Accessing array of shared_ptr

Question

#include <memory>

int main()
{
   std::shared_ptr<double> array (new double [256], [](double * d){
      delete [] d;
   });
}

I made a shared_ptr pointing into an array of doubles which has its own custom deleter.

Now how can I access the array? Let's say I wish to access the array at index 1. I tried the usual "bracket method" but I get errors.

The word array points to its first element by default, but what if I want to access the 2nd element? Using increments and bracket gives me the "no match for operator" error.

Can someone explain to me what's happening under the hood?

I am asking this for research purposes, despite being aware that unique_ptr and vector will do a better job.

Sorry to ask but shouldn't it be `std::shared_ptr array (new double [256], [](double * d){ delete [] d; });` or `std::shared_ptr array (new double [256]);` in c++17 or even `std::shared_ptr array = std::make_shared(256);`? — krjw, Feb 05 '19 at 15:50

score 11 · Accepted Answer · edited May 23 '17 at 10:34

11

The bracket notation is defined to work with pointer types (and you're right that, given array array, the expression array decays to an expression with such a type which points to the first element) but, despite its function, std::shared_ptr is not a pointer type.

You would have to obtain the raw pointer first:

array.get()[n];

Where n is, of course, a valid array subscript.

This is also the case with std::unique_ptr (though note that, in that case, you do not need to supply your own deleter!).

edited May 23 '17 at 10:34

Community

1
1

answered Jun 11 '15 at 12:07

Lightness Races in Orbit

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@CarloBrew Yes it is. – Cory Kramer Jun 11 '15 at 12:08
`(*array)[n]` this one gives me error btw. invalid type for subscript. – Carlos Miguel Colanta Jun 11 '15 at 12:10
1

@CarloBrew yes and no. with unique_ptr you van specify the template parameter as `double[]` and not `double` and you don't have to provide destrcutor - the unique ptr will call `delete[]` – David Haim Jun 11 '15 at 12:13
@LightnessRacesinOrbit here's what i tried `(*array)[0] = 50.0;` `std::cout<<(*array)[0];` – Carlos Miguel Colanta Jun 11 '15 at 12:14
@David: Yes my answer says that :) – Lightness Races in Orbit Jun 11 '15 at 12:14
@CarloBrew: Sorry, `(*array)[n]` was me being stupid. That's a `double`, not `double*`, of course. [You could use `(&*array)[n]`](http://coliru.stacked-crooked.com/a/ff1acab24fcd7e5a), but I've decided to just remove that variant from the answer. – Lightness Races in Orbit Jun 11 '15 at 12:15

score 0 · Answer 2 · answered Feb 25 '20 at 04:19

in C++17, support for array of shared_ptr like in unique_ptr( c++11).

int main()
{

   std::shared_ptr<double[]> array1 (new double [3]{4,5,6}, [](double * d){
   delete [] d;});

   std::unique_ptr<double[]> array2 (new double [3]{4,5,6}, [](double * d){
   delete [] d });
}

Accessing array of shared_ptr

2 Answers2

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