unique_ptr behaviour while passing through function

Question

In below code snippet i am getting segmentation fault while passing unique_ptr as as value. usually this is known issue with auto_ptr as due to ownership issue (Assignee pointer becomes NULL) it can't be accessed after assignment. but why i am facing the same issue with unique_ptr even though it is having move semantics.

what i understood auto_ptr use copy constructor and unique_ptr use move function to transfer the ownership. But in both cases assignee pointer becomes null then what is the significance of having move semantics here. I believe compiler should flash the error if unique_ptr is passed as a value like below statement as copy constructor is private in unique_ptr. could you throw some light on this regarding this behaviour.

unique_ptr<Test> a = p;//deleted function (copy constructor).

Here is the code snippet:

#include<iostream>

#include<memory>

using namespace std;

class Test
{
public:
    Test(int a = 0 ) : m_a(a)
    {
    }
    ~Test( )
    {
        cout<<"Calling destructor"<<endl;
    }

    int m_a;
};


//***************************************************************
void Fun(std::unique_ptr<Test> p1 )
{
    cout<<p1->m_a<<endl;
}
//***************************************************************
int  main( )
{
    std::unique_ptr<Test> p( new Test(5) );
    Fun(std::move(p));
    cout<<p->m_a<<endl;

    return 0;
}

With unique_ptr and move, your have to be explicit about the ownership transfer. — Jarod42
You are moving ownership of your object into the function (indicated by the explicit move), leaving the original pointer empty (nullptr) and then you dereference that pointer. What did you expect would happen and why are you passing a unique ptr into the function at all? A normal or would do just fine. — MikeMB

Chris Drew Chris Drew · Accepted Answer · 2015-08-02T10:36:23

By writing std::move(p) you are casting p to an r-value reference. unique_ptr has a constructor that takes an r-value reference (a move constructor) so it constructs the by-value parameter using this move constructor and there is no compilation error.

You should not pass a unique_ptr by value unless you want to transfer ownership.

If you transfer ownership then the unique_ptr in the outer scope does not own anything anymore and it is undefined behaviour to dereference it.

In this case you should pass Test by const reference:

void fun(const Test& t) {
    std::cout << t.m_a << std::endl;
} 

int main(){
    auto p = std::make_unique<Test>(5);
    fun(*p);
}

See here for more information about how to pass smart pointers in general.

unique_ptr behaviour while passing through function

3 Answers