and practically may pun types. Essentially you passed content of structure as raw values and format string described how to treat them. Luckily, in this case struct has no padding, so first five integers were read accordingly. The behaviour in this case is unspecified and depends on implementation of compiler and ABI.