std::accumulate数据上溢，求解？

Question

std::vector<uchar> vChannel(hsv.rows * hsv.cols);
// std::vector<double> ivChannel(hsv.rows * hsv.cols);
for (int y = 0; y < hsv.rows; ++y) {
      for (int x = 0; x < hsv.cols; ++x) {
            vChannel[y * hsv.cols + x] = hsv.at<cv::Vec3b>(y, x)[2];
            // ivChannel[y * hsv.cols + x] = (double)vChannel[y * hsv.cols + x];
       }
}
double vAvg = static_cast<double>(std::accumulate(vChannel.begin(), vChannel.end(), 0)) / vChannel.size();
// double vAvg = static_cast<double>(std::accumulate(ivChannel.begin(), ivChannel.end(), 0)) / ivChannel.size();

 std::cout << vAvg << "\t ";

不论是用上面未注释的代码，还是注释掉的代码，vAvg都会返回负值。
已验证vChannel中所有均为正数。
求问为啥上溢？std::accumulate有隐式数据类型转换？

Answer 1

已解决，是我忘记修改accumulate的第三个参数了，uchar应该改成0LL，double应该改成0.0