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Asked: 2026-05-10T23:23:08+00:00

I was wondering why shared_ptr doesn’t have an implicit constructor. The fact it doesn’t

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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 */
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1 Answer

  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.

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