Object array initialization without default constructor

Question

#include 
class Car
{
private:
  Car(){};
  int _no;
public:
  Car(int no)
  {
    _no=no;
  }
  void printNo()
  {
    std::cout<<_no<

        

Answer 1

I don't think there's type-safe method that can do what you want.

Answer 2

No, there isn't. New-expression only allows default initialization or no initialization at all.

The workaround would be to allocate raw memory buffer using operator new[] and then construct objects in that buffer using placement-new with non-default constructor.

Answer 3

You can use placement-new like this:

class Car
{
    int _no;
public:
    Car(int no) : _no(no)
    {
    }
};

int main()
{
    void *raw_memory = operator new[](NUM_CARS * sizeof(Car));
    Car *ptr = static_cast<Car *>(raw_memory);
    for (int i = 0; i < NUM_CARS; ++i) {
        new(&ptr[i]) Car(i);
    }

    // destruct in inverse order    
    for (int i = NUM_CARS - 1; i >= 0; --i) {
        ptr[i].~Car();
    }
    operator delete[](raw_memory);

    return 0;
}

Reference from More Effective C++ - Scott Meyers:
Item 4 - Avoid gratuitous default constructors

Answer 4

You can always create an array of pointers , pointing to car objects and then create objects, in a for loop, as you want and save their address in the array , for example :

#include <iostream>
class Car
{
private:
  Car(){};
  int _no;
public:
  Car(int no)
  {
    _no=no;
  }
  void printNo()
  {
    std::cout<<_no<<std::endl;
  }
};
void printCarNumbers(Car *cars, int length)
{
    for(int i = 0; i<length;i++)
         std::cout<<cars[i].printNo();
}

int main()
{
  int userInput = 10;
  Car **mycars = new Car*[userInput];
  int i;
  for(i=0;i<userInput;i++)
      mycars[i] = new Car(i+1);

note new method !!!

  printCarNumbers_new(mycars,userInput);


  return 0;
}    

All you have to change in new method is handling cars as pointers than static objects in parameter and when calling method printNo() for example :

void printCarNumbers_new(Car **cars, int length)
{
    for(int i = 0; i<length;i++)
         std::cout<<cars[i]->printNo();
}

at the end of main is good to delete all dynamicly allocated memory like this

for(i=0;i<userInput;i++)
  delete mycars[i];      //deleting one obgject
delete[] mycars;         //deleting array of objects

Hope I helped, cheers!

Answer 5

Nope.

But lo! If you use std::vector<Car>, like you should be (never ever use new[]), then you can specify exactly how elements should be constructed*.

*Well sort of. You can specify the value of which to make copies of.

---

Like this:

#include <iostream>
#include <vector>

class Car
{
private:
    Car(); // if you don't use it, you can just declare it to make it private
    int _no;
public:
    Car(int no) :
    _no(no)
    {
        // use an initialization list to initialize members,
        // not the constructor body to assign them
    }

    void printNo()
    {
        // use whitespace, itmakesthingseasiertoread
        std::cout << _no << std::endl;
    }
};

int main()
{
    int userInput = 10;

    // first method: userInput copies of Car(5)
    std::vector<Car> mycars(userInput, Car(5)); 

    // second method:
    std::vector<Car> mycars; // empty
    mycars.reserve(userInput); // optional: reserve the memory upfront

    for (int i = 0; i < userInput; ++i)
        mycars.push_back(Car(i)); // ith element is a copy of this

    // return 0 is implicit on main's with no return statement,
    // useful for snippets and short code samples
} 

With the additional function:

void printCarNumbers(Car *cars, int length)
{
    for(int i = 0; i < length; i++) // whitespace! :)
         std::cout << cars[i].printNo();
}

int main()
{
    // ...

    printCarNumbers(&mycars[0], mycars.size());
} 

Note printCarNumbers really should be designed differently, to accept two iterators denoting a range.

Answer 6

My way

Car * cars;

// else were

extern Car * cars;

void main()
{
    // COLORS == id
    cars = new Car[3] {
        Car(BLUE),
            Car(RED),
            Car(GREEN)
    };
}