使用 libjpeg 在 C++ 中旋转 JPEG 图像

我正在尝试根据 EXIF 元数据中存在的“方向”参数，使用 libjpeg v9 在 C++ 中旋转 JPEG 图像。我能够获取“方向”参数，并在此基础上，我还能够将图像旋转到另一个文件中，以便旋转后的图像对应于“方向”值 1。

请参阅我从“jpegtran.c”文件中获取的代码并且工作正常（不存在读取 EXIF 元数据代码）：

#include <iostream>
#include <jpeglib.h>
#include <jerror.h>

#include "transupp.h"

void setTransformation(jpeg_transform_info *transformObj, JXFORM_CODE transformation){
    transformObj->perfect = FALSE;
    transformObj->trim = FALSE;
    transformObj->force_grayscale = FALSE;
    transformObj->crop = FALSE;
    transformObj->transform = transformation;
}

void releaseRes(j_decompress_ptr srcPtr, j_compress_ptr destPtr){
    
    jpeg_finish_compress(destPtr);
    jpeg_destroy_compress(destPtr);

    (void) jpeg_finish_decompress(srcPtr);
    jpeg_destroy_decompress(srcPtr);
}

void rotateImage(const char *inputFilename, const char *outputFilename, JXFORM_CODE transformVal){

    FILE *inputFile = fopen(inputFilename, "r");
    if(inputFile==NULL){
        std::cerr<<"ERROR: cannot open input file\n";
        return;
    }

    struct jpeg_decompress_struct srcObj;
    struct jpeg_error_mgr srcErrMgr;
    
    struct jpeg_compress_struct destObj;
    struct jpeg_error_mgr destErrMgr;

    jvirt_barray_ptr *srcCoefArr;
    jvirt_barray_ptr *destCoefArr;

    //transformation object
    jpeg_transform_info transformObj;

    //set error handler
    srcObj.err = jpeg_std_error(&srcErrMgr);
    jpeg_create_decompress(&srcObj);

    destObj.err = jpeg_std_error(&destErrMgr);
    jpeg_create_compress(&destObj);

    //set the transformation properties
    setTransformation(&transformObj, transformVal);

    jpeg_stdio_src(&srcObj, inputFile);
    JCOPY_OPTION copyOpt = JCOPYOPT_DEFAULT;

    jcopy_markers_setup(&srcObj, copyOpt);

    (void) jpeg_read_header(&srcObj, TRUE);

    if(!jtransform_request_workspace(&srcObj, &transformObj)){
        std::cerr<<"Transformation is not perfect\n";
        return;
    }

    srcCoefArr = jpeg_read_coefficients(&srcObj);
    jpeg_copy_critical_parameters(&srcObj, &destObj);

    destCoefArr = jtransform_adjust_parameters(&srcObj, &destObj, srcCoefArr, &transformObj);

    FILE *outputFile = fopen(outputFilename, "wb");
    if(outputFile==NULL){
        std::cerr<<"ERROR: cannot open output file\n";
        fclose(inputFile);    
        releaseRes(&srcObj, &destObj);
        return;
    }

    jpeg_stdio_dest(&destObj, outputFile);

    jpeg_write_coefficients(&destObj, destCoefArr);

    jcopy_markers_execute(&srcObj, &destObj, copyOpt);

    jtransform_execute_transformation(&srcObj, &destObj, srcCoefArr, &transformObj);

    releaseRes(&srcObj, &destObj);
    //close files
    fclose(inputFile);
    fclose(outputFile);
}

但是，我不想将旋转图像存储到另一个文件中，而是想就地旋转到缓冲区或使用临时缓冲区，但不进行压缩，如上面的代码所示。

下面是将解压后的数据放入缓冲区的代码：

void rotateImage(const char *filename){
    FILE *file = fopen(filename, "r");
    if(!file){
        std::cerr<<"Error in reading file\n";
        return;
    }

    struct jpeg_decompress_struct info;
    struct jpeg_error_mgr jerr;
    
    info.err = jpeg_std_error(&jerr);

    jpeg_CreateDecompress(&info, JPEG_LIB_VERSION, (size_t) sizeof(struct jpeg_decompress_struct));
        
    jpeg_stdio_src(&info, file);

    (void) jpeg_read_header(&info, TRUE);
    jpeg_start_decompress(&info);

    uint32_t channels = 3;
    uint32_t rowStride = info.output_width * channels;
    uint64_t dataSize = rowStride * info.output_height;

    unsigned char *buffer = new unsigned char[dataSize];
    unsigned char *rowData[1];

    while(info.output_scanline < info.output_height){
        //initial value of output_Scanline state var is 0
        rowData[0] = buffer + info.output_scanline * rowStride; 
        jpeg_read_scanlines(&info, rowData, 1);
    }

    /*Now, i want to rotate this buffer (or with other temp buffer without compression as in 
     first code) as per "orientation", either 90, 180, 270*/

    /* here */

    jpeg_finish_decompress(&info);
    jpeg_destroy_decompress(&info);
    fclose(file);
    delete buffer;
}

不过，我尝试使用临时缓冲区（类似于非方矩阵的矩阵旋转）通过以下代码旋转缓冲区 90 度：

//90 degree clockwise
unsigned char *tmpBuf = new unsigned char[dataSize];
int row = info.output_height;
int col = info.output_width;
for(int i=0; i<row; i+=1){
    for(int j=0;j<col; j+=1){
        //copied 3 bytes as each pixed takes 3 bytes for RGB
        memcpy(tmpBuf + (j*row + row-i-1)*3, buffer + (i*col + j)*3, 3);
    }
}

但是，我相信，这不是旋转 JPEG 的正确方法，因为我将此数据发送到的应用程序不接受旋转的数据（仅供参考，我按照应用程序尊重的“方向”旋转它）。这让我相信这不是旋转 JPEG 图像的正确方法。与第一种方法一样，我要向其发送数据的应用程序接受首先旋转为压缩数据，然后再次解压缩到缓冲区。但是，我认为，这不是更好的方法。

所以，我需要你的帮助。请让我知道实现它所需的步骤。任何代码示例或教程也会有所帮助。

Thanks