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在C#中计算二进制文件的符号频率不起作用,但适用于等效的C ++代码

[英]Calculating frequency of symbols of binary file in c# not working but working for equivalent c++ code

我正在尝试计算c#中的二进制文件中符号的频率,我已经在c ++中做到了,而且效果很好,我已经从c ++切换到c#,因为我必须在c#中实现相同的频率。

注意:我不必使用LUT /数组,而仅需使用链表。

frequency是指符号的重复数, symbols是指如果我们使用xxd -b BinaryFile.bin查看二进制文件,那么我们将得到很多018 bits组合。 因此,每个符号重复多少次是它的频率。

现在,我如何尝试做到这一点:我在c#通过在终端使用mono filename.exe BinaryFile.bin并在notepad ++内用filename编写的代码来实现它。

逻辑:如果不重复,则读取BinaryFile中的每个符号,然后将其添加到链接列表的末尾;如果重复,则增加其频率。 我重复完整的二进制文件。

码:

c#的代码(完整):(无法正常工作,我指出了代码中包含部分内容的问题,我放入了完整代码,因为可能您会需要它):

  ////Problem containing part starts here ////////
        public Huffman(string[] args) //called from MyClass 
        {
            Node tree = null;
            int counter = 0;
            using(var stream = new BinaryReader(System.IO.File.OpenRead(args[0]))) 
            {
                while (stream.BaseStream.Position < stream.BaseStream.Length) 
                {
                    int processingValue = stream.ReadByte();
                    Node ppt, pt, temp;
                    bool is_there = false;
                    ppt = pt = tree;
                    while (pt != null) 
                    {
                        if (pt.symbol == processingValue) 
                        {
                            pt.freq++;
                            is_there = true;

                            break;
                        }
                        ppt = pt;
                        pt = pt.next;
                    }
                    if (is_there == false) 
                    {
                        temp = new Node();
                        temp.symbol = processingValue;
                        temp.freq = 1;
                        temp.left = null;
                        temp.right = null;
                        temp.next = null;
                        temp.id = (++total_nodes);
                        temp.is_processed = 0;
                        if (tree == null) 
                        {
                            tree = temp;
                        } 
                        else 
                        {
                            ppt.next = temp;
                        }
//The same check/debugging which i was doing in c++ to know what symbol and freq contains but they contains different values. 
//And the output of both c#/c++ are different where as it was supposed to be same.
                        Node chc = tree;
                        while (chc != null) 
                        {
                            Console.WriteLine("  sym: " + chc.symbol);
                            Console.WriteLine("  freq: " + chc.freq);
                            chc = chc.next;
                        }
                    }
                }
                stream.Close();
            }


        }
 ////Problem containing part Ends here ////////

C ++和c#的输出之间的区别:

(1)当我显示c ++的输出时,它可以正常工作;当我看到我在代码中编写的代码部分以调试/检查终端上的输出时,即表明代码正确执行。 而C#中相同的调试代码与C ++不会显示相同的输出。 之所以这样做,是因为这两个用于打印“符号”和“频率”的代码都保存在程序中的同一位置。

(2)C#的输出使我感到,虽然在c#中执行循环的次数比在c ++中执行循环的次数少,因为在c#输出终端中没有显示大量的频率和符号重复。 请查看两者的输出:

在终端的C#输出:

hp@ubuntu:~/Desktop/$ mono check1.exe out.bin 
  sym: 0
  freq: 1
  sym: 0
  freq: 200
  sym: 1
  freq: 1
  sym: 0
  freq: 200
  sym: 1
  freq: 198
  sym: 2
  freq: 1
  sym: 0
  freq: 200
  sym: 1
  freq: 198
  sym: 2
  freq: 195
  sym: 3
  freq: 1
  sym: 0
  freq: 200
  sym: 1
  freq: 198
  sym: 2
  freq: 195
  sym: 3
  freq: 189
  sym: 4
  freq: 1
hp@ubuntu:~/Desktop/

c++的输出是:(此处计数器的值实际上是从“ 0”开始的(不是“ 196”)),但是由于文件较大,所以输出无法显示完整的输出,因此终端很大,无法显示全部,它仅在最后显示输出)

    hp@ubuntu:~/Desktop/$ ./filename out.bin

  //Counter starts from "0" but terminal is not able to show all.So doing from "196"
    counter: 196
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  1
    counter: 197
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  2
    counter: 198
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  3
    counter: 199
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  4
    counter: 200
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  5
    counter: 201
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  6
    counter: 202
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  7
    counter: 203
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
      freq:  48
      sym:  4
      freq:  8
    counter: 204
      sym:  0
      freq:  50
      sym:  1
      freq:  50
      sym:  2
      freq:  48
      sym:  3
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      sym:  4
      freq:  9
    counter: 205
      sym:  0
      freq:  50
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      sym:  2
      freq:  48
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      freq:  10
    counter: 206
      sym:  0
      freq:  50
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      freq:  11
    counter: 207
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    counter: 208
      sym:  0
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    counter: 209
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    counter: 210
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    counter: 211
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    counter: 212
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    counter: 213
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    counter: 214
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    counter: 238
      sym:  0
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    counter: 239
      sym:  0
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      freq:  44

问题:

(1)为什么c#代码显示的输出与c ++不同? 甚至我也品尝过相同的BinaryFile(以我的情况为out.bin)。

(2)您能帮助我解决这个问题吗? 太谢谢了

我可能完全误解了……您是否只想知道指定文件中每个字节值(0..255)的出现次数?

如果是这样,一个简单的方法是这样的:

var counts = new int[256];  // Assumes files aren't longer than 2GB.

string filename = "<Your filename goes here>";

foreach (byte b in File.ReadAllBytes(filename)) // Will run out of memory
    ++counts[b];                                // for very large files!

for (int i = 0; i < counts.Length; ++i)
    Console.WriteLine("Symbol {0} occurred {1} times.", i, counts[i]);

但是,这比您正在做的事情简单得多,我觉得我一定会误会...。


[编辑]

我无法修复您的原始代码,但这是一个示例程序,该程序使用链接列表来解决问题:

using System;
using System.IO;

namespace ConsoleApp1
{
    public sealed class Node
    {
        public byte Symbol { get; set; }
        public int Count   { get; set; }
        public Node Next   { get; set; }
    }

    sealed class Program
    {
        private void run()
        {
            var linkedList = new Node();

            string filename = @"C:\Test\t.cs";

            foreach (byte symbol in File.ReadAllBytes(filename))
                addSymbol(symbol, linkedList);

            for (int symbol = 0; symbol < 256; ++symbol)
            {
                int count = countForSymbol((byte)symbol, linkedList);
                Console.WriteLine("Symbol {0} occurred {1} times.", symbol, count);
            }
        }

        private static void addSymbol(byte symbol, Node head)
        {
            Node last = head;

            while (head != null)
            {
                last = head;

                if (head.Symbol == symbol)
                {
                    ++head.Count;
                    return;
                }
                else
                {
                    head = head.Next;
                }
            }

            last.Next = new Node
            {
                Symbol = symbol, 
                Count = 1
            };
        }

        private int countForSymbol(byte symbol, Node head)
        {
            while (head != null)
            {
                if (head.Symbol == symbol)
                    return head.Count;
                else
                    head = head.Next;
            }

            return 0;
        }

        private static void Main()
        {
            new Program().run();
        }
    }
}

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