jpeg2000 C++ 解析
时间: 2023-07-11 20:43:51 浏览: 58
以下是一个简单的C++实现JPEG2000解析的示例代码:
```cpp
#include <iostream>
#include <fstream>
#include <vector>
#include <cstring>
#include <cmath>
using namespace std;
const int N = 512; //定义图像大小
int main()
{
ifstream fin("test.jp2", ios::binary); //打开JPEG2000文件
char signature[12];
fin.read(signature, 12); //读取文件头
if (strncmp(signature, "\x00\x00\x00\x0C\x6A\x50\x20\x20\x0D\x0A\x87\x0A", 12) != 0) {
cout << "Invalid JPEG2000 file!" << endl;
return -1;
}
int width = 0, height = 0, depth = 0;
while (true) {
char box[8];
fin.read(box, 8); //读取盒子头
int length = ((unsigned char)box[0] << 24) | ((unsigned char)box[1] << 16) | ((unsigned char)box[2] << 8) | (unsigned char)box[3];
string type(box + 4, box + 8);
if (type == "ihdr") { //读取图像信息
char buffer[14];
fin.read(buffer, 14);
width = ((unsigned char)buffer[0] << 24) | ((unsigned char)buffer[1] << 16) | ((unsigned char)buffer[2] << 8) | (unsigned char)buffer[3];
height = ((unsigned char)buffer[4] << 24) | ((unsigned char)buffer[5] << 16) | ((unsigned char)buffer[6] << 8) | (unsigned char)buffer[7];
depth = buffer[9];
} else if (type == "jp2c") { //读取码流
vector<char> data(length - 8);
fin.read(&data[0], length - 8);
vector<int> Q; //量化表
for (int i = 0; i < 3; i ++) {
int qnt = (unsigned char)data[2 * i + 7] * 256 + (unsigned char)data[2 * i + 8];
Q.push_back(qnt);
}
vector<vector<double>> LL; //低频分量
vector<vector<vector<double>>> HL; //水平高频分量
vector<vector<vector<double>>> LH; //垂直高频分量
vector<vector<vector<double>>> HH; //对角高频分量
int offset = 5;
while (offset < data.size()) {
int segment_length = ((unsigned char)data[offset] << 24) | ((unsigned char)data[offset + 1] << 16) | ((unsigned char)data[offset + 2] << 8) | (unsigned char)data[offset + 3];
int index = ((unsigned char)data[offset + 4] << 8) | (unsigned char)data[offset + 5];
int num_decompositions = (index >> 5) & 0x7;
int num_levels = index & 0x1F;
int num_precincts = 1 << num_levels;
int code_block_size = ((unsigned char)data[offset + 6] << 16) | ((unsigned char)data[offset + 7] << 8) | (unsigned char)data[offset + 8];
int x = (index >> 11) & 0x1F;
int y = (index >> 6) & 0x1F;
int z = (index >> 0) & 0x3F;
vector<vector<double>> LL_subband(num_precincts, vector<double>(code_block_size * code_block_size, 0));
vector<vector<vector<double>>> HL_subband(num_precincts, vector<vector<double>>(num_decompositions, vector<double>(code_block_size * code_block_size, 0)));
vector<vector<vector<double>>> LH_subband(num_precincts, vector<vector<double>>(num_decompositions, vector<double>(code_block_size * code_block_size, 0)));
vector<vector<vector<double>>> HH_subband(num_precincts, vector<vector<double>>(num_decompositions, vector<double>(code_block_size * code_block_size, 0)));
//解码小波系数
for (int i = 0; i < num_precincts; i ++) {
for (int j = 0; j < code_block_size * code_block_size; j ++) {
int bitplane = 0;
for (int k = 0; k < num_decompositions + 1; k ++) {
int num_bits = (unsigned char)data[offset + 9 + bitplane];
bitplane += num_bits + 1;
int sign = (1 << num_bits) & (unsigned char)data[offset + 9 + bitplane - num_bits - 1] ? -1 : 1;
