刽子手游戏我试图投掷" ArgumentOutOfRangeException"当我尝试显示下划线时,在for循环中。我尝试重写代码,但没有任何效果。
List<Label> underline;
int left = 300;
int top = 275;
for (int i = 0; i < data.solution.Length; i++) //data is the object that contains the word that is the solution
{
underline = new List<Label>();
underline[i].Parent = this;
underline[i].Text = "_";
underline[i].Size = new Size(35, 35);
underline[i].Location = new Point(left, top);
left += 30;
underline[i].Font = new Font(new FontFamily("Microsoft Sans Serif"), 20, FontStyle.Bold);
}
我不明白这有什么问题。当我点击新游戏时,它立即抛出它。
答案 0 :(得分:4)
您在此处的每个for
循环迭代中实例化一个新的标签列表:
for (int i = 0; i < data.solution.Length; i++)
{
underline = new List<Label>();
...
}
您可能希望实现的是创建标签列表并对其进行修改。为此,您可以创建一个列表,然后创建标签并将它们逐个添加到此列表中:
var underline = new List<Label>(data.solution.Length);
for (int i = 0; i < data.solution.Length; i++)
{
var lbl = new Label();
lbl.Parent = ...
...
underline.Add(lbl);
}
另一个解决方案是使用LINQ:
var underline = data.solution.Select(x =
{
left += 30;
return new Label
{
Parent = this,
Text = "_",
...
}
}).ToList();
答案 1 :(得分:3)
首先不要为for循环中的每个迭代创建一个新列表,在循环之外创建它。其次,您必须先将新的Label
实例添加到列表中,然后才能通过它的索引访问它:
List<Label> underline = new List<Label>();
int left = 300;
int top = 275;
for (int i = 0; i < data.solution.Length; i++) //data is the object that contains the word that is the solution
{
underline.Add(new Label());
underline[i].Parent = this;
underline[i].Text = "_";
underline[i].Size = new Size(35, 35);
underline[i].Location = new Point(left, top);
left += 30;
underline[i].Font = new Font(new FontFamily("Microsoft Sans Serif"), 20, FontStyle.Bold);
}