问个简单的 这次

Question

receive[i ]  这个数组里有数据 0x37 0x37 0x35 0x36想转为字符串 为 77 56  该怎么搞？就是把 ASC 转为  字符串

Answer 1

string st = "";
                        for (j = 0; j < 4; j++)
                        {
                                      st =st + receive[j].ToString ;
                             
                        }    
这样么？

Answer 2

怎么会不明白呢？
ASC 为的0x37   不就是  十进制的  7么？
 如果是 0x65 就转为 A 
如果是 0x97 就转为 a就是这个意思

Answer 3

string st = "";
                        for (j = 0; j < 4; j++)
                        {
                                      st =st + (receive[j]-48).ToString ;
                             
                        }    

Answer 4

string st = "";
                        for (j = 0; j < 4; j++)
                        {
                                      st =st + (receive[j]%16).ToString ;
                             
                        }    

Answer 5

string st = "";
                        for (j = 0; j < 4; j++)
                        {
                                      st =st + (receive[j]-48).ToString ;
                             
                        }    

Answer 6

0的Asc值是48，所以得先减掉48.同上。

Answer 7

agree 用个tostring 就搞定了

Answer 8

st =st + (receive[j]-48).ToString ;请问你 编译了没？Error 1 Operator '+' cannot be applied to operands of type 'string' and 'method group' C:\Documents and Settings\Administrator\Desktop\TestUDP\TestUDP\Form1.cs 405 34 TestUDP

Answer 9

string st = "";
                        for (j = 0; j < 4; j++)
                        {
                                      st =st + (receive[j]-48).ToString ;
                             
                        }    

Answer 10

string st = "";
                        for (j = 0; j < 4; j++)
                        {
                                      st =st + (receive[j]-48).ToString ;
                             
                        }    不可以的 难道没别的方法了

Answer 11

st =st + (receive[j]-48).ToString 
不好意思，写错了，改成ToString();就可以了。
我测试了如下：
string st = "";
            byte[] receive = new byte[4];
            receive[0] = 0x37;
            receive[1] = 0x37;
            receive[2] = 0x35;
            receive[3] = 0x36;
            int j;
            for (j = 0; j < 4; j++)
            {
                st = st + (receive[j] - 48).ToString();
            }
            Console.WriteLine(st);输出为“7756”

Answer 12

            string str;
            ASCIIEncoding s = new ASCIIEncoding();
            byte[] a = new byte[4];
            a[0] =0x37;
            a[1] =0x37;
            str = s.GetString(a );
            textBox1.Text = str;
唉 还是MSDN 好啊  看看帮助 还是有用的 
靠人不如靠自己啊

Answer 13

TO
liujia_0421()
 兄弟我试了你那个 是没问题但如果是0x68  0X69 就不对了 不是我想要的 D 和 E 了 
而我的哪个程序就可以了