Question

我有两个应用程序需要以快速的方式相互传递值，并且需要保留一些值（当我重新启动计算机时它仍然存在），所以我需要创建一个文件，现在我知道如何与int：

有关

    using (BinaryWriter writer = new BinaryWriter(new FileStream(@"C:\TEST", FileMode.Open)))
    {
        writer.Write(0);  //00 00 00 00
        writer.Write(1);  //01 00 00 00
        writer.Write(2);  //02 00 00 00
        writer.Write(3);  //03 00 00 00 
        writer.Write(int.MaxValue); //FF FF FF 7F
    }

    byte[] test = new byte[4];
    using (BinaryReader reader = new BinaryReader(new FileStream(@"C:\TEST", FileMode.Open)))
    {

        reader.BaseStream.Seek(8, SeekOrigin.Begin);
        reader.Read(test, 0, 4);
        Console.WriteLine(BitConverter.ToInt32(test, 0));     //2

        reader.BaseStream.Seek(16, SeekOrigin.Begin);
        reader.Read(test, 0, 4);
        Console.WriteLine(BitConverter.ToInt32(test, 0));     //2147483647    

        Console.Read();
    }

但如何处理double？

Answer 1

就像

一样简单

writer.Write((double)int.MaxValue);

BinaryWriter Class

Write(Double)将一个8字节的浮点值写入当前流并将流位置提前八个字节

至于阅读

reader.ReadDouble()

BinaryReader.ReadDouble Method ()

如何使用BinaryWriter写双值？

1 个答案: