c++ - 私有(private)继承和隐式转换

Question

我有一个从std::string 私有(private)继承的类，并添加了一些函数。我希望能够像 std::string 一样使用此类，因此我试图定义一个隐式转换运算符 (operator string())。但是，我不断收到 inaccessible base 错误。

#include <iostream>
#include <string>

using namespace std;
class Test:private string {
    int _a;
    public:
    operator string() {
        return "hello";
    }
};

int main() {
    Test t;
    if(t == "hello") {
        cout<<"world\n";
    }
}

错误:

trial.cpp: In function ‘int main()’:
trial.cpp:15:13: error: ‘std::basic_string<char>’ is an inaccessible base of ‘Test’
 if(t == "hello") {
         ^

问题:

1. 定义这样的转换是不是一个坏主意？这是否违反了任何推荐的编程实践？
2. 我怎样才能让它发挥作用？

编辑:Clang 更有帮助

trial.cpp:8:5: warning: conversion function converting 'Test' to its base class 'std::basic_string<char>' will never be used
    operator string() {
    ^
trial.cpp:15:8: error: cannot cast 'Test' to its private base class 'basic_string<char, std::char_traits<char>, std::allocator<char> >'
    if(t == "hello") {
       ^
trial.cpp:5:12: note: declared private here
class Test:private string {
       ^~~~~~~~~~~~~~

Answer 1

私有(private)基类的转换函数违背了私有(private)继承的目的。这就是标准禁止这种行为的部分原因:

A conversion function is never used to convert a (possibly cv-qualified) object to the (possibly cv-qualified) same object type (or a reference to it), to a (possibly cv-qualified) base class of that type (or a reference to it), or to (possibly cv-qualified) void.

如果你想访问基类，你应该公开继承。这使代码对于可能需要维护它的其他人来说是可读的和全面的。