CODEWITHSUNDEEP
c++unique-ptr
Ramy Al Zuhouri
Ramy Al Zuhouri

Reputation: 21966

unique_ptr pointing always to the same memory area

Consider this code:

using namespace std;

int* get()
{
    unique_ptr<int> p (new int[4]);
    return p.get();
}
int main(int argc, char **argv)
{
    int *arr1=get();
    int* arr2=get();
    for(int i=0;i<4;i++)
    {
        arr1[i]=i;
        arr2[i]=i*2;
    }
    for(int i=0;i<4;i++)
        cout << arr1[i];
    return 0;
}

arr1 and arr2 point to the same area of memory. So they share the same values. I don't understand why, when I call arr2=get() :

unique_ptr<int> p (new int[4]);

This object shouldn't be created again? It isn't deleted because still reachable by arr1. How to get two arrays of different memory areas?

Upvotes: 1

Views: 468

Answers (2)

111111
111111

Reputation: 16148

I am fairly sure you are playing with undefined behavior which is bad.

the data being pointed to was destroyed when the unique pointer was destroyed, the fact the values are the same, and the same slot was chosen is luck.

for pointers to array type use a vector

std::vector<int> get()
{
    return std::vector<int>(4);
}

int main()
{ 
    std::vector<int> arr1=get();
    std::vector<int> arr2=get();
    return 0;
}

for normal single value pointers then you can return a unique_ptr;

std::unique_ptr<int> get(){
    return std::unique_ptr<int>(new int(0));
}
:::
std::unique_ptr<int> ptr=get();

Upvotes: 5

AlexTheo
AlexTheo

Reputation: 4164

There are a memory leak in such kind of usage of smart pointers. unique_ptr will use operator delete in order to free the allocated memory, but you need delete []. Also the return value of get function will be an invalid ptr because the unique_ptr will free the allocated area. If you need dynamically allocated arrays then use std::vector.

Upvotes: 0

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