霍夫曼解码功能反复解压缩一个字符

Question

I have a program that produces a Huffman tree based on ASCII character frequency read in a text input file. The Huffman codes are stored in a string array of 256 elements, empty string if the character is not read. This program also encodes and compresses an output file.

I am now trying to decompress and decode my current output file which is opened as an input file and a new output file is to have the decoded message identical to the original text input file.

My thought process for this part of the assignment is to recreate a tree with huffman codes and then while reading 8 bits at a time, traverse through tree until I reach a leaf node where I will have updated an empty string(string answer) and then output it to my output file.

My problem: After writing this function I see that only one character in between all of the other characters of my original input file gets output repeatedly. I am confused as to why this is the case because I am expecting the output file to be identical to the original input file.

Any guidance or solution to this problem is appreciated.

(For encodedOutput function, fileName is the input file parameter, fileName2 is the output file parameter)

(For decodeOutput function, fileName2 is the input file parameter, fileName 3 is output file parameter)

code[256] is a parameter for both of these functions and holds the Huffman code for each unique character read in the original input file, for example, the character 'H' being read in the input file may have a code of "111" stored in the code array for code[72] at the time it is being passed to the functions.

freq[256] holds the frequency of each ascii character read or holds 0 if it is not in original input file.

void encodeOutput(const string & fileName, const string & fileName2, string code[256]) {
    ifstream ifile;
    ifile.open(fileName, ios::binary);
    if (!ifile)
    {
        die("Can't read again");
    }
    ofstream ofile;
    ofile.open(fileName2, ios::binary);
    if (!ofile) {
        die("Can't open encoding output file");
    }
    int read;
    read = ifile.get();
    char buffer = 0, bit_count = 0;
    while (read != -1) {
        for (unsigned b = 0; b < code[read].size(); b++) { // loop through bits (code[read] outputs huffman code)
            buffer <<= 1;
            buffer |= code[read][b] != '0';
            bit_count++;
            if (bit_count == 8) {
                ofile << buffer;
                buffer = 0;
                bit_count = 0;
            }
        }
        read = ifile.get();
    }

    if (bit_count != 0)
        ofile << (buffer << (8 - bit_count));

    ifile.close();
    ofile.close();
}
// Work in progress
void decodeOutput(const string & fileName2, const string & fileName3, string code[256], const unsigned long long freq[256]) {
    ifstream ifile;
    ifile.open(fileName2, ios::binary);
    if (!ifile)
    {
        die("Can't read again");
    }
    ofstream ofile;
    ofile.open(fileName3, ios::binary);
    if (!ofile) {
        die("Can't open encoding output file");
    }
    priority_queue < node > q;
    for (unsigned i = 0; i < 256; i++) {
        if (freq[i] == 0) {
            code[i] = "";
        }
    }

    for (unsigned i = 0; i < 256; i++)
        if (freq[i])
            q.push(node(unsigned(i), freq[i]));

    if (q.size() < 1) {
        die("no data");
    }

    while (q.size() > 1) {
        node *child0 = new node(q.top());
        q.pop();
        node *child1 = new node(q.top());
        q.pop();
        q.push(node(child0, child1));
    } // created the tree
    string answer = "";
    const node * temp = &q.top(); // root 
    for (int c; (c = ifile.get()) != EOF;) {
        for (unsigned p = 8; p--;) { //reading 8 bits at a time 
            if ((c >> p & 1) == '0') { // if bit is a 0
                temp = temp->child0; // go left
            }
            else { // if bit is a 1
                temp = temp->child1; // go right
            }
            if (temp->child0 == NULL && temp->child1 == NULL) // leaf node
            {
                ans += temp->value;
                temp = &q.top();
            }
            ofile << ans;
        }
    }
}

Answer 1

(c >> p & 1) == '0'

Will only return true when (c >> p & 1) equals 48, so your if statement will always follow the else branch. The correct code is:

(c >> p & 1) == 0