Question

我在c ++中有一个随机森林的实现，我在matlab中通过mex运行。它运行平稳，直到它到达下面的功能，它会卡住并开始消耗内存，直到计算机冻结。

void MyFunction(
  const IDataPointCollection& data,
  std::vector<std::vector<int> >& leafNodeIndices,
  ProgressStream* progress=0 ) const
{
  ProgressStream defaultProgressStream(std::cout, Interest);
  progress = (progress==0)?&defaultProgressStream:progress;

  leafNodeIndices.resize(TreeCount());

  tbb::parallel_for<int>(0,TreeCount(),[&](int t)
  {
    leafNodeIndices[t].resize(data.Count());

    (*progress)[Interest] << "\rApplying tree " << t << "...";
    trees_[t]->Apply(data, leafNodeIndices[t]);
  });
  (*progress)[Interest] << "STUCK HERE" << std::endl;
  return;
}

通过上面trees_[t]->Apply()的代码，我能够将其缩小到下面的递归函数：

void ApplyNode(
  int nodeIndex,
  const IDataPointCollection& data,
  std::vector<unsigned int>& dataIndices,
  int i0,
  int i1,
  std::vector<int>& leafNodeIndices,
  std::vector<float>& responses_)
{
std::cout<<"applying node"<<std::endl;
  assert(nodes_[nodeIndex].IsNull()==false);

  Node<F,S>& node = nodes_[nodeIndex];

  if (node.IsLeaf())
  {
    for (int i = i0; i < i1; i++)
      leafNodeIndices[dataIndices[i]] = nodeIndex;
    return;
  }

  else if (i0 == i1)   // No samples left
    return;

  else 
  {
        for (int i = i0; i < i1; i++)
            responses_[i] = node.Feature.GetResponse(data, dataIndices[i]);

        int ii = Partition(responses_, dataIndices, i0, i1, node.Threshold);

        // Recurse for child nodes.
        ApplyNode(nodeIndex * 2 + 1, data, dataIndices, i0, ii, leafNodeIndices, responses_);
        ApplyNode(nodeIndex * 2 + 2, data, dataIndices, ii, i1, leafNodeIndices, responses_);
        return;
  }
    }

对递归函数的每次调用都有不同的计算时间，具体取决于node.Feature.GetResponse()函数。如果我使所有递归调用的计算时间相同（通过更改GetResponse()），代码将顺利运行。

    float AxisAlignedFeatureResponse::GetResponse(const IDataPointCollection& data, int index) const {

    double retArg;
    // retrieve DataManager object
    const DataManager& concreteData = (const DataManager&)(data);

//     // retrieve data point at index 
    DataPoint currDataPoint         = concreteData.getDataPoint(index);
//     
//     // get coordinates of data point
    Coordinate currCoordinates      = currDataPoint.getOrigPos();
//     
//     // get intensity image of the respective data point
    int imgIndex    = currDataPoint.getImageIndex();  
    Image currImg   = concreteData.getImage(imgIndex);    
    Image currFeatureImg = concreteData.getFeatureImage(imgIndex);
    // return respective feature
    int featureNumber = (int)(this->axis*(double)concreteData.getNumberOfFeatures());    
    if(featureNumber>=concreteData.getNumberOfFeatures()){
        cout<<"warning! trying to reach a feature that is not there!"<<endl;
        featureNumber=concreteData.getNumberOfFeatures()-1;
    }

    std::vector<Coordinate> feature = concreteData.getFeature(featureNumber);
    Coordinate tmp=currCoordinates+feature[0];
    if(feature[1].x == 0) {
        retArg = currCoordinates.x*feature[0].x+currCoordinates.y*feature[0].y+currCoordinates.z*feature[0].z;
    //retArg = 0;      //DOING THIS runs the code smoothly
    } 

    else if(feature[1].x == 2) {
       retArg = currFeatureImg.getValue(feature[0]);
    } else {
        retArg = currImg.mean(tmp,feature[1]);
    }
    return (float)(retArg);   
        //return (float) 0;
}

Answer 1

这看起来像是valgrind的工作。除非您的程序突然终止，否则我看不出任何导致内存泄漏的原因。内存泄漏通常是由未释放的堆变量引起的（“new”后面没有后续的“delete”）。

Valgrind将为您提供罪魁祸首和行号，您可以设置断点并逐步使用gdb查看究竟发生了什么。也许发布运行“valgrind -v your_program”的详细输出。不要忘记使用-g选项进行编译，以便为valgrind和gdb提供完整的调试数据。

内存泄漏可能是由于递归函数

1 个答案: