VB6 的 Base64 编码的 C# 版本

Question

我需要使用 Base64 对字符串进行编码。我有一个 VB6 算法，我需要准确地编写一个 C# 版本。我能答对 99%，但最后一个字符是错误的，我不确定到底出了什么问题。

这是要编码的字符串:

¼9ÀPʨ!Œaøc4vøÝFãgïÒ_£¡˜è/[t•Ôý56wBpïcÃ.Cð`

VB6 编码给出以下字符串:

vDnAUMqoFiGMYfhjNHb43UbjZ+/SX6MBoZjoL1t0ldT9NTZ3QnDvYwMTww4uQ/Ag

C# 编码给出以下字符串:

vDnAUMqoFiGMYfhjNHb43UbjZ+/SX6MBoZjoL1t0ldT9NTZ3QnDvYwMTww4uQ/Ax

如您所见，除了最后一个字母之外，所有内容都是相同的。

VB6 算法:

Const sBASE_64_CHARACTERS = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/"


Function Base64encode(ByVal asContents)
    Dim lnPosition
    Dim lsResult
    Dim Char1
    Dim Char2
    Dim Char3
    Dim Char4
    Dim Byte1
    Dim Byte2
    Dim Byte3
    Dim SaveBits1
    Dim SaveBits2
    Dim lsGroupBinary
    Dim lsGroup64

    If Len(asContents) Mod 3 > 0 Then asContents = asContents & String(3 - (Len(asContents) Mod 3), " ")
    lsResult = ""

    For lnPosition = 1 To Len(asContents) Step 3
        lsGroup64 = ""
        lsGroupBinary = Mid(asContents, lnPosition, 3)

        Byte1 = Asc(Mid(lsGroupBinary, 1, 1)): SaveBits1 = Byte1 And 3
        Byte2 = Asc(Mid(lsGroupBinary, 2, 1)): SaveBits2 = Byte2 And 15
        Byte3 = Asc(Mid(lsGroupBinary, 3, 1))

        Char1 = Mid(sBASE_64_CHARACTERS, ((Byte1 And 252) \ 4) + 1, 1)
        Char2 = Mid(sBASE_64_CHARACTERS, (((Byte2 And 240) \ 16) Or (SaveBits1 * 16) And &HFF) + 1, 1)
        Char3 = Mid(sBASE_64_CHARACTERS, (((Byte3 And 192) \ 64) Or (SaveBits2 * 4) And &HFF) + 1, 1)
        Char4 = Mid(sBASE_64_CHARACTERS, (Byte3 And 63) + 1, 1)
        lsGroup64 = Char1 & Char2 & Char3 & Char4

        lsResult = lsResult + lsGroup64
    Next

    Base64encode = lsResult
End Function

这是 C# 的等效内容:

public const string sBASE_64_CHARACTERS = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
    public static string Base64encode(string asContents)
    {
        int lnPosition;
        string lsResult;
        string Char1;
        string Char2;
        string Char3;
        string Char4;
        int Byte1;
        int Byte2;
        int Byte3;
        int SaveBits1;
        int SaveBits2;
        string lsGroupBinary = null;
        string lsGroup64 = null;

        if (Microsoft.VisualBasic.Strings.Len(asContents) % 3 > 0)
            asContents = asContents + (3 - (Microsoft.VisualBasic.Strings.Len(asContents) % 3));
        lsResult = "";

        for (lnPosition = 1; lnPosition <= Microsoft.VisualBasic.Strings.Len(asContents); lnPosition += 3)
        {
            lsGroup64 = "";
            lsGroupBinary = Microsoft.VisualBasic.Strings.Mid(asContents, lnPosition, 3);

            Byte1 = Microsoft.VisualBasic.Strings.Asc(Microsoft.VisualBasic.Strings.Mid(lsGroupBinary, 1, 1));
            SaveBits1 = Byte1 & 3;
            Byte2 = Microsoft.VisualBasic.Strings.Asc(Microsoft.VisualBasic.Strings.Mid(lsGroupBinary, 2, 1));
            SaveBits2 = Byte2 & 15;
            Byte3 = Microsoft.VisualBasic.Strings.Asc(Microsoft.VisualBasic.Strings.Mid(lsGroupBinary, 3, 1));

            Char1 = Microsoft.VisualBasic.Strings.Mid(sBASE_64_CHARACTERS, ((Byte1 & 252) / 4) + 1, 1);
            Char2 = Microsoft.VisualBasic.Strings.Mid(sBASE_64_CHARACTERS, (((Byte2 & 240) / 16) | (SaveBits1 * 16) & 0xff) + 1, 1);
            Char3 = Microsoft.VisualBasic.Strings.Mid(sBASE_64_CHARACTERS, (((Byte3 & 192) / 64) | (SaveBits2 * 4) & 0xff) + 1, 1);
            Char4 = Microsoft.VisualBasic.Strings.Mid(sBASE_64_CHARACTERS, (Byte3 & 63) + 1, 1);
            lsGroup64 = Char1 + Char2 + Char3 + Char4;

            lsResult = lsResult + lsGroup64;
        }

        return lsResult;
    }
}

我做错了什么？

Answer 1

原始代码用空格填充字符串，使长度为 3 的整数倍。翻译后的代码没有正确执行此操作。此行翻译不正确:

If Len(asContents) Mod 3 > 0 Then asContents = asContents & String(3 - (Len(asContents) Mod 3), " ")

你拥有它

if (Microsoft.VisualBasic.Strings.Len(asContents) % 3 > 0)
    asContents = asContents + (3 - (Microsoft.VisualBasic.Strings.Len(asContents) % 3));

但应该是

if (Microsoft.VisualBasic.Strings.Len(asContents) % 3 > 0)
    asContents = asContents + Microsoft.VisualBasic.Strings.String(3 - (Microsoft.VisualBasic.Strings.Len(asContents) % 3), " ");

您拥有的代码不是附加空格，而是附加一个数字。