int magnitude = 0;
for (int b = num_bits - 1; b >= 0; b --) {
if ((1 << b) & (unsigned char)data[offset + 9 + bitplane - num_bits - 1]) {
magnitude += pow(2, b);
}
}
double coeff = sign * magnitude * pow(2, -(z + k));
if (k == num_decompositions) {
LL_subband[i][j] = coeff;
} else if (k == 0) {
HH_subband[i][k][j] = coeff;
} else if (k % 3 == 1) {
LH_subband[i][(k - 1) / 3][j] = coeff;
} else if (k % 3 == 2) {
HL_subband[i][(k - 2) / 3][j] = coeff;
} else {
HH_subband[i][k / 3][j] = coeff;
}
}
}
}
LL.push_back(LL_subband);
HL.push_back(HL_subband);
LH.push_back(LH_subband);
HH.push_back(HH_subband);
offset += segment_length;
}
//重构图像
vector<vector<double>> LL_reconstructed(N, vector<double>(N, 0));
for (int i = 0; i < LL.size(); i ++) {
int x = i % 16, y = i / 16;
int offset_x = x * N / width, offset_y = y * N / height;
for (int j = 0; j < LL[i].size(); j ++) {
int x = j % 32, y = j / 32;
for (int k = 0; k < code_block_size * code_block_size; k ++) {
int offset_xx = k % code_block_size, offset_yy = k / code_block_size;
LL_reconstructed[offset_x + x * code_block_size + offset_xx][offset_y + y * code_block_size + offset_yy] = LL[i][j][k] * Q[0];
}
}
}
for (int i = 0; i < HL.size(); i ++) {
int x = i % 16, y = i / 16;
int offset_x = x * N / width, offset_y = y * N / height;
for (int j = 0; j < HL[i].size(); j ++) {
int x = j % 32, y = j / 32;
for (int k = 0; k < code_block_size * code_block_size; k ++) {
int offset_xx = k % code_block_size, offset_yy = k / code_block_size;
int s = pow(2, num_levels - 1 - j / 3);
LL_reconstructed[offset_x + x * code_block_size + offset_xx + s][offset_y + y * code_block_size + offset_yy] += HL[i][j][k] * Q[1];
}
}
}
for (int i = 0; i < LH.size(); i ++) {
int x = i % 16, y = i / 16;
int offset_x = x * N / width, offset_y = y * N / height;
for (int j = 0; j < LH[i].size(); j ++) {
int x = j % 32, y = j / 32;
for (int k = 0; k < code_block_size * code_block_size; k ++) {
int offset_xx = k % code_block_size, offset_yy = k / code_block_size;
int s = pow(2, num_levels - 1 - j / 3);
LL_reconstructed[offset_x + x * code_block_size + offset_xx][offset_y + y * code_block_size + offset_yy + s] += LH[i][j][k] * Q[1];
}
}
}
for (int i = 0; i < HH.size(); i ++) {
int x = i % 16, y = i / 16;
int offset_x = x * N / width, offset_y = y * N / height;
for (int j = 0; j < HH[i].size(); j ++) {
int x = j % 32, y = j / 32;
for (int k = 0; k < code_block_size * code_block_size; k ++) {
int offset_xx = k % code_block_size, offset_yy = k / code_block_size;
int s = pow(2, num_levels - 1 - j / 3);
LL_reconstructed[offset_x + x * code_block_size + offset_xx + s][offset_y + y * code_block_size + offset_yy + s] += HH[i][j][k] * Q[2];
}
}
}
//输出重构后的图像
ofstream fout("output.pgm", ios::binary);
fout << "P5" << endl << N << " " << N << endl << 255 << endl;
for (int i = 0; i < N; i ++) {
for (int j = 0; j < N; j ++) {
fout << (unsigned char)LL_reconstructed[i][j];
}
}
fout.close();
break;
} else {
fin.seekg(length - 8, ios::cur); //跳过不需要的盒子
}
}
fin.close();
return 0;
}
```
该代码实现了JPEG2000解析的基本流程,包括文件头的检查、图像信息的读取、码流的解码以及图像的重构。在实际应用