shared_ptr assignment notation implicit conversion

Question

considering following code , Why I can’t use the assignment notation here , Why that is considered to be an implicit conversion.

shared_ptr<string> pNico = new string("nico"); // ERROR implicit conversion
shared_ptr<string> pNico{new string("nico")};   // OK

Perhaps construction from a raw pointer is made explicit for `shared_ptr` so that `foo(&obj);` won't accidentally take ownership of `obj` — Piotr Skotnicki, May 27 '16 at 18:31
The assignment operator is not overloaded for `T*`, only for other smart pointers — xvan, May 27 '16 at 18:34
@sleeptightpupper Question isn't here that we use make_shared , iknow that make_shared way of creation is faster and safer because it uses one instead of two allocations: one for the object and one for the shared data the shared pointer uses to control the object — Adib, May 27 '16 at 18:43
@PiotrSkotnicki , i find your Answer correct , could you please explain it in details — Adib, May 27 '16 at 18:45
@Adib the syntax `T x = y;` is called *copy-initialization*, as opposed to `T x(y);` which is *direct-initialization*. Copy-initialization applies not only to `T x = y;`, but also for passing arguments to a function, `f(y);` with e.g. `void f(T x) {}`. the `explicit` specifier prevents implicit conversions that could happen in copy-initialization. I believe it was made intentionally for `shared_ptr`, for the reason I mentioned — Piotr Skotnicki, May 27 '16 at 18:47

score 4 · Accepted Answer · answered May 27 '16 at 18:45

The constructor is explicit to prevent somebody from doing something like this:

void foo(std::shared_ptr<std::string> s) { }

int main()
{
   std::string s;
   foo(&s);
}

If it was implicit, the shared_ptr could take ownership of a stack allocated variable and try to delete it..which wouldn't make sense.

xvan · Answer 2 · 2016-05-27T18:53:14.153

2

ie

template< class Y >
explicit shared_ptr( Y* ptr );

the explicit keyword prevents the copy initialization.

Only Converting constructors can be used on copy initialization.

edited May 27 '16 at 18:53

answered May 27 '16 at 18:45

xvan

2 Answers2