2维的字符数组定义问题！就是 char[,] = {} ?

Question

我想定义一个2维的字符数组，如果c语言应该是这样定义：char[][] ch = 
{  
   "",
   "",
   "ABC",
}请注意，前2行的值我要定义为空。       public static char[,] ch = 
       {
            {}, //错误
            {'A','B','C'},
            {'D','E','F'},
            {'G','H','I'},
       };在c#里我定义不了空的字符，请问应该怎么样定义？谢谢！

Answer 1

我试过{' '},这样 也是提示错误信息
       public static char[,] ch = 
       {
            {' '},
            {'A','B','C'},
            {'D','E','F'},
            {'G','H','I'},
       };

Answer 2

       public static char[,] ch = 
       {
            {' ',' ',' '},
            {'A','B','C'},
            {'D','E','F'},
            {'G','H','I'}
       };
或
       public static char[,] ch = 
       {
            {'\0','\0','\0'},
            {'A','B','C'},
            {'D','E','F'},
            {'G','H','I'}
       };

Answer 3

好像找到原因了，我这样定义可以：       public static char[,] ch = 
       {
            {' ',' ',' '},
            {' ',' ',' '},
            {'A','B','C'},
            {'D','E','F'},
            {'G','H','I'},
            {'J','K','L'},
       }这样的话每个字符的长度是一样的，但是我不想每个字符长度一样，该怎么处理？

Answer 4

是不是每一行的字符数不一样？可以试试嵌套数组
public static char[][] ch = new char[3][];
ch[2] = new char[]{...};
一二行用的时候再定义.

Answer 5

不一样为什么不用string[]呢？

Answer 6

        char[][] ch = new Char[5][];        ch[0] = new Char[]{' '};
        ch[1] = new Char[]{' ',' ',' '};
        ch[2] = new Char[]{'A','B','C'};
        ch[3] = new Char[]{'D','E','F'};
        ch[4] = new Char[]{'G','H','I'};

Answer 7

char[,] ch = {{' ',' '},{'A','B','C'},{'D','E','F'},{'G','H','I'}};