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std::shared_ptr<T>::get

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< cpp‎ | memory‎ | shared ptr
 
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std::shared_ptr
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shared_ptr::get
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T* get() const noexcept;
 (直到 C++17)
element_type* get() const noexcept;
 (自 C++17 起)

返回存储的指针。

内容

[编辑] 参数

(无)

[编辑] 返回值

存储的指针。

[编辑] 注解

shared_ptr 可能在存储指向另一个对象的指针的同时，共享对象的所有权。get() 返回存储的指针，而不是管理的指针。

[编辑] 示例

运行此代码
#include <iostream>
#include <memory>
#include <string_view>
 
int main()
{
    auto output = [](std::string_view msg, int const* pInt)
    {
        std::cout << msg << *pInt << " in " << pInt << '\n';
    };
 
    int* pInt = new int(42);
    std::shared_ptr<int> pShared = std::make_shared<int>(42);
 
    output("Naked pointer: ", pInt);
//  output("Shared pointer: ", pShared); // compiler error
    output("Shared pointer: ", &*pShared); // OK, calls operator*, then takes addr
    output("Shared pointer with get(): ", pShared.get());
 
    delete pInt;
 
    std::cout << "\nThe shared_ptr's aliasing constructor demo.\n";
    struct Base1 { int i1{}; };
    struct Base2 { int i2{}; };
    struct Derived : Base1, Base2 { int i3{}; };
 
    std::shared_ptr<Derived> p(new Derived());
    std::shared_ptr<Base2> q(p, static_cast<Base2*>(p.get()));
    std::cout << "q shares ownership with p, but points to Base2 subobject:\n"
              << "p.get(): " << p.get() << '\n'
              << "q.get(): " << q.get() << '\n'
              << "&(p->i1): " << &(p->i1) << '\n'
              << "&(p->i2): " << &(p->i2) << '\n'
              << "&(p->i3): " << &(p->i3) << '\n'
              << "&(q->i2): " << &(q->i2) << '\n';
}

可能的输出

Naked pointer: 42 in 0xacac20
Shared pointer: 42 in 0xacac50
Shared pointer with get(): 42 in 0xacac50
 
The shared_ptr's aliasing constructor demo.
q shares ownership with p, but points to Base2 subobject:
p.get(): 0xacac20
q.get(): 0xacac24
&(p->i1): 0xacac20
&(p->i2): 0xacac24
&(p->i3): 0xacac28
&(q->i2): 0xacac24

[编辑] 参见

 解引用存储的指针
(公共成员函数) [编辑]
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