为什么 shared_ptr 有一个显式构造函数 [英] Why shared_ptr has an explicit constructor

问题描述

我想知道为什么 shared_ptr 没有隐式构造函数.这里提到的事实并非如此:为此获得 boost::shared_ptr

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

(我找到了原因，但认为无论如何发布这个问题都会很有趣.)

(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 */

推荐答案

在这种情况下，shared_ptr 将尝试释放分配给您的栈的 int.你不会想要那样的，所以显式构造函数让你考虑一下.

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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