我如何用C重新present二进制数++（用于连接霍夫曼codeR）？ [英] How do I represent binary numbers in C++ (used for Huffman encoder)?

问题描述

我写我自己的霍夫曼EN codeR ，到目前为止我已经创建哈夫曼树用minHeap弹出了两个频率最低的节点，并链接到他们一个节点，然后推新节点退一（起泡，冲洗，重复，直到只有一个节点）。

I am writing my own Huffman encoder, and so far I have created the Huffman tree by using a minHeap to pop off the two lowest frequency nodes and make a node that links to them and then pushing the new node back one (lather, rinse, repeat until only one node).

所以，现在我已经创建了树，但我需要使用此树codeS分配给每个字符。我的问题是我不知道怎么了一些二进制重新presentation存储在C ++中。我记得读取无符号的字符是一个字节的标准，但我不确定。

So now I have created the tree, but I need to use this tree to assign codes to each character. My problem is I don't know how to store the binary representation of a number in C++. I remember reading that unsigned char is the standard for a byte, but I am unsure.

我知道我必须recusively遍历树，每当我打了一个叶子结点，我必须指派任何code是目前重新presenting路径相应的字符。

I know I have to recusively traverse the tree and whenever I hit a leaf node I must assign the corresponding character whatever code is current representing the path.

下面是我到目前为止有：

Here is what I have so far:

void traverseFullTree(huffmanNode* root, unsigned char curCode, unsigned char &codeBook){

    if(root->leftChild == 0 && root->rightChild == 0){ //you are at a leaf node, assign curCode to root's character
        codeBook[(int)root->character] = curCode;
    }else{ //root has children, recurse into them with the currentCodes updated for right and left branch
        traverseFullTree(root->leftChild, **CURRENT CODE SHIFTED WITH A 0**, codeBook );
        traverseFullTree(root->rightChild, **CURRENT CODE SHIFTED WITH A 1**, codeBook);
    }

    return 0;
}

codeBook是我的阵列有最多256个字符的codeS（为ASCII每个可能的字符）的地方，但我只打算实际分配codeS到出现值在树。

CodeBook is my array that has a place for the codes of up to 256 characters (for each possible character in ASCII), but I am only going to actually assign codes to values that appear in the tree.

我不知道这是否是我的遍历哈夫曼树的corrent方式，但这是什么，立刻似乎工作（虽然我没有测试过）。另外我怎么叫没有零或一整棵树的根的摆频功能（树的顶部）？

I am not sure if this is the corrent way to traverse my Huffman tree, but this is what immediately seems to work (though I haven't tested it). Also how do I call the traverse function of the root of the whole tree with no zeros OR ones (the very top of the tree)?

我应该使用一个字符串，而不是和追加到字符串，要么是零或1？

Should I be using a string instead and appending to the string either a zero or a 1?

推荐答案

由于计算机是二进制...在C / C ++的所有号码都已经二进制格式。

Since computers are binary ... ALL numbers in C/C++ are already in binary format.

int a = 10;

变量 A 是二进制数。

您想看看什么是位操作，运营商，如＆安培; | ＆LT;＆LT; ＆GT;方式＆gt;

What you want to look at is bit manipulation, operators such as & | << >>.

随着霍夫曼编码，您将下将数据打包成字节数组。

With the Huffman encoding, you would pack the data down into an array of bytes.

这是因为我已经编写的C很长一段时间，所以这是一个现成的袖口伪code ...

It's been a long time since I've written C, so this is an "off-the-cuff" pseudo-code...

完全未经测试 - 但应该给你正确的想法。

Totally untested -- but should give you the right idea.

char buffer[1000]; // This is the buffer we are writing to -- calc the size out ahead of time or build it dynamically as go with malloc/ remalloc.

void set_bit(bit_position) {
  int byte = bit_position / 8;
  int bit = bit_position % 8;

  // From http://stackoverflow.com/questions/47981/how-do-you-set-clear-and-toggle-a-single-bit-in-c
  byte |= 1 << bit;
}

void clear_bit(bit_position) {
  int byte = bit_position / 8;
  int bit = bit_position % 8;

  // From http://stackoverflow.com/questions/47981/how-do-you-set-clear-and-toggle-a-single-bit-in-c
 bite &= ~(1 << bit);
}


// and in your loop, you'd just call these functions to set the bit number.
set_bit(0);
clear_bit(1);

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