Question

我试图扭转一种方法，但实际上我已经被困了几天了。我不知怎的再也无法绕过这段代码...似乎有些愚蠢的心理障碍。

public string HexValueToStringValue(string myString, int pattern = 196) {
  pattern = pattern % 0x100;

  string myReturnString = "";
  for (int x = 1; x <= myString.Length; x += 2)
  {
    myReturnString = myReturnString + Conversions.ToString(Strings.Chr(Convert.ToInt32(Strings.Mid(myString, x, 2), 0x10) ^ pattern));

    int test = Convert.ToInt32(Strings.Mid(myString, x, 2), 0x10) + 1;
    pattern = test % 0x100;
  }

  return myReturnString; 
}

我只想要与给定方法完全相反。我想使用给定的算法从常规字符串创建十六进制字符串。任何帮助将不胜感激。

猜猜现在是时候对这种方法进行一些重构了 - 但实际上我真的失去了对它的追踪......

Answer 1

我重构了你的代码并编写了相反的方法。

正如我在评论A ^ B == B ^ A和A ^ B ^ B == A中所写，所以test = test ^ pattern; test = pattern ^ test;没用，因为它等同于test = test ^ pattern; test = test ^ pattern，这相当于到test = test ^ pattern ^ pattern。

public static string HexValueToStringValue(string hex, int pattern = 196)
{
    pattern = pattern % 0x100;

    var sb = new StringBuilder(hex.Length / 2);

    for (int i = 0; i < hex.Length; i += 2)
    {
        string h = hex.Substring(i, 2);
        int val = Convert.ToInt32(h, 16);
        char ch = (char)(val ^ pattern);
        sb.Append(ch);

        pattern = (val + 1) % 0x100;
    }

    return sb.ToString();
}

public static string StringToHexValue(string str, int pattern = 196)
{
    pattern = pattern % 0x100;

    var sb = new StringBuilder(str.Length * 2);

    for (int i = 0; i < str.Length; i++)
    {
        char ch = str[i];
        int val = ch ^ pattern;
        string h = val.ToString("X2");
        sb.Append(h);

        pattern = (val + 1) % 0x100;
    }

    return sb.ToString();
}

反转HexValueToStringValue转换方法

1 个答案: