C++ `wstring_convert`std:：wstring_convert导致错误C2280:尝试引用已删除的函数

我试图实现一个字符串类，该类同时处理std:：String和std:：wstring，以减少转换代码

在这个类中，我决定使用延迟更新策略。每次只需修改需要更新的字符串。 如果call operator+=const wchar\u t，那么我应该更新wstr:wstring，如果wstring根据wstrUpdated:bool不正确，那么通过调用wstring\u convert使用str:string更新它，然后执行该操作

此时，wstring_convert将无法工作，并告诉错误C2280。 我很想知道是什么导致了这个问题。我该如何解决它

这是我的密码：

#pragma once
// MutableString.h
#include<string>
#include<codecvt>
#include<locale>

class MutableString
{
public:
    MutableString();
    ~MutableString();

    MutableString(const MutableString& other);
    MutableString(const std::string& other);
    MutableString(const std::string&& other);
    MutableString(const std::wstring& other);
    MutableString(const std::wstring&& other);
    MutableString(const char* other);
    MutableString(const wchar_t* other);

    void clear();

    MutableString& operator = (const MutableString& other);
    MutableString& operator += (const wchar_t ch);
    MutableString& operator += (const char ch);

    std::string getMultiByteString();
    std::wstring getWideString();
private:
    bool strUpdated;
    bool wstrUpdated;
    std::string str;
    std::wstring wstr;
    std::wstring_convert<std::codecvt_utf8_utf16<wchar_t>> converter;
};

//MutableString.cpp
#include "MutableString.h"


MutableString::MutableString()
{
    str = "";
    wstr = L"";
    strUpdated = false;
    wstrUpdated = false;
}


MutableString::~MutableString()
{
    ;
}
MutableString::MutableString(const MutableString& other)
{
    strUpdated = other.strUpdated;
    wstrUpdated = other.wstrUpdated;
    str = other.str;
    wstr = other.wstr;
}

MutableString::MutableString(const std::string& other)
{
    strUpdated = true; 
    wstrUpdated = false;
    str = other;
}
MutableString::MutableString(const std::string&& other)
{
    strUpdated = true;
    wstrUpdated = false;
    str = other;
}
MutableString::MutableString(const std::wstring& other)
{
    strUpdated = false;
    wstrUpdated = true;
    wstr = other;
}
MutableString::MutableString(const std::wstring&& other)
{
    strUpdated = false;
    wstrUpdated = true;
    wstr = other;
}
MutableString::MutableString(const char* other)
{
    strUpdated = true;
    wstrUpdated = false;
    str = other;
}
MutableString::MutableString(const wchar_t* other)
{
    strUpdated = false;
    wstrUpdated = true;
    wstr = other;
}

void MutableString::clear()
{
    strUpdated = true;
    wstrUpdated = true;
    wstr.clear();
    str.clear();
}

MutableString& MutableString::operator = (const MutableString& other)
{
    str = other.str;
    wstr = other.wstr;
    strUpdated = other.strUpdated;
    wstrUpdated = other.wstrUpdated;
    return *this;
}

MutableString& MutableString::operator += (const wchar_t ch)
{
    if (strUpdated == true)
    {
        if (wstrUpdated == false)
        {
            wstr = converter.from_bytes(str.c_str());
        }
        wstrUpdated = true;
        wstr += ch;
        strUpdated = false;
    }
    else
    {
        if (wstrUpdated == true)
        {
            wstr += ch;
            strUpdated = false;
        }
    }
    return *this;
}

MutableString& MutableString::operator += (const char ch)
{
    if (wstrUpdated == true)
    {
        if(strUpdated == false)
        {
            str = converter.to_bytes(wstr.c_str());
        }
        strUpdated = true;
        str += ch;
        wstrUpdated = false;
    }
    else
    {
        if (strUpdated == true)
        {
            str += ch;
            wstrUpdated = false;
        }
    }
    return *this;
}

std::string MutableString::getMultiByteString()
{
    if (strUpdated == false && wstrUpdated == true)
    {
        //change
        strUpdated = true;
        str = converter.to_bytes(wstr.c_str());
        return str;
    }
    else
    {
        if (strUpdated == false)
            return 

---

查看完整错误可能会有所帮助，但我猜wstring\u convert不可复制。@RetiredInja是的，我在本地代码范围中创建了“wstring\u convert”，而不是您所说的类成员，它工作得很好。但是仍然令人困惑。MutableString:：MutableStringconst std:：string&&other是毫无意义的。把它扔掉。您可以添加MutableString:：MutableStringstd:：string&&other，但您需要学习移动语义以充分利用它。@R.MartinhoFernandes是的，您是对的。我认为，如果我犯了错误并修改了“std:：string&&other”，那么添加“const”会使代码更安全。。但是当我开始思考时…即使修改了右值也不会对任何东西产生影响。。