thinking in cpp 中 的一个小问题
#include <iostream>
using namespace std;

class Dog {
public:
int run(int i) const {
cout < < "run\n ";
return i;
}
int eat(int i) const {
cout < < "eat\n ";
return i;
}
int sleep(int i) const {
cout < < "ZZZ\n ";
return i;
}
typedef int (Dog::*PMF)(int) const;
// operator-> * must return an object
// that has an operator():
class FunctionObject {
Dog* ptr;
PMF pmem;
public:
// Save the object pointer and member pointer
FunctionObject(Dog* wp, PMF pmf)
: ptr(wp), pmem(pmf) {
cout < < "FunctionObject constructor\n ";
}
// Make the call using the object pointer
// and member pointer
int operator()(int i) const {
cout < < "FunctionObject::operator()\n ";
return (ptr-> *pmem)(i); // Make the call
}
};
FunctionObject operator-> *(PMF pmf) {
cout < < "operator-> * " < < endl;
return FunctionObject(this, pmf);
}
};

int main() {
Dog w;
Dog::PMF pmf = &Dog::run;
cout < < (w-> *pmf)(1) < < endl;
pmf = &Dog::sleep;
cout < < (w-> *pmf)(2) < < endl;
pmf = &Dog::eat;
cout < < (w-> *pmf)(3) < < endl;
} ///:~


typedef int (Dog::*PMF)(int) const; 这句代码是什么意思
还有int operator()(int i) const {
cout < < "FunctionObject::operator()\n ";
return (ptr-> *pmem)(i); // Make the call
}


FunctionObject operator-> *(PMF pmf) {

cout < < "operator-> * " < < endl;
return FunctionObject(this, pmf);
}
这两段代码不懂

[解决办法]
C++ Primer上找“指向成员的指针”