How can I pass an array of objects?

Question

Here I have a very simple program. My aim is to let b equal c, that is to copy all the content of c into b. But I don't know how. The getdata() function returns a pointer pointing to array of objects c, but how can it be used to put c into b?

#include<iostream>
#include<stdlib.h>
using namespace std;
class A
{
    public:
    A(int i,int j):length(i),high(j){}
    int length,high;
};

class B
{
    private:
    A c[3] = {A(9,9),A(9,9),A(9,9)};
    public:
    A* getdata()
    {
        return c;
    }
};

int main()
{
    A b[3]={A(0,0),A(0,0),A(0,0)};
    B *x = new B();
    cout<< x->getdata() <<endl;
    cout << b[1].length<<endl;
    return 0;
}

Use std::array. It has natural copying semantics. There's also no reason whatsoever to use new here. — Qaz
– Qaz, Commented Aug 30, 2014 at 21:48
Really thank you! I have figured out myself. One of the easy ways is to use A* z=x->getdata(), passing the data of c to z. Then we can use z to fill b. This might not be a good answer since its efficiency is low. — Ebrahim True
– Ebrahim True, Commented Sep 1, 2014 at 0:23

Mr.C64 · Accepted Answer · 2014-08-30 22:24:58Z

1

In modern C++, make yourself a favor and use a convenient container class to store your arrays, like STL std::vector (instead of using raw C-like arrays).

Among other features, std::vector defines an overload of operator=(), which makes it possible to copy a source vector to a destination vector using a simple b=c; syntax.

#include <vector>  // for STL vector
....

std::vector<A> v;  // define a vector of A's

// use vector::push_back() method or .emplace_back()
// or brace init syntax to add content in vector...

std::vector<A> w = v;  // duplicate v's content in w

That's a possible partial modification of your code, using std::vector (live here on codepad):

#include <iostream>
#include <vector>
using namespace std;

class A
{
public:
    A(int l, int h) : length(l), high(h) {}
    int length, high;
};

class B
{
private:
    vector<A> c;

public:
    const vector<A>& getData() const
    {
        return c;
    }

    void setData(const vector<A>& sourceData)
    {
        c = sourceData;
    }
};

int main()
{
    vector<A> data;
    for (int i = 0; i < 3; ++i) // fill with some test data...
        data.push_back(A(i,i));

    B b;
    b.setData(data);

    const vector<A>& x = b.getData();
    for (size_t i = 0; i < x.size(); ++i) // feel free to use range-for with C++11 compilers
        cout << "A(" << x[i].length << ", " << x[i].high << ")\n";
}

edited Aug 30, 2014 at 22:24

answered Aug 30, 2014 at 21:54

Mr.C64

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2 Comments

Ebrahim True Over a year ago

Thank you! I'm a beginner of C++, I will try to use this in future situations.

Mr.C64 Over a year ago

@EbrahimChu: You're welcome. And don't forget that everyone of us once was a "beginner of C++" ;)

Arpit · Accepted Answer · 2014-08-30 21:48:22Z

0

Instead of creating an array of A i.e. 'b' in main, create a pointer to A. And then initialize it by calling the getdata().

A *b;

B *x = new B();

b = x->getdata();

answered Aug 30, 2014 at 21:48

Arpit

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1 Comment

Ebrahim True Over a year ago

Thank you, this is the easiest way I can think.

Vlad from Moscow · Accepted Answer · 2014-08-30 21:52:09Z

0

Here is an example

#include <iostream>
#include <algorithm>

class A
{
public:
    A( int i, int j ) : length( i ), high( j ){}
    int length, high;
};

class B
{
private:
    A c[3] = {A(9,9),A(9,9),A(9,9)};
public:
    A* getdata()
    {
        return c;
    }
};

int main() 
{
    A b[3] = { A(0,0), A(0,0), A(0,0) };

    B *x = new B();
    A *c = x->getdata();

    std::copy( c, c + 3, b );


    for ( const A &a : b ) std::cout << a.length << '\t' << a.high << std::endl;

    delete []x;

    return 0;
}

The output is

9   9
9   9
9   9

Instead of standard algorithm std::copy you may use an ordinary loop. For example

for ( size_t i = 0; i < 3; i++ ) b[i] = c[i];

answered Aug 30, 2014 at 21:52

Vlad from Moscow

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1 Comment

Ebrahim True Over a year ago

Thank you as well! You all give better solutions than mine!

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