c - 读取位置位置 0x1D5C4C2F 访问冲突

Question

该函数在读取原始像素值时引发访问冲突，我不明白为什么。 可以将其视为我正在运行的代码的唯一部分，我已经单独运行了此代码并得到了相同的结果。

string filenames[]={"firstclick.raw", "secondclick.raw","thirdclick.raw","fourthclick.raw","fifthclick.raw","sixthclick.raw","seventhclick.raw","eighthclick.raw"};
FILE *file;
int height= 750, width = 453, bbp=3;

unsigned char ****images;
images = (unsigned char ****)malloc(sizeof(unsigned char ***)*8);
for(int j = 0; j<8; j++){
    images[j] = (unsigned char ***)malloc(sizeof(unsigned char**)*height);
    for(int i = 0; i<height; i++){
        images[j][i]= (unsigned char **)malloc(sizeof(unsigned char*)*width);
        for(int k = 0; k<bbp; k++)
            images[j][i][k]= (unsigned char *)malloc(sizeof(unsigned char)*bbp);
    }
}

for (int i = 0; i<8; i++){
    if (!(file=fopen(filenames[i].c_str(),"rb"))){
        cout << "Cannot open file: "<<filenames[i].c_str() <<endl;
        exit(1);
    }
    fread(images[i], sizeof(unsigned char), height*width*bbp, file);
    fclose(file);
}

Answer 1

这里的问题是您已将数组的每个元素分配为单独的数组(内存中的其他位置，其位置保留为指针)。但是当您读入时，您会假设它是一个连续的 block 。您将覆盖所有这些指针，并溢出缓冲区以启动。

如果您希望图像成为一组离散的内存块，请像这样分配:

unsigned char ** images;
int i;

images = malloc( sizeof(unsigned char *) * 8 );
for( i = 0; i < 8; i++ ) {
    images[i] = malloc( width * height * bpp );
}

请注意，sizeof(unsigned char) 由标准定义为始终为 1。您无需始终乘以 sizeof(unsigned char) .

现在，要获取图像中的像素地址，您需要相乘(通常是行优先):

unsigned char * pixel = images[i] + (y * width + x) * bpp;
unsigned char r = pixel[0];
unsigned char g = pixel[1];
unsigned char b = pixel[2];