计算c#中二进制文件符号的频率不起作用但是对等效的c ++代码起作用

Sss*_*Sss 1 c# c++ algorithm binaryfiles linked-list

我正在尝试计算二进制文件中符号的频率c#,我已经在c ++中完成了,并且工作正常,我已经从c ++切换到c#因为我必须在c#中实现相同的.

注意:我不必使用LUT/Arrays只能使用链表.

通过frequency我的意思是符号的重复,并以数字symbols我的意思是如果我们看到使用的二进制文件xxd -b BinaryFile.bin,然后我们会得到很多8 bits的组合01.那么每个符号重复的次数是它的频率.

现在,我是如何尝试这样做的:c#通过mono filename.exe BinaryFile.bin在终端上用notepad ++中的文件名编写代码来实现它.

逻辑: 我读取BinaryFile中的每个符号,如果没有重复,那么我将它添加到链表的尾部,如果它重复,那么我增加它的频率.我重复完整的二进制文件.

码:

代码为c#(完整):(哪些不能正常工作,我已经指出问题包含代码的一部分,我把完整的代码,因为可能你需要它):

  ////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 ////////
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c ++和c#的输出之间的差异:

(1)当我显示c ++的输出时它正常工作,当我看到我在代码中编写的代码部分来调试/检查终端上的输出时,显示代码的正确执行.而c#中的相同调试代码与c ++没有显示相同的输出.它应该是这样做的,因为打印"symbol"和"freq"的这些代码都保存在程序中的相同位置.

(2)c#的输出让我觉得循环在c#中执行的次数少于c ++,因为在c#输出终端中没有显示大量重复的freq和符号.请看两者的输出:

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/
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而输出c++是:(这里计数器的值实际上从"0"开始(不是"196")但由于文件较大而无法显示完整输出,因此输出大的终端无法显示全部,它只显示最后的输出)

    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
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      freq:  50
      sym:  2
      freq:  48
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      sym:  4
      freq:  9
    counter: 205
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      freq:  50
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      freq:  48
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      freq:  10
    counter: 206
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      freq:  48
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      freq:  11
    counter: 207
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      freq:  48
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      freq:  12
    counter: 208
      sym:  0
      freq:  50
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      freq:  48
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    counter: 209
      sym:  0
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问题:

(1)为什么c#代码显示的输出与c ++不同?即使我尝过相同的BinaryFile(在我的情况下也是out.bin).

(2)你能帮我解决这个问题吗?十分感谢

Mat*_*son 5

我可能完全误解了这个......你只是想知道指定文件中每个字节值(0..255)的出现次数吗?

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]);
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然而,这比你正在做的事情简单得多,我觉得我一定是误会......


[编辑]

我无法修复您的原始代码,但这是一个使用链表来解决问题的示例程序:

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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  • @ user234839我明白了 - 这是作业呢?我根本不会使用链表来解决这类问题; 你可以做LB2建议的事情,但它有点作弊;)你无法在C#中学习指针 - 它不会使用它们(除非在特殊和不寻常的情况下).C#使用引用代替. (2认同)