为什么Guid.ToString（" n"）与从相同guid的字节数组生成的十六进制字符串相同？ [英] Why isn't Guid.ToString("n") the same as a hex string generated from a byte array of the same guid?

问题描述

  [TestMethod] 
 public void TestByteToString（）
 { 
 var guid = new Guid（61772f3ae5de5f4a8577eb1003c5c054）; 
 var guidString = guid.ToString（n）; 
 var byteString = ToHexString（guid.ToByteArray（））; 
 
 Assert.AreEqual（guidString，byteString）; 
 
 $ b private String ToHexString（Byte [] bytes）
 {
 var hex = new StringBuilder（bytes.Length * 2）; 
 foreach（var b以字节为单位）
 {
 hex.AppendFormat（{0：x2}，b）; 
} 
 return hex.ToString（）; 
} 
  

结果如下：

Assert.AreEqual失败。预期：其中61772f3ae5de5f4a8577eb1003c5c054取代。实际：< 3a2f7761dee54a5f8577eb1003c5c054>。

解决方案

好了，他们是一样的，字节。前四个是相同的，只是以相反的顺序。

基本上，当从字符串创建时，它被假定为big-endian格式：最高字节在左边。但是，当存储在内部（在Intel-ish机器上）时，字节排序为little-endian：最高位字节在右边。

Consider the following unit test:

    [TestMethod]
    public void TestByteToString()
    {
        var guid = new Guid("61772f3ae5de5f4a8577eb1003c5c054");
        var guidString = guid.ToString("n");
        var byteString = ToHexString(guid.ToByteArray());

        Assert.AreEqual(guidString, byteString);
    }

    private String ToHexString(Byte[] bytes)
    {
        var hex = new StringBuilder(bytes.Length * 2);
        foreach(var b in bytes)
        {
            hex.AppendFormat("{0:x2}", b);
        }
        return hex.ToString();
    }

Here's the result:

Assert.AreEqual failed. Expected:<61772f3ae5de5f4a8577eb1003c5c054>. Actual:<3a2f7761dee54a5f8577eb1003c5c054>.

解决方案

Well, they are the same, after the first 4 bytes. And the first four are the same, just in the reverse order.

Basically, when created from the string, it's assumed to be in "big-endian" format: Highest byte to the left. However, when stored internally (on an Intel-ish machine), the bytes are ordered "little-endian": highest order byte to the right.

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