双函数分割故障返回双重

时间:2012-10-29 12:41:45

标签: c++ function segmentation-fault double

我有双重功能

double Grid::getAverageNeighborhoodFitnessEvaluation(int agentPositionX, int agentPositionY)
{
  GetNeighbourhood(agentPositionX, agentPositionY,neighborhoodEvaluations);
  int neighborscount = 0;
  double fitnesssum = 0;
  double neighborfitness;
  double value;
  for (size_t i = 0; i < neighborhoodEvaluations.size(); ++i)
  {
    if ((*(neighborhoodEvaluations.at(i))) != NULL)
    {
      neighborfitness = (*(neighborhoodEvaluations.at(i)))->GetFitness();
      if(neighborfitness<0)
          neighborfitness=0;
      fitnesssum+=neighborfitness;
      neighborscount++;
    }
  }
  value = fitnesssum/neighborscount;
  return value;
}

GetNeighbourhood将一个定义类型(代理)的数组分配给neighborhoodEvaluations

*(neighborhoodEvaluations.at(i)))->GetFitness();返回一个double,表示数组中该点的值。这些都已经使用过,没有任何问题。

从我的main调用时(RealX和RealY是两个整数)

int currentFitness = getAverageNeighborhoodFitnessEvaluation(RealX,RealY);

始终有效

double currentFitness = getAverageNeighborhoodFitnessEvaluation(RealX,RealY);

导致分段错误

有谁知道可能会导致这种可能性和/或int可以采用什么值,但似乎不能达到双倍?

到目前为止,我已将错误追溯到我们的代理实施

Agent.cpp

#include "Agent.h"

Agent::Agent(void)
{
    m_age = 0;
    m_fitness = -1;
}

Agent::~Agent(void)
{
}

int Agent::GetAge()
{
    return m_age;
}

double Agent::GetFitness()
{
    return m_fitness;
}

void Agent::IncreaseAge()
{
    m_age++;
}

AgentType Agent::GetType()
{
    return m_type;
}

Agent.h

#ifndef AGENT_H
#define AGENT_H

enum AgentType { candidateSolution, cupid, reaper, breeder};

class Agent
{
public:
    Agent(void);
    virtual ~Agent(void);

    double GetFitness();
    int GetAge();
    void IncreaseAge();
    AgentType GetType();
    virtual void RandomizeGenome() = 0;

protected:
    double m_fitness;
    AgentType m_type;
private:
    int m_age;
};
#endif // !AGENT_H

似乎无法找到确切的问题

2 个答案:

答案 0 :(得分:0)

使用本文http://www.cs.cmu.edu/~gilpin/tutorial/快速开始使用gdb调试器。然后告诉我们哪条线产生分段故障。

答案 1 :(得分:0)

根据您对gdb调试器答案的评论,我看到您在空对象(GetFitness)上调用Agent::GetFitness (this=0x0)方法。这意味着neighborhoodEvaluations.at(i)返回一个空指针。 at()只检查越界,但如果开始放入数组的是空指针,at()将无法帮助你。为了防范这一点,你应该改变

if ((*(neighborhoodEvaluations.at(i))) != NULL)

if (neighborhoodEvaluations.at(i) != NULL)

如果neighborEvaluations不应包含空指针,则必须追踪getNeighborhood()将它们放在那里的原因。也许你正在为你的点集边缘的元素寻找越界的邻居?