使用此代码时,在c#中索引越界

Question

如果我有1到100位数,我应该得到输出

1--100 
2--99
 3--98 
.  
.. 
.. 
49---50 

代码低于其给定索引的范围,数组不需要很多维度

static void Main(string[] args)
{
    //// A. 2D array of strings.
    string[][] a = new string[100][];

    int bound0 = a.GetUpperBound(0);
    int bound1 = a.GetUpperBound(1);
    for (int i = 0; i <= bound0; i++)
    {
        for (int x = 100; x <= bound1; x--)
        {
            string s1 = a[i][x];
            Console.WriteLine(s1);
        }
    }
    Console.WriteLine();
    Console.ReadKey();
}

Answer 1

您需要为数组提供第二个维度.在内部循环中,您将递减loop变量而不是递增,这也会导致超出范围的异常.您可能需要知道锯齿状和二维数组之间的区别.该职位将解释这一点.这个语句int bound1 = a.GetUpperBound(1); 由于尚未声明第二个维度,因此给出了异常.

使用锯齿状数组.

string[][] a = new string[100][];
int bound0 = a.GetUpperBound(0);
for(int i = 0; i <= bound0; i++)
a[i] = new string[3];

for (int i = 0; i <= bound0; i++)
{
        int bound1 = a[i].GetUpperBound(0);
        for (int x = 0; x <= bound1; x++)
        {
            a[i][x] = (i + x).ToString();
            string s1 = a[i][x];
            Console.WriteLine(s1);
        }
 }

使用二维数组.

string[,] a = new string[100,4];
int bound0 = a.GetUpperBound(0);
int bound1 = a.GetUpperBound(1);
for (int i = 0; i < bound0; i++)
{
    for (int x = 0; x < bound1; x++)
    {
        a[i,x] = (i+x).ToString();
        string s1 = a[i,x];
        Console.WriteLine(s1);
    }
}
Console.WriteLine();
Console.ReadKey();

根据更新进行编辑

string[][] a = new string[100][];
int bound0 = a.GetUpperBound(0);
for(int i = 0; i <= bound0; i++)
a[i] = new string[100];

for (int i = 0; i <= bound0; i++)
{
        int bound1 = a[i].GetUpperBound(0);
        for (int x = bound1; x >= 0; x--)
        {
            a[i][x] = (i+1).ToString() +"--"+ (x+1).ToString();
            string s1 = a[i][x];
            Console.WriteLine(s1);
        }
 }