何时释放一个unique_ptr？

Question

使用此代码：

void f(std::unique_ptr<int> q)
{
}

void g()
{
    std::unique_ptr<int> p{new int{42}};
    f(std::move(p));
}

在哪一行p被释放？我想在f函数的出口处说一下，因为它是使用std :: move移到那里的，但是我不确定这个答案也不是很确定。

Answer 1

在哪一行p被释放？

Answer 2

p分配在哪一行？

Answer 3

您正在[[转让]]所有权q，因此p现在拥有资源。因此，一旦q被破坏，即在函数f的末尾，您的整数就会被破坏。[唯一指针本身在作用域的结尾处被破坏，即，p在g()返回时被销毁，而在q返回时被销毁f()。

Answer 4

为了清楚地更改您的代码段，请按照以下方式进行

#include <iostream>
#include <memory>
#include <utility>

struct A
{
    ~A() { std::cout << "~A()\n"; }
};

void f(std::unique_ptr<A> q)
{
    std::cout << "f() begins\n";
}

void g()
{
    std::cout << "g() begins\n";
    std::unique_ptr<A> p{new A };
    f( std::move( p ) );
    std::cout << "p == nullptr is " << ( p == nullptr ) << '\n';
    std::cout << "g() endss\n";
}

int main() 
{
    std::cout << "Within main()\n";
    g();

    return 0;
}

程序输出为

Answer 5

p将在退出g函数时被销毁，就像其他具有局部作用域的变量一样。当时，它不再保存数据了。要理解，实际上您不需要单独的功能，只需考虑一下：

int* n = new int(42);

{
    std::unique_ptr<int> p(n); // p how owns the value pointed to by n
                               // for brevity, I'll say p owns n from now on,
                               // although technically not correct...

    { // opening another scope! (corresponds to function call)
        std::unique_ptr<int> q; // another variable, nothing else is a
                                // function parameter either...

        q = std::move(p);       // this happens, too, when calling a function
                                // at this point, the ownership is transferred to q
                                // q now owns n, p is left with just nothing
                                // (i. e. now holds a null-pointer)
    } // at this point, q is going out of scope; as it is the current
      // owner of n, it will delete it

    // q still is in scope, but it has transferred ownership, remember?

} // at this point, q is destroyed; as it doesn't own anything at all any more
  // it dies without doing anything...