Why isn't Guid.ToString(“n”) the same as a hex string generated from a byte array of the same guid?

问题

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>.

回答1:

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.

回答2:

If you compare the results, you can see that the first three groups are reversed:

61 77 2f 3a   e5 de   5f 4a   8577eb1003c5c054
3a 2f 77 61   de e5   4a 5f   8577eb1003c5c054

That's because in the GUID structure, these 3 groups are defined as DWORD and two WORDs rather than bytes:

{0x00000000,0x0000,0x0000,{0x00,0x00,0x00,0x00,0x00,0x00,0x00,0x00}}

so in memory, an Intel processor stores them in Little-endian order (the most significant byte the last).

回答3:

A GUID is structured as follows:

int a
short b
short c
byte[8] d

So for the part represented by a your code gets the bytes reversed. All other parts are transformed correctly.

来源：https://stackoverflow.com/questions/3320501/why-isnt-guid-tostringn-the-same-as-a-hex-string-generated-from-a-byte-arra

标签

guid

hex