如果我编译下面的代码,我会得到一个
“错误C2064:术语不评估为采用2个参数的函数”
doOpp()
方法中的。如果我为q = f(p, i);
更改q = add(p, i);
,则所有内容编译都很正常(add()
方法中的doOpp()
是第二个参数)。我尝试了几种解决方案但似乎没有任何解决方案。
template <class T = double>
class Points {
public:
Points(void);
Points(Point3d<T> *, int);
Points(T *, int);
Points(Points &);
~Points(void);
void clear();
int size();
void push_back(Point3d<T> &);
typename vector<Point3d<T>>::iterator begin(void){return pnts.begin();};
typename vector<Point3d<T>>::iterator end(void){return pnts.end();};
void assign(Point3d<T> *, int);
void assign(T *, int);
void assign(Points &);
Points valid(void);
Points valid(int *);
bool setValid(int, bool);
int validSize();
// Pointer to operator function prototype
typedef Point3d<T> (__thiscall Points<T>::*oppFunc) (Points<T> &, int);
Points operator=(Points &);
// + Operator method
Points operator+(Points &p){return doOpp(p, &Points<T>::add);};
Points operator-(Points &){return doOpp(p, &Points<T>::sub);};
Points operator*(Points &){return doOpp(p, &Points<T>::mlt);};;
private:
Points(int);
void assign(int);
int assign(Points &, Points &);
vector<Point3d<T>> pnts;
bool isSameSize(Points<T> &, char *);
// General purpose operator function
Points doOpp(Points<T> &, oppFunc);
// Addition method called by doOpp
Point3d<T> add(Points &p, int i) {return pnts[i] + p.pnts[i];}
Point3d<T> sub(Points &p, int i) {return pnts[i] - p.pnts[i];}
Point3d<T> mlt(Points &p, int i) {return pnts[i] * p.pnts[i];}
};
template <class T> int
Points<T>::assign(Points &p1, Points &p2) {
int s = min(p1.size(), p2.size());
assign(s);
return s;
}
template <class T> bool
Points<T>::isSameSize(Points &p, char *s) {
bool same = (pnts.size() == p.pnts.size());
#ifndef POINTS_NO_THROW
if (!same) {
PointsException e;
e.setMessage((string) s + " source sizes not equal");
throw e;
}
#endif
return same;
}
template <class T> Points<T>
Points<T>::doOpp(Points &p, oppFunc f) {
isSameSize(p, __FUNCTION__);
Points<T> r;
int s = r.assign(*this, p);
Point3d<T> q;
for (int i = 0; i < s; i++) {
q = f(p, i); // Causes error C2064 but q = add(p, i); works fine
}
return r;
}
int
main () {
Points<> p, q;
q.assign(a, 4);
p = q + q;
return 1;
}
答案 0 :(得分:7)
由于oppFunc f
是成员函数指针,您应该使用特殊语法通过成员函数指针进行调用:
q = (this->*f)(p, i);
答案 1 :(得分:0)
应该这样:
Points operator-(Points &){return doOpp(p, &Points<T>::sub);};
不是:
Points operator-(Points &p){return doOpp(p, &Points<T>::sub);}; // note the p argument