问题描述
我从某个博客中了解到，默认构造的(空) shared_ptr 被自动初始化为 nullptr .但是在标准中找不到任何这样的明确声明.
I had read from some blog that a default-constructed (empty) shared_ptr is automatically initialized to nullptr. But could not find any such explicit statement in the statndards.
我写了一个小片段(Linux编译)来确认这一点:
I wrote a small snippet (Linux Compiled) to confirm this:
#include <iostream>
#include <memory>
struct Base;
int main()
{
std::shared_ptr<Base> p;
Base* b;
if (p == nullptr) {
std::cout << "p IS NULL \n";
}
else {
std::cout << "p NOT NULL \n";
}
if (b == nullptr) {
std::cout << "b IS NULL \n";
}
else {
std::cout << "b NOT NULL \n";
}
return 0;
}
p IS NULL
b NOT NULL
由此我看到，在声明时智能指针被隐式分配了 nullptr .有人可以确认这种行为吗?使用 shared_ptr 而不手动为其分配 nullptr 是否安全?
From this I see that the smart pointers are implicitly assigned nullptr at the time of Declaration. Can somebody confirm this behavior? Is it safe to use a shared_ptr without manually assigning a nullptr to it?
推荐答案
是的， cppreference 告诉我们，默认构造函数与只将 nullptr 传递给构造函数相同:
Yes, cppreference tells us that the default constructor is identical to just passing a nullptr to the constructor:
constexpr shared_ptr() noexcept; (1)
constexpr shared_ptr( std::nullptr_t ) noexcept; (2)
也来自 C ++标准草案2017 :
...
constexpr shared_ptr()noexcept;
2 效果:构造一个空的 shared_ptr 对象.
3 后置条件: use_count()== 0&&get()== nullptr .
constexpr shared_ptr() noexcept;
2 Effects: Constructs an empty shared_ptr object.
3 Postconditions: use_count() == 0 && get() == nullptr.
这篇关于是否将默认构造的(空)shared_ptr自动初始化为nullptr?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！