Why shared_ptr has an explicit constructor

问题

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 */

回答1:

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.

回答2:

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

回答3:

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).

回答4:

int main() {

    int foo = 5;
    fun(&foo);

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

    return 0;
}

回答5:

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.

来源：https://stackoverflow.com/questions/304093/why-shared-ptr-has-an-explicit-constructor

标签

c++

boost

refcounting