Using shared_ptr for different types

Question

I am afraid I think I am missing something here and would like some help. I have searched online (perhaps not very well) but am not finding information which is helping me.

I have a scenario where I want to create three separate vectors all containing shared_ptrs to different types. So I want a a vector containing shared_ptrs of an Enemy object and a vector containing shared_ptrs to an Enemy2 object.

I have successfully used a vector of shared_ptrs before but if I remember correctly, I only ever needed to store the same type so did not have any issues. I am currently trying to implement it like this:

// Defining my vectors
std::vector<std::shared_ptr<Enemy1>> enemyVec1;
std::vector<std::shared_ptr<Enemy2>> enemyVec2;

// Creating new enemy objects and putting in to vector
std::shared_ptr<Enemy1> enemy1(new Enemy());
std::shared_ptr<Enemy2> enemy2(new Enemy());

enemyVec1.push_back(enemy1);
enemyVec2.push_back(enemy2);

Here is the error message:

c:\program files (x86)\microsoft visual studio 10.0\vc\include\memory(1705): error C2664: 'std::tr1::_Ptr_base<_Ty>::_Reset0' : cannot convert parameter 1 from 'Enemy *' to 'Enemy2 *'

I am vaguely familiar with what might be going on here but I am not 100% sure. I am even less sure of how I could go about coding this the "right" way as a lot of examples I look at never seem to cover the use of shared_ptrs for different types.

If I take out all references to Enemy2 class and class and only ever use shared_ptrs for one type exclusively then the code works fine. I thought I knew how shared_ptrs worked but I clearly do not so I need to read up some more about them.

But in the mean time, could someone explain to me what I have done wrong here?

Thanks

Shouldn't it be new Enemy2() instead of the second new Enemy()? — Messa
– Messa, Commented Feb 27, 2014 at 14:46
Even if Enemy1 and Enemy2 inherit from Enemy, you can't implicitly downcast to one particular derived class. It's safe to upcast a pointer from derived class to base class, but not visa versa. If you really need to cast a base to a derived class, you usually need dynamic_cast or std::dynamic_pointer_cast. — user180247
– user180247, Commented Feb 27, 2014 at 14:47
It only will work if Enemy inherits from both Enemy1 and Enemy2 classes — hidrargyro
– hidrargyro, Commented Feb 27, 2014 at 14:48

Angew is no longer proud of SO · Accepted Answer · 2014-02-27 14:48:26Z

2

You're initialising both enemy1 and enemy2 with new Enemy(). This will only work if Enemy is derived from both Enemy1 and Enemy2. I doubt that's the case, hence the error.

I believe you should actually be initialising them with instances of the appropriate types Enemy1 and Enemy2. You should also consider using std::make_shared:

auto enemy1 = std::make_shared<Enemy1>();
auto enemy2 = std::make_shared<Enemy2>();

answered Feb 27, 2014 at 14:48

Angew is no longer proud of SO

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1 Comment

Chris Over a year ago

You have got to be kidding me! Time and time again, I make these silly mistakes when I should know better! :( Thank you

Wojtek Surowka · Accepted Answer · 2014-02-27 14:45:56Z

0

It looks like the pointer to Enemy2 is not a pointer to Enemy, which most probably means that Enemy2 does not inherit from Enemy. Everything looks like working for Enemy1 though.

answered Feb 27, 2014 at 14:45

Wojtek Surowka

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