将数组转换为 IEnumable < T >

假设您有这样一个基本的 Employee类:

class Employee
{
public string Name;
public int Years;
public string Department;
}

然后(在一个单独的类中)我得到了以下代码片段(我想我只理解最后一个) :

我相信下面的代码片段可以工作，因为数组初始化程序创建了一个 Employee 对象数组，这些对象的类型与分配给它的工作量变量的类型相同。

Employee[] workforceOne = new Employee[] {
new Employee() { Name = "David", Years = 0, Department = "software" },
new Employee() { Name = "Dexter", Years = 3, Department = "software" },
new Employee() { Name = "Paul", Years = 4, Department = "software" } };

然后我有以下代码片段。我相信这是可行的，因为 Employee对象的数组是实现 IEnumerable的 Array ()类的实现。因此，我相信这就是为什么数组可以被分配给 IEnumable？

IEnumerable workforceTwo = new Employee[] {
new Employee() { Name = "David", Years = 0, Department = "software" },
new Employee() { Name = "Dexter", Years = 3, Department = "software" },
new Employee() { Name = "Paul", Years = 4, Department = "software" } };

然后我有这个代码片段:

IEnumerable<Employee> workforceThree = new Employee[] {
new Employee() { Name = "David", Years = 0, Department = "software" },
new Employee() { Name = "Dexter", Years = 3, Department = "software" },
new Employee() { Name = "Paul", Years = 4, Department = "software" } };

我不确定为什么这个代码片段有效？IEnumerable<Employee>继承自 IEnumerable(并重载(或重载?)GetEnumerator()方法) ，但是我不应该因此需要上述工作的铸造:

//The cast does work but is not required
IEnumerable<Employee> workforceFour = (IEnumerable<Employee>)new Employee[] {
new Employee() { Name = "David", Years = 0, Department = "software" },
new Employee() { Name = "Dexter", Years = 3, Department = "software" },
new Employee() { Name = "Paul", Years = 4, Department = "software" } };