c++ - "default definition would be ill-formed"是什么意思？

Question

我是 C++ 的初学者，我尝试创建一个对象，但遇到错误，而且我不明白哪里出了问题。这是我收到错误时的 hpp 文件 + cpp 文件:

Manager::Manager(const Manager &manager ) :
Worker(manager.name,manager.id,manager.salary){
   this->workers=manager.workers;
}

class Manager:public Worker {
    private:
    std::vector<Worker> workers;         
    public:
        Manager(const char* name, int id, int salary);
        Manager(const Manager &manager );      
};

错误:

In file included from Manager.hpp:5:0,
                 from Manager.cpp:1:
Worker.hpp:7:7: note: ‘Worker& Worker::operator=(const Worker&)’ is implicitly deleted because the default definition would be ill-formed:
 class Worker{

但是当我这样做时它起作用了:

Manager::Manager(const Manager &manager ) :
    Worker(manager.name,manager.id,manager.salary), workers(manager.workers){
               }

谁能告诉我为什么？

---

编辑:

这是 worker 类(Class)的代码:

这是worker.hpp

  class Worker{
        protected:
            const std::string name;
            const int id;
            int salary;

        public:
            Worker(const std::string& name, int id, int salary);
            Worker(const Worker& worker );
    };

这是 worker.cpp:

#include "Worker.hpp"

Worker::Worker(const std::string &names, int ids, int salarys) :
name(names), id(ids), salary(salarys)
{
}

Worker::Worker(const Worker &worker): 
name(worker.name), id(worker.id), salary(worker.salary)
{

}

Worker::~Worker() {

}

std::string Worker::toString(){
    std::string s="Worker id:" + id ;
    return s;
}

Answer 1

编译器通过对每个字段执行赋值生成的默认赋值运算符。因此，如果其中之一是 const 限定的或不可复制的，编译器将无法发出赋值运算符。 Worker 类(您应该在问题中发布)很可能包含一个或多个此类字段。

请注意，调用 this->workers=manager.workers; 将为存储在 worker 中的每个 Worker 调用复制赋值运算符，同时调用workers(manager.workers) 将调用定义明确的复制构造函数。