default constructors and object copying

Question

I'm wondering why I need to declare a default constructor in this case. For one thing doesn't the compiler do that automatically if i leave it out? And regardless, I still don't see why its necessary. Also, I get the error even if I omit 'obj_B = origin.obj_B;'

class B
{
public:
    bool theArray[5] ;

    B(bool x) {theArray[1] = x;};
    //B(){};    
};

class A
{
public:
    B obj_B;

    A() : obj_B(1) {};
    A(A const &origin) {obj_B = origin.obj_B;}; //error:no matching function for call 
                                                      //to B::B()
}; 

int main () 
{
    std::vector <A> someAs;
    for(int q=0;q<10;q++)
        someAs.push_back(A());

    for(int q=0;q<10;q++)
        std::cout << someAs[q].obj_B.theArray[1] << std::endl;
}

Wheatevo · Accepted Answer · 2011-02-21 03:35:15Z

6

The compiler only creates a default constructor if you don't specify an alternate constructor.

Because you made:

B(bool x) {theArray[1] = x;}

The default constructor will not be created for you.

The specific error you're getting is because A(A const &origin) doesn't specify the constructor to use for obj_B explicitly.

The following code would work:

A(A const &origin) : obj_B(1) {obj_B = origin.obj_B;}

By the way, you don't need a trailing semicolon on a function definition.

edited Feb 21, 2011 at 3:35

answered Feb 21, 2011 at 3:28

Wheatevo

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5 Comments

Matt Munson Over a year ago

@ Wheatevo Cool. I know its not an issue in this code, but is there any reason that using a default copy constructor or default assignement-operator could fail to copy the contents of a non-dynamically allocated array (eg theArray5)?

Wheatevo Over a year ago

In the vast majority of cases, the default copy constructor, default assignment operator, and default destructor will be sufficient if you don't use dynamically allocated memory.

Wheatevo Over a year ago

Anyway, there's no reason that the default copy or assignment operator would fail to copy the contents of a typical array. You only need to make your own copy constructor if you're dealing with dynamically allocated memory or something like reference counters. The default copy will automatically copy each member variable of one object to another (excluding static members).

Matt Munson Over a year ago

@Wheatevo Alright. That's what everyone's been saying. But that's the problem I've been having and I can't for the life of me figure out what's wrong. What about the tiny minority of cases?

Wheatevo Over a year ago

I would suggest posting a new question with the code you're having trouble with :)

Jerry Coffin · Accepted Answer · 2011-02-21 03:30:12Z

2

If you don't define any ctors for a class, the compiler will synthesize a default constructor. If you define another constructor though (e.g., one that takes an argument) then the compiler does not synthesize one for you, and you have to define one yourself.

In case you care, C++ 0x adds an "=default;" declaration to tell the compiler to provide a ctor that it would have provided by default, even if you have defined another ctor.

answered Feb 21, 2011 at 3:30

Jerry Coffin

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Comments

aschepler · Accepted Answer · 2011-02-21 03:36:45Z

2

To define a copy constructor for A that does not require a default constructor for B, use member initializer syntax:

class A {
public:
    A(A const& origin) : obj_B(origin.obj_B) {}
    //...
};

answered Feb 21, 2011 at 3:36

aschepler

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Comments

Lou · Accepted Answer · 2011-02-21 04:32:35Z

0

To make one final point...

Assuming you hadn't defined a non-default constructor, failing to define a default constructor would've resulted in the elements of theArray[] being undefined. That is a bad habit to get into usually leading to bugs down the road.

answered Feb 21, 2011 at 4:32

Lou

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