为了复试，重新从零翻开了一遍C++程序设计基础，看到函数指针，才想到这个地方我一开始学的时候就没学懂，趁这次机会把它搞懂。直接上代码，注释很详细：

#include 
        
         using namespace std;typedef int (*funType)(int, int);typedef int funType2(int, int);typedef int *(*pfunType)(int, int);typedef int *(pfunType2)(int, int);int funA(int a, int b){	cout << "调用了 int funA()" << endl;	return a + b;}int *funB(int a, int b){	cout << "调用了int *funB()" << endl;	int *p = new int;	*p = a - b;	return p;}void callFunA(funType* cfun, int a, int b)//这里传递的是funType类型的指针{	cout << (*cfun)(a, b) << endl;// ok,这里是解指针调用	//cout << cfun(a, b) << endl; //error 明显调用表达式前的括号必须具有（指针）函数类型}void callFunA2(funType cfun, int a, int b)//这里传递的是funType类型变量{	cout << (*cfun)(a, b) << endl;//ok,这里是解指针调用	cout << cfun(a, b) << endl;//ok,这里直接调用}void callFunB(pfunType* pcfun, int a, int b){	int *p = (*pcfun)(a, b);	//int *p = pcfun(a, b);//error 明显调用表达式前的括号必须具有（指针）函数类型	cout << *p<< endl;	delete p;}void callFunB2(pfunType2 pcfun, int a, int b){	int *p1 = (*pcfun)(a, b);//ok	int *p2 = pcfun(a, b);//ok	cout << *p1 << endl;	cout << *p2 << endl;	delete p1;	delete p2;}int main(){	int a = 5, b = 8;	funType fT = funA; //ok	//funType* fT_p = &funA;//error 类型参数不兼容	//funType2 fT2 = funA;//error 未能初始化fT2，怎么能初始化fT2？，难道funType2 就不能用来声明函数对象？	pfunType pfT = funB;//0k	//pfunType2 pfT2 = funB;//error 同fT2	callFunA(&fT, a, b);//0k	//callFunA(funA, a, b);//error 类型参数不兼容，尝试直接传递funA函数入口地址失败	callFunA2(fT, a, b);//ok		callFunB(&pfT, a, b);//ok	//callFunB(funB, a, b);//error 类型参数不兼容	callFunB2(pfT, a, b);//ok	callFunB2(funB, a, b);//ok	system("pause");	return 0;}