如何将列表从表单发送到另一个表单

Question

I have two forms in a Windows Forms project: Form1 and aCurso .

I'm trying to send a List with objects of a class called curso (I mean: List<curso> ) from Form1 to aCurso .

But Visual Studio show this:

Accessibility inconsistency: The type of parameter List<curso> is less accessible than the method aCurso.aCurso(List<curso>) .

So, here is the code from Form1 :

using System;
using System.Collections.Generic;
using System.ComponentModel;
using System.Data;
using System.Drawing;
using System.Linq;
using System.Text;
using System.Threading.Tasks;
using System.Windows.Forms;

namespace _18_05_18
{
    public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
        }
        List<curso> cursos = new List<curso>();
        private void btnAC_Click(object sender, EventArgs e)
        {
            Form f = new aCurso(cursos);
            f.Show();
        }
    }
}

---

Here's code from aCurso :

 using System; using System.Collections.Generic; using System.ComponentModel; using System.Data; using System.Drawing; using System.Linq; using System.Text; using System.Threading.Tasks; using System.Windows.Forms; namespace _18_05_18 { public partial class aCurso : Form { List<curso> cursos = new List<curso>(); public aCurso(List<curso> cursos) { InitializeComponent(); this.cursos = cursos; } } } 

---

Here's code from class curso :

 using System; using System.Collections.Generic; using System.Linq; using System.Text; using System.Threading.Tasks; namespace _18_05_18 { class curso { private string nombre; public curso(string nombre) { this.nombre = nombre; } } } 

Answer 1

You cannot expose a public method signature where some of the parameter types of the signature are not public. It wouldn't be possible to call the method from outside since the caller couldn't construct the parameters required.

All you have to do is make curso class public

public class curso
{
    private string nombre;
    public curso(string nombre)
    {
        this.nombre = nombre;
    }
}