C#中如何读取一个任意文件然后以二进制的形式输出为文本

[fonlan]

一个WinForm的编程，界面上就一个简单的文件打开对话框和一个RichText窗口，现在要求把对话框中打开的任意文件转换为原始的二进制编码（即0101的形式）后在RichText中显示出来，我是个初学者，不知道怎么搞哦，大家帮下我啊。

[王集鹄]

参考如下代码：

//using System.IO;

private void button1_Click(object sender, EventArgs e)
{
    switch (openFileDialog1.ShowDialog())
    {
        case DialogResult.Yes:
        case DialogResult.OK:
            break;
        default:
            return;
    }

    FileStream fileStream = new FileStream(openFileDialog1.FileName, 
        FileMode.Open, FileAccess.Read);
    byte[] buffer = new byte[fileStream.Length];
    fileStream.Read(buffer, 0, buffer.Length);
    StringBuilder result = new StringBuilder(0x1000);
    fileStream.Close();

    for (int i = 0; i < buffer.Length; i++)
    {
        result.Append(Convert.ToString(buffer[i] + 0x100, 2).Substring(1, 8));
        result.Append(" ");
    }
    richTextBox1.Text = result.ToString();
}

[etmsg]

不知道你这样显示的意义是什么，以下代码供你参考
private string ByteToBit(byte d)
{
string sTmp = "";
byte b = d;

for (int i = 0; i < 8; i++)
{
if ((b & 0x80) == 0x80)
sTmp += "1 ";
else
sTmp += "0 ";
b <<= 1;
}
return sTmp;

}

private void File()
{

byte[] buf;
int dLen = 0;
string sTmp = "";
FileStream fs = new FileStream(@"c:\aa.txt", FileMode.Open, FileAccess.Read, FileShare.Read);
BinaryReader br = new BinaryReader(fs);

try
{
dLen = (int)fs.Length;
buf = br.ReadBytes(dLen);

for (int i = 0; i < buf.Length; i++)
{
sTmp += ByteToBit(buf[i])+"\t"+Convert.ToString(buf[i],16).PadLeft(2,'0');
this.textBox1.Text += sTmp;
this.textBox1.Text += System.Environment.NewLine;
sTmp = "";
}

}
finally
{
br.Close();
fs.Close();
}

}

[fonlan]

请问那如何只取文件起始的4096个二进制字符，并把它们写入一个4096位的数组中呢？

[王集鹄]

byte[] buffer = new byte[4096];