如何在LINQ C#编程中同时使用Take和Skip运算符
我们正在创建Employee类的两个实例e和e1。e分配给e1。这两个对象都指向同一个参考，因此对于所有Equals，我们将得到预期的输出。
在第二种情况下，即使属性值相同，我们也可以观察到。等于返回false。本质上，当参数引用不同的对象时。等于不检查值，并且始终返回false。
例子1
class Program{
static void Main(string[] args){
Employee e = new Employee();
e.Name = "Test";
e.Age = 27;
Employee e2 = new Employee();
e2 = e;
var valueEqual = e.Equals(e2);
Console.WriteLine(valueEqual);
//第二种情况
Employee e1 = new Employee();
e1.Name = "Test";
e1.Age = 27;
var valueEqual1 = e.Equals(e1);
Console.WriteLine(valueEqual1);
Console.ReadLine();
}
}
class Employee{
public int Age { get; set; }
public string Name { get; set; }
}输出结果
True False
例子2
class Program{
static void Main(string[] args){
Employee e = new Employee();
e.Name = "Test";
e.Age = 27;
Employee e2 = new Employee();
e2 = e;
var valueEqual = e.Equals(e2);
Console.WriteLine(valueEqual);
Employee e1 = new Employee();
e1.Name = "Test";
e1.Age = 27;
var valueEqual1 = e.Equals(e1);
Console.WriteLine(valueEqual1);
Console.ReadLine();
}
}
class Employee{
public int Age { get; set; }
public string Name { get; set; }
public override bool Equals(object? obj){
if (obj == null)
return false;
if (this.GetType() != obj.GetType()) return false;
Employee p = (Employee)obj;
return (this.Age == p.Age) && (this.Name == p.Name);
}
public override int GetHashCode(){
return Age.GetHashCode() ^ Name.GetHashCode();
}
}输出结果
True True