类函数模板 用函数做参数提示 student::bujige”: 非静态成员函数的非法调用

恒大蓝狮

#include <iostream>
#include <string>
class student;
template <typename T, class fun>
int my_count_if(T *a,int n,fun f) //f 是判断特征的函数
{
int count = 0;
for ( int i=0; i<n; ++i )
{
  if (a[i].f)
  {
   count ++;
  }
}
return count ;
}


class student
{
private:
int score;
std::string  name;
public:
student(int a=0,char* s="")
  :score(a),name(s){}
student (const student& s)
{
  if (this!=&s)
  {
   this->name=s.name;
   this->score=s.score;
  }
}
student &operator=(const student&rhs)
{
  if (this!=&rhs)
  {
  this->name=rhs.name;
  this->score=rhs.score;
  }
  return *this;
}
bool bujige() //定义不及格的函数
{
  return score<60;
}
  
};
int main()
{
student a[]={student(50,"tom"),student(49,"jerry"),student(91,"henry")};
int n=sizeof(a)/sizeof(*a);
int num=my_count_if(a,n,student::bujige);
}

求大神！！

恒大蓝狮

其中bujige是作为一个判定性质的函数