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 figured out the reason but thought it would be a fun question to post anyway.)

#include <boost/shared_ptr.hpp>
#include <iostream>

using namespace boost;
using namespace std;

void fun(shared_ptr<int> ptr) {
    cout << *ptr << endl;
}

int main() {
    int foo = 5;
    fun(&foo);
    return 0;
}

/* shared_ptr_test.cpp: In function `int main()':
 * shared_ptr_test.cpp:13: conversion from `int*' to non-scalar type `
 *  boost::shared_ptr<int>' requested */

score 10 · Accepted Answer · answered Nov 20 '08 at 02:30

10

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.

answered Nov 20 '08 at 02:30

cadabra

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score 6 · Answer 2 · answered Oct 07 '11 at 14:38

6

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

answered Oct 07 '11 at 14:38

curiousguy

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score 2 · Answer 3 · edited Nov 20 '08 at 02:16

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).

edited Nov 20 '08 at 02:16

Ferruccio

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answered Nov 20 '08 at 02:05

Alex Lim

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score -1 · Answer 4 · edited Nov 20 '08 at 03:30

-1

int main() {

    int foo = 5;
    fun(&foo);

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

    return 0;
}

edited Nov 20 '08 at 03:30

Evan Teran

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answered Nov 20 '08 at 02:37

user35978

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1

I don't see how it is fundamentally different from `delete &foo;` – curiousguy Oct 07 '11 at 14:40

score -3 · Answer 5 · answered Mar 25 '11 at 07:06

-3

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.

answered Mar 25 '11 at 07:06

V_V

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@vBx: in mentioned example [1](http://stackoverflow.com/questions/304093/why-shared-ptr-has-an-explicit-constructor/304183#304183) offset address operator (&) has been used to pass parameter by pointer to function. Should use reference instead. – V_V Jun 04 '11 at 10:19
`operator&` gives the address, not the "offset" (whatever that means) of a lvalue. – curiousguy Oct 07 '11 at 14:34

Why shared_ptr has an explicit constructor

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