我的代码的内联函数不存储字符串字符串值

Question

我正在关注一个与我的代码非常相似但不完全相同的在线教程。我已经对其进行了修改，以便更好地理解它，但是我在从内联函数获取/返回字符串值时遇到了问题。一旦编译器退出最后一个大括号Another.cpp，字符串值就会变为空。

有什么方法可以在不更改文件结构和代码的情况下解决我的问题？特别是，我不想更改 Another.h 的任何内容。请，如果有人可以在这里解释内联函数背后的原因，我将不胜感激。

另一个.h

#pragma once

#include<iostream>
#include"Temple.h"
#include<memory>

namespace NamespaceAn
{
    class Test
    {
    public:
        static void Init();
        inline static std::shared_ptr<Outside::Inside>& GetString() { return s_String; }
    private:
        static std::shared_ptr<Outside::Inside> s_String;

    };
}

另一个.cpp

#include "Another.h"
#include<string>

namespace NamespaceAn 
{
    std::shared_ptr<Outside::Inside>Test::s_String;

    void Test::Init()
    {
        //This s_String value is getting disapprear as soon as the compiler get out of the curly bracket

        auto s_String = Outside::Inside("asdf");

        //below all are commented out for the Another.cpp
        //std::string  Ast = "asdf";

        /*std::string Ast = "asdf";
        auto s_String = new Outside::Inside(Ast);*/

        /*Outside::Inside* ob = new Outside::Inside(Ast);
        auto s_String = &(ob->Get());*/
        //                               *****    Attention ****
        //no operator "=" matches the operands and binary '=' no operator found which takes a right-hand ooperand of type std::string*(no acceptable conversion)
        //s_String = &(ob->Get());



        //auto s_String = Ast;

        //auto s_String = Outside::Inside("asdf");
        
        //auto s_String = new Outside::Inside("asdf");

        //s_String = reinterpret_cast<Ast>;
        //*s_String = &(Outside::Inside("asdf"));
        //s_String = reinterpret_cast<Outside::Inside*>(&Outside::Inside("asdf"));

    }
}

Temple.h

#pragma once
#include <string>

namespace Outside
{
    class Inside
    {
        std::string m_String = "tesi";
    public:
        Inside(std::string x) 
        {
            m_String = x;
        }
        std::string Get() 
        {
            return m_String;
        }

        /*std::string* m_String;
    public:
        Inside(std::string &x)
        {
            m_String = &x;
        }
        std::string Get()
        {
            return 
            *m_String;
        }*/


        /*template<typename T>
        static T Set(T x) 
        {
            return T;
        }*/
    };

}

ma​​in.cpp

#pragma once

#include<iostream>
#include"Another.h"
#include<memory>



int main() 
{
    NamespaceAn::Test::Init();

        /*auto asdf = NamespaceAn::Test::GetString();
        std::cout << asdf << std::endl;*/

        std::cout << NamespaceAn::Test::GetString();
    
    std::cin.get();
}

额外

在Another.cpp里面，我也试过下面的代码：

std::string  Ast = "asdf";
Outside::Inside* ob = new Outside::Inside(Ast);
auto s_String = &(ob->Get());

Answer 1

在Test::Init，这一行：

auto s_String = Outside::Inside("asdf");

声明一个局部变量s_String，它隐藏静态成员变量。所以当你退出函数时，你会看到静态成员变量的原始值，它从未被修改过。

相反，只需这样做：

s_String = Outside::Inside("asdf");

分配给静态成员变量。由于s_String 是shared_ptr，因此您要查找的可能是：

s_String = std::make_shared(Outside::Inside("asdf"));