我们可以在 C 中使用双函数指针吗?
can we have a double function pointer in C?
我想知道与双指针不同 (int**)
,我们可以有双函数指针吗?
我的意思是函数指针指向另一个函数指针的地址?
我想要类似的东西
int add(int A , int B){
return A+B;
}
int main(void){
int (*funcpointerToAdd)(int,int) = add; // single function pointer pointing to the function add
printf("%d \n",funcpointerToAdd(2,3));
int (**doubleFuncPointerToAdd)(int,int) = &funcpointerToAdd;
printf("%d \n",doubleFuncPointerToAdd(2,3));
return 0;
}
但这给了我一个错误called object ‘doubleFuncPointerToAdd’ is not a function or function pointer
无论如何都可以做这件事吗?
您可以使用指向函数指针的指针,但您必须先对它们进行一次引用:
int add(int A , int B){
return A+B;
}
int main(void){
int (*funcpointerToAdd)(int,int) = &add;
//By the way, it is a POINTER to a function, so you need to add the ampersand
//to get its location in memory. In c++ it is implied for functions, but
//you should still use it.
printf("%d \n",funcpointerToAdd(2,3));
int (**doubleFuncPointerToAdd)(int,int) = &funcpointerToAdd;
printf("%d \n",(*doubleFuncPointerToAdd)(2,3));
//You need to dereference the double pointer,
//to turn it into a normal pointer, which you can then call
return 0;
}
其他类型也是如此:
struct whatever {
int a;
};
int main() {
whatever s;
s.a = 15;
printf("%d\n",s.a);
whatever* p1 = &s;
printf("%d\n",p1->a); //OK
//x->y is just a shortcut for (*x).y
whatever** p2 = &p1;
printf("%d\n",p2->a); //ERROR, trying to get value (*p2).a,
//which is a double pointer, so it's equivalent to p1.a
printf("%d\n",(*p2)->a); //OK
}
我想知道与双指针不同 (int**)
,我们可以有双函数指针吗?
我的意思是函数指针指向另一个函数指针的地址?
我想要类似的东西
int add(int A , int B){
return A+B;
}
int main(void){
int (*funcpointerToAdd)(int,int) = add; // single function pointer pointing to the function add
printf("%d \n",funcpointerToAdd(2,3));
int (**doubleFuncPointerToAdd)(int,int) = &funcpointerToAdd;
printf("%d \n",doubleFuncPointerToAdd(2,3));
return 0;
}
但这给了我一个错误called object ‘doubleFuncPointerToAdd’ is not a function or function pointer
无论如何都可以做这件事吗?
您可以使用指向函数指针的指针,但您必须先对它们进行一次引用:
int add(int A , int B){
return A+B;
}
int main(void){
int (*funcpointerToAdd)(int,int) = &add;
//By the way, it is a POINTER to a function, so you need to add the ampersand
//to get its location in memory. In c++ it is implied for functions, but
//you should still use it.
printf("%d \n",funcpointerToAdd(2,3));
int (**doubleFuncPointerToAdd)(int,int) = &funcpointerToAdd;
printf("%d \n",(*doubleFuncPointerToAdd)(2,3));
//You need to dereference the double pointer,
//to turn it into a normal pointer, which you can then call
return 0;
}
其他类型也是如此:
struct whatever {
int a;
};
int main() {
whatever s;
s.a = 15;
printf("%d\n",s.a);
whatever* p1 = &s;
printf("%d\n",p1->a); //OK
//x->y is just a shortcut for (*x).y
whatever** p2 = &p1;
printf("%d\n",p2->a); //ERROR, trying to get value (*p2).a,
//which is a double pointer, so it's equivalent to p1.a
printf("%d\n",(*p2)->a); //OK
}