我正在尝试打印以下代码返回的值:
Agent** Grid::GetAgent(int x, int y)
{
return &agents[x][y];
}它返回一个双指针,并打印
std::cout << *grid.GetAgent(j, k) << endl; 给出了内存位置,但当我尝试
std::cout << **grid.GetAgent(j, k) << endl; 我得到了错误
main.cpp:53: error: no match for ‘operator<<’ in ‘std::cout << * * grid.Grid::GetAgent(j, k)’如何打印*grid.GetAgent(j,k)中的值?
下面是Agent.h
#ifndef AGENT_H
#define AGENT_H
enum AgentType { candidateSolution, cupid, reaper, breeder};
class Agent
{
public:
Agent(void);
~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和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;
}发布于 2012-10-15 19:56:53
您需要定义一个函数ostream& operator<<(ostream&, const Agent&)
ostream& operator<<(ostream& out, const Agent& x)
{
// your code to print x to out here, e.g.
out << (int)x.GetType() << ' ' << x.GetFitness() << ' ' << x.GetAge() << '\n';
return out;
}C++不能神奇地打印Agent,您必须告诉它是如何打印的。
https://stackoverflow.com/questions/12895058
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