Question

我想使用参数

将列表传递给函数

List<Employee> iList = new List<Employee>();
public void GetEmployeeData()
{
    for (int i = 0; i < 10; i++)
    {
         iList[i].age = 12;
         iList[i].Name = "Sandhya";
         iList[i].cadder = "A+";
         PassListData(iList[i].Name, iList[i].age, iList[i].cadder);
    }
}

void PassListData(string Name, int age, string Cadder)
{
    Console.WriteLine(Name);
    Console.WriteLine(age);
    Console.WriteLine(Cadder);
}

Answer 1

您可以直接将列表传递给下面的函数。首先将所有员工详细信息添加到列表中。然后将其作为列表传递

public void GetEmployeeData()
{
    var iList = new List<Employee>();
    for (int i = 0; i < 10; i++)
    {
         var employee = new Employee();             
         employee.age = 12;
         employee.Name = "Sandhya";
         employee.cadder = "A+";
         iList.Add(employee);
    }
    PassListData(iList);
}

void PassListData(List<Employee> employees)
{
    // loop through here
}

Answer 2

简单地：

List<Employee> iList = new List<Employee>();
public void GetEmployeeData()
{
    for (int i = 0; i < 10; i++)
    {
         iList[i].age = 12;
         iList[i].Name = "Sandhya";
         iList[i].cadder = "A+";
    }
    PassListData(iList);
}

void PassListData(List<Employee> list)
{
    foreach(var e in list)
    {
        Console.WriteLine(e.Name);
        Console.WriteLine(e.age);
        Console.WriteLine(e.cadder);
    }
}

Answer 3

工作example

如果列表是班级列表，那么你可以这样做

using System;
using System.Collections.Generic;

public class Program
{
public static void Main()
{

    List<Student> list =  new List<Student>();
    list.Add(new Student{name= "name1", age= 21, city= "city1"});
    list.Add(new Student{name= "name2", age= 22, city= "city2"});
    list.Add(new Student{name= "name3", age= 24, city= "city3"});

    getList(list);
}

static void getList(List<Student> list)
{
  foreach(var s in list)
  {
    Console.WriteLine("name = "+s.name + ", age= " + s.age +", city = "+ s.city);
  }

  }
}

public class Student{

    public string name  {set; get;}
    public int age  {set; get;}
    public string city  {set; get;}

}

结果

name = name1, age= 21, city = city1
name = name2, age= 22, city = city2
name = name3, age= 24, city = city3

如何将列表作为参数传递给函数

3 个答案: