CODEWITHSUNDEEP
c++c++11pointers
VirtualUser
VirtualUser

Reputation: 215

In one case shared_ptr cause runtime error, in another doesn't

Why

struct X{};

X x;
X *q = &x;
std::shared_ptr<X> p(&x);

causes error and

X x;
X *q = &x;
std::shared_ptr<X> p0 = std::make_shared<X>(x);

doesn't cause runtime error?

I read that in the first case I have "two different pointers, pointing to the same data and one of them is shared"

But I think that there is the same situation in the second case?

runtime error

free(): invalid pointer

Upvotes: 3

Views: 342

Answers (2)

lubgr
lubgr

Reputation: 38315

In the first case, the object x is deleted although it's lifetime already ends with the end of its scope. Here, you create an instance on the stack and in your scope :

X x; // lifetime automatically ends at the end of the scope

Then, you ask the std::shared_ptr to control the lifetime of the address of x, too.

std::shared_ptr<X> p(&x); // calls delete at the end of the scope

Hence, you free the same variable twice, which is undefined behavior. In the second case, you create a std::shared_ptr to a copied instance that is independent of the x instance.

std::shared_ptr<X> p0 = std::make_shared<X>(x); // Copies x, then deletes the copy

This is no problem, as both objects are only deleted once.

Note the semantics of std::make_shared here: the arguments you pass to it are forwarded to the constructor of its template argument. Hence you trigger the compiler-generated copy constructor X(const X&) to be invoked. This is very different from creating a std::shared_ptr from an already existing pointer to via std::shared_ptr p(&x);.

Upvotes: 9

David Schwartz
David Schwartz

Reputation: 182865

In the first case, p takes ownership of x. When p goes out of scope, it deletes x. But x was not dynamically allocated. Oops.

You don't have that situation in the second case. The make_shared function dynamically allocates a new, shared object. When p0 goes out of scope, it destroys that newly-created object. Note that, in this case, *p0 is not x but the new object.

Upvotes: 7

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