저는 C++에서 함수 포인터에 대해 배우고 있습니다. 다음 예제는 모두 예상 결과를 컴파일하고 반환하지만 예제 3은 갈 길이라고 배웠습니다. 왜 다른 예제들이 여전히 효과가 있습니까?C++의 함수 포인터 구문
이상한 것처럼 보인 다른 예가 위의 예제와 달리 모든 예제가 작동하는 f,g,h,i
입니다. 예제 1-8과 비교하면 왜 작동하지 않습니까?
int executeOperator1(int a, int b, int f(int,int)){
return f(a,b);
}
int executeOperator2(int a, int b, int f(int,int)){
return (*f)(a,b);
}
int executeOperator3(int a, int b, int (*f)(int,int)){
return f(a,b);
}
int executeOperator4(int a, int b, int (*f)(int,int)){
return (*f)(a,b);
}
int op(int x, int y){
return x+y;
}
int main(int argc, char *argv[])
{
int a = 2, b=3;
//the following 8 examples compile nicely:
cout << "a=" << a << " b=" << b << " res=" << executeOperator1(a,b,op) <<endl; //1
cout << "a=" << a << " b=" << b << " res=" << executeOperator2(a,b,op) <<endl; //2
cout << "a=" << a << " b=" << b << " res=" << executeOperator3(a,b,op) <<endl; //3
cout << "a=" << a << " b=" << b << " res=" << executeOperator4(a,b,op) <<endl; //4
cout << "a=" << a << " b=" << b << " res=" << executeOperator1(a,b,&op) <<endl; //5
cout << "a=" << a << " b=" << b << " res=" << executeOperator2(a,b,&op) <<endl; //6
cout << "a=" << a << " b=" << b << " res=" << executeOperator3(a,b,&op) <<endl; //7
cout << "a=" << a << " b=" << b << " res=" << executeOperator4(a,b,&op) <<endl; //8
//int f(int,int) = op; //does not compile
int (*g)(int,int) = op; //does compile
//int h(int,int) = &op; //does not compile
int (*i)(int,int) = &op;//does compile
return 0;
}