C 语言中可以有双函数指针吗？

Question

我想知道与双指针不同(int**)，我们可以有双函数指针吗？

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我的意思是函数指针指向另一个函数指针的地址？

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我想要类似的东西

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int add(int A , int B){\n    return A+B;\n}\n\nint main(void){\n\n    int (*funcpointerToAdd)(int,int) = add; // single function pointer pointing to the function add\n    printf("%d \\n",funcpointerToAdd(2,3));\n  \n\n    int (**doubleFuncPointerToAdd)(int,int) = &funcpointerToAdd;\n    printf("%d \\n",doubleFuncPointerToAdd(2,3));\n\n    return 0;\n}\n

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但这给了我一个错误called object \xe2\x80\x98doubleFuncPointerToAdd\xe2\x80\x99 is not a function or function pointer

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无论如何这有可能做这件事吗？

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Answer 1

您可以使用指向函数指针的指针，但必须首先引用它们一次：

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;
}

对于其他类型也是如此：

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
}