构造函数已声明为explicit 为什么复制构造函数里还是可以隐式转换哪？解决方案

#include <iostream>
#include <string>
using namespace std;

class Employee
{
private:
    string name;
    int ID;
    static int cnt;
public:
    Employee() {}
 explicit Employee(string nam)
    {
        cnt++;
        name=nam;
        ID=cnt;
    }
    Employee(const Employee& obj)
    {
        name=obj.name;
        ID=obj.ID;
    }
    Employee& operator=(const Employee &rhs)
    {
        name=rhs.name;
        ID=rhs.ID;
        return *this;
    }
    void printf()
    {
        cout<<ID<<" "<<name<<endl;
    }
    
};

int Employee::cnt=0;

int main()
{
    Employee emp("adam");
    Employee emp1=emp;
    string na="bill";
    Employee emp2(na);
    
    emp.printf();
    emp1.printf();
    emp2.printf();
    return 0;
}

void TestFunc(const Employee& e)
{
    //...
}
string name = "tom";
TestFunc(name);//编译不通过，Employee的转换构造函数必须要显式使用
TestFunc(Employee(name));//编译通过