我正在开展一个小型图像处理项目 . 我想运行一个CUDA程序来进行图像相减 . 所以你有图像背景和背景相同的图像，但上面还有其他一些东西 . 一旦你减去了图像，你将得到剩下的东西 . 这两个图像都是480 * 360，我的gpu是GTX780 . 我的程序抛出错误 ./main': free(): invalid next size (normal): 0x000000000126bd70 *** Aborted (core dumped) 并且输出图像错误 . 我一直在竭尽全力解决这个问题 . 这是代码：
内核：
__global__ void add(unsigned char* a, unsigned char* b, unsigned char* c, int numCols, int numWidth) {
int i = blockIdx.x * blockDim.x + threadIdx.x; //Column
int j = blockIdx.y * blockDim.y + threadIdx.y; //Row
if(i < numWidth && j < numCols)
{
int idx = j * numCols + i;
c[idx] = b[idx] - a[idx];
}
}
和主要功能：
int main() {
CImg<unsigned char> img1("1.bmp");
CImg<unsigned char> img2("2.bmp");
//both images have the same size
int width = img1.width();
int height = img1.height();
int size = width * height * 3; //both images of same size
dim3 blockSize(16, 16, 1);
dim3 gridSize((width + blockSize.x - 1) / blockSize.x, (height + blockSize.y - 1) / blockSize.y, 1);
unsigned char *dev_a, *dev_b, *dev_c;
cudaMalloc((void**)&dev_a, size * (sizeof(unsigned char)));
cudaMalloc((void**)&dev_b, size * (sizeof(unsigned char)));
cudaMalloc((void**)&dev_c, size * (sizeof(unsigned char)));
cudaMemcpy(dev_a, img1, size * (sizeof(unsigned char)), cudaMemcpyHostToDevice);
cudaMemcpy(dev_b, img2, size * (sizeof(unsigned char)), cudaMemcpyHostToDevice);
add<<<gridSize, blockSize>>>(dev_a, dev_b, dev_c, height, width);
cudaMemcpy(img2, dev_c, size * (sizeof(unsigned char)), cudaMemcpyDeviceToHost);
img2.save("out.bmp");
cudaFree(dev_a);
cudaFree(dev_b);
cudaFree(dev_c);
return 0;
}
图像加载CImg库 .
1 回答
问题在于在主机代码中错误地使用cimg容器 . 根据documentation，图像数据指针通过
data()方法访问，这意味着主机代码中的cudaMemcpy调用应该提供img1.data()和img2.data().[这个答案来自评论并作为社区维基条目添加]