在您的卷积中，有两件事可能不会导致崩溃。第一个是样式：您使用x 迭代图像的行，我将其更多地描绘为y 位移，反之亦然。第二个是当您计算总和时，您不会在评估每个像素的内核（内部两个循环）之前重置变量sum = 0。相反，您在所有像素上累积 sum，可能最终导致整数溢出。虽然严格来说这是 UB，可能会导致崩溃，但这不是您面临的问题。
如果您愿意确认崩溃发生在第一个像素上（x = ksize/2、y = ksize/2），那么由于崩溃发生在从内核读取的第一个系数处，我怀疑您可能通过了“错误的事情” " 作为kernel。如前所述，kernel 是int**。对于 3x3 的内核大小，这意味着要正确调用此函数，您必须在堆或堆栈上分配 int* 的数组，其中存储了 3 个指向 int 数组的指针，每个数组有 3 个系数。如果您改为传递int[3][3] 数组，卷积函数将尝试将数组中的第一个或两个int 解释为指向int 的指针，如果不是，并尝试取消引用它以拉入系数。这很可能会导致段错误。
我也不知道您为什么要退回累计金额。这不是卷积的“传统”输出，但我猜你对输出图像的平均亮度感兴趣，这是合法的；在这种情况下，您应该使用单独且更宽的整数累加器（long 或 long long），最后将其除以输出中的像素数。
您可能从 Internet 上找到了 PGM 数据结构，例如 here。请允许我放弃这个最佳实践建议。在我的领域（计算机视觉）中，选择的计算机视觉库 OpenCV 确实不将矩阵表示为row 指向col 元素缓冲区的指针数组。相反，分配了一大块内存，在这种情况下，大小至少为image->row * image->col * sizeof(int)，但通常image->row * image->step * sizeof(int)，其中image->step 是image->col，四舍五入到下一个4 或16 的倍数。然后，只有一个保留单个指针，指向整个图像的基础的指针，但如果图像不连续，则必须保留一个额外的字段（步骤）。
因此我会修改你的代码：
/* Includes */
#include <stdlib.h>
/* Defines */
#define min(a, b) (((a) < (b)) ? (a) : (b))
#define max(a, b) (((a) > (b)) ? (a) : (b))
/* Structure */
/**
* Mat structure.
*
* Stores the number of rows and columns in the matrix, the step size
* (number of elements to jump from one row to the next; must be larger than or
* equal to the number of columns), and a pointer to the first element.
*/
typedef struct Mat{
int rows;
int cols;
int step;
int* data;
} Mat;
/* Functions */
/**
* Allocation. Allocates a matrix big enough to hold rows * cols elements.
*
* If a custom step size is wanted, it can be given. Otherwise, an invalid one
* can be given (such as 0 or -1), and the step size will be chosen
* automatically.
*
* If a pointer to existing data is provided, don't bother allocating fresh
* memory. However, in that case, rows, cols and step must all be provided and
* must be correct.
*
* @param [in] rows The number of rows of the new Mat.
* @param [in] cols The number of columns of the new Mat.
* @param [in] step The step size of the new Mat. For newly-allocated
* images (existingData == NULL), can be <= 0, in
* which case a default step size is chosen; For
* pre-existing data (existingData != NULL), must be
* provided.
* @param [in] existingData A pointer to existing data. If NULL, a fresh buffer
* is allocated; Otherwise the given data is used as
* the base pointer.
* @return An allocated Mat structure.
*/
Mat allocMat(int rows, int cols, int step, int* existingData){
Mat M;
M.rows = max(rows, 0);
M.cols = max(cols, 0);
M.step = max(step, M.cols);
if(rows <= 0 || cols <= 0){
M.data = 0;
}else if(existingData == 0){
M.data = malloc(M.rows * M.step * sizeof(*M.data));
}else{
M.data = existingData;
}
return M;
}
/**
* Convolution. Convolves input by the given kernel (centered) and stores
* to output. Does not handle boundaries (i.e., in locations near the border,
* leaves output unchanged).
*
* @param [in] input The input image.
* @param [in] kern The kernel. Both width and height must be odd.
* @param [out] output The output image.
* @return Average brightness of output.
*
* Note: None of the image buffers may overlap with each other.
*/
int convolution(const Mat* input, const Mat* kern, Mat* output){
int i, j, x, y;
int coeff, data;
int sum;
int avg;
long long acc = 0;
/* Short forms of the image dimensions */
const int iw = input ->cols, ih = input ->rows, is = input ->step;
const int kw = kern ->cols, kh = kern ->rows, ks = kern ->step;
const int ow = output->cols, oh = output->rows, os = output->step;
/* Kernel half-sizes and number of elements */
const int kw2 = kw/2, kh2 = kh/2;
const int kelem = kw*kh;
/* Left, right, top and bottom limits */
const int l = kw2,
r = max(min(iw-kw2, ow-kw2), l),
t = kh2,
b = max(min(ih-kh2, oh-kh2), t);
/* Total number of pixels */
const int totalPixels = (r-l)*(b-t);
/* Input, kernel and output base pointers */
const int* iPtr = input ->data;
const int* kPtr = kern ->data + kw2 + ks*kh2;
int* oPtr = output->data;
/* Iterate over pixels of image */
for(y=t; y<b; y++){
for(x=l; x<r; x++){
sum = 0;
/* Iterate over elements of kernel */
for(i=-kh2; i<=kh2; i++){
for(j=-kw2; j<=kw2; j++){
data = iPtr[j + is*i + x];
coeff = kPtr[j + ks*i ];
sum += data * coeff;
}
}
/* Compute average. Add to accumulator and store as output. */
avg = sum / kelem;
acc += avg;
oPtr[x] = avg;
}
/* Bump pointers by one row step. */
iPtr += is;
oPtr += os;
}
/* Compute average brightness over entire output */
if(totalPixels == 0){
avg = 0;
}else{
avg = acc/totalPixels;
}
/* Return average brightness */
return avg;
}
/**
* Main
*/
int main(int argc, char* argv[]){
/**
* Coefficients of K. Binomial 3x3, separable. Unnormalized (weight = 16).
* Step = 3.
*/
int Kcoeff[3][3] = {{1, 2, 1}, {2, 4, 2}, {1, 2, 1}};
Mat I = allocMat(1920, 1080, 0, 0);/* FullHD 1080p: 1920x1080 */
Mat O = allocMat(1920, 1080, 0, 0);/* FullHD 1080p: 1920x1080 */
Mat K = allocMat( 3, 3, 3, &Kcoeff[0][0]);
/* Fill Mat I with something.... */
/* Convolve with K... */
int avg = convolution(&I, &K, &O);
/* Do something with O... */
/* Return */
return 0;
}
参考：多年计算机视觉经验。