如何在C＃中的十六进制数和十进制数之间转换？

解决方案

回答

String stringrep = myintvar.ToString("X");

int num = int.Parse("FF", System.Globalization.NumberStyles.HexNumber);

回答

看来你可以说

Convert.ToInt64(value, 16)

从十六进制获取小数。

另一种方法是：

otherVar.ToString("X");

回答

从Geekpedia：

// Store integer 182
int decValue = 182;

// Convert integer 182 as a hex in a string variable
string hexValue = decValue.ToString("X");

// Convert the hex string back to the number
int decAgain = int.Parse(hexValue, System.Globalization.NumberStyles.HexNumber);

回答

十六进制->十进制：

Convert.ToInt64(hexValue, 16);

十进制->十六进制

string.format("{0:x}", decValue);

回答

要将十进制转换为十六进制，请执行以下操作...

string hexValue = decValue.ToString("X");

要将十六进制转换为十进制，请执行以下任一操作：

int decValue = int.Parse(hexValue, System.Globalization.NumberStyles.HexNumber);

或者

int decValue = Convert.ToInt32(hexValue, 16);

回答

static string chex(byte e)                  // Convert a byte to a string representing that byte in hexadecimal
    {
        string r = "";
        string chars = "0123456789ABCDEF";
        r += chars[e >> 4];
        return r += chars[e &= 0x0F];
    }           // Easy enough...

    static byte CRAZY_BYTE(string t, int i)     // Take a byte, if zero return zero, else throw exception (i=0 means false, i>0 means true)
    {
        if (i == 0) return 0;
        throw new Exception(t);
    }

    static byte hbyte(string e)                 // Take 2 characters: these are hex chars, convert it to a byte
    {                                           // WARNING: This code will make small children cry. Rated R.
        e = e.ToUpper(); // 
        string msg = "INVALID CHARS";           // The message that will be thrown if the hex str is invalid

        byte[] t = new byte[]                   // Gets the 2 characters and puts them in seperate entries in a byte array.
        {                                       // This will throw an exception if (e.Length != 2).
            (byte)e[CRAZY_BYTE("INVALID LENGTH", e.Length ^ 0x02)], 
            (byte)e[0x01] 
        };

        for (byte i = 0x00; i < 0x02; i++)      // Convert those [ascii] characters to [hexadecimal] characters. Error out if either character is invalid.
        {
            t[i] -= (byte)((t[i] >= 0x30) ? 0x30 : CRAZY_BYTE(msg, 0x01));                                  // Check for 0-9
            t[i] -= (byte)((!(t[i] < 0x0A)) ? (t[i] >= 0x11 ? 0x07 : CRAZY_BYTE(msg, 0x01)) : 0x00);        // Check for A-F
        }           

        return t[0x01] |= t[0x00] <<= 0x04;     // The moment of truth.
    }

回答

这并不是真正最简单的方法，但是此源代码使我们可以纠正任何类型的八进制数，即23.214、23和0.512等。希望这能够帮到你..

public string octal_to_decimal(string m_value)
    {
        double i, j, x = 0;
        Int64 main_value;
        int k = 0;
        bool pw = true, ch;
        int position_pt = m_value.IndexOf(".");
        if (position_pt == -1)
        {
            main_value = Convert.ToInt64(m_value);
            ch = false;
        }
        else
        {
            main_value = Convert.ToInt64(m_value.Remove(position_pt, m_value.Length - position_pt));
            ch = true;
        }

        while (k <= 1)
        {
            do
            {
                i = main_value % 10;                                        // Return Remainder
                i = i * Convert.ToDouble(Math.Pow(8, x));                   // calculate power
                if (pw)
                    x++;
                else
                    x--;
                o_to_d = o_to_d + i;                                        // Saving Required calculated value in main variable
                main_value = main_value / 10;                               // Dividing the main value 
            }
            while (main_value >= 1);
            if (ch)
            {
                k++;
                main_value = Convert.ToInt64(Reversestring(m_value.Remove(0, position_pt + 1)));
            }
            else
                k = 2;
            pw = false;
            x = -1;
        }
        return (Convert.ToString(o_to_d));
    }