Why shared_ptr has an explicit constructor

Question

I was wondering why shared_ptr doesn\'t have an implicit constructor. The fact it doesn\'t is alluded to here: Getting a boost::shared_ptr for this

(I f

Answer 1

I think there is no reason to have explicit in this constructor.

Mentioned examples with incorrect using of offset address operator (&) make no sense since there is no place to use such operator in modern C++. Except only such idiomatic code in assignment/comparision operator as 'this == &other' and maybe some test code.

Answer 2

Long time lurker, and a 3rd year soft eng student here, Haphazard guess would be, to stop you from attempting to convert a 'natural' pointer to a shared_ptr, then deallocing the pointed object, without the shared_ptr knowing about the dealloc.

(Also, reference counting problems blah blah).

Answer 3

int main() {

    int foo = 5;
    fun(&foo);

    cout << foo << endl; // ops!!

    return 0;
}

Answer 4

In this case, the shared_ptr would attempt to free your stack allocated int. You wouldn't want that, so the explicit constructor is there to make you think about it.

Answer 5

The logical reason is that:

• calling the delete operator is not implicit in C++
• the creation of any owning smart pointer (shared_whatever, scoped_whatever, ...) is really a (delayed) call to the delete operator