数字转换器（以字符串格式大整数） [英] Numeric Converter (big integers in string format)

问题描述

我试图写一个类在C＃中，一个大数目（字符串格式）转换为任意数量的系统，我发现这个code在<一个href="http://stackoverflow.com/questions/2652760/how-to-convert-a-gi-normous-integer-in-string-format-to-hex-format-c">How到GI-normous整数（字符串格式）转换为十六进制格式？ （C＃）

I'm trying to write a class in C# that converts a big number (string format) to any number system, I found this code at How to convert a gi-normous integer (in string format) to hex format? (C#)

    var s = "843370923007003347112437570992242323";
    var result = new List<byte>();
    result.Add( 0 );
    foreach ( char c in s )
    {
        int val = (int)( c - '0' );
        for ( int i = 0 ; i < result.Count ; i++ )
        {
            int digit = result[i] * 10 + val;
            result[i] = (byte)( digit & 0x0F );
            val = digit >> 4;
        }
        if ( val != 0 )
            result.Add( (byte)val );
    }

    var hex = "";
    foreach ( byte b in result )
        hex = "0123456789ABCDEF"[ b ] + hex;

这code也适用于任何数字系统（2 ^ n个基本）与一些修改到code。

This code also works for any numeric system (2^n base) with a few modifications to the code.

现在的问题是，我不知道该算法的逻辑（for语句）。 是否有人可以解释一下这部分的code：

The problem is that I do not understand the logic of the algorithm (the for statement). Can someone please explain this part of the code:

for ( int i = 0 ; i < result.Count ; i++ )
{
    int digit = result[i] * 10 + val;
    result[i] = (byte)( digit & 0x0F );
    val = digit >> 4;
}
if ( val != 0 )
    result.Add( (byte)val );

为了使这个code转换，例如，一个字符串十进制为一个字符串的base64，我需要改变的面膜，因此可以计算出六位，而不是仅仅四为十六进制的系统，然后由6右移位到剩余添加到下一个字节。

In order to make this code to convert, for example, a string decimal to a string base64, I need to change the mask so it can calculate six bits, instead of just four for the hex system, and then right-shift digit by 6 to add the remaining to the next byte.

 result[i] = (byte)( digit & 0x03F );
 val = digit >> 6;  // 2^6 = 64

和最后只是改变它的外表表打印结果

and finally just change the look-up table to print the result

hex =
"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/" [ b ] + hex;

这是此行中的for循环的一个我不完全理解

It is this line in the for-loop the one I don't totally understand

int digit = result[i] * 10 + val;

这是这条线，随着循环，这样做对每个迭代结果的每个字节？而且最重要的，为什么呢？

What is this line, along with the loop, doing on each iteration to every byte of result? and most importantly, why?

推荐答案

也许，code是比较容易，如果你比较算法的简化版本，其中的整数可以重新psented在<$ P $明白code> INT 。然后结果将是一个32位整数，你会初始化为0，并为每一个数字，你会成倍的结果将现有的值 10，并添加下一个数字：

Perhaps the code is easier to understand if you compare the algorithm to the simpler version where the integer can be represented in an int. Then result would be a 32 bit integer which you would initialize to 0 and for each digit you would multiply the existing value of result with 10 and add the next digit:

int result = 0;
foreach ( char c in s ) {
  int val = (int)( c - '0' );
  result = 10*result + val;
}

有关大整数，你需要一个字节数组重新present要了解整数和循环是一种用10乘以未知长度的字节数组中的价值，同时还加入了数字值。由于乘以10不是一个简单的移位你需要额外的code。

For big integers you need a byte array to represent the integer and the loop that you want to understand is a way to multiply the value in the byte array of unknown length with 10 while also adding the digit value. Because multiplying by 10 is not a simple bit shift you need that extra code.

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