c++ - 返回函数的函数是怎么个形式？

Question

书上说

函数可以直接返回一个可调用对象

可调用对象包括函数，函数指针等

所以函数可以直接返回一个函数。我怎么好像从来没见过？你们能举个例子给我看看吗？
还有函数返回lambda的例子

Answer 1

1. A pointer to a function, such as

   int (*p)(int, int);

   p is a pointer to a function. This function has two parameters of type int and returns int.

2. A function that returns a function pointer can be defined in the following two ways.

3. • Direct definition

     int (*foo(char c))(int, int);

   • Use typedef

     typedef int OP(int, int);
     OP* foo(char c);

     The two methods are equivalent, and the second one is easier to see clearly.

1. returns a pointer to the function.

4. Application examples

   int foo_plus(int a, int b)
   {
     return a + b;
   }
   
   int foo_multiply(int a, int b)
   {
     return a * b;
   }
   
   typedef int OP(int, int);
   OP* foo(char c)
   {
      OP* p;
      switch(c) {
        case 'p':
          p = foo_plus;
          break;
        case 'm':
          p = foo_multiply;
          break;
        default:
          p = foo_multiply;
      }
      return p;
   }
   
   int main()
   {
     OP* f;
     int n;
     f = foo('p');
     n = f(1, 2);
     printf("%d", n); // 3
     f = foo('m');
     n = f(1, 2);
     printf("%d", n); // 2
     
     return 0;
   }

Supplement:
The function returns an object, which can be called.

A function object, that is, an object that overloads the bracket operator "()".
e.g.

class A{

public:
    int operator()(int x， int y)
    {
        return x + y;
    }

};
A a;
a(1, 2); has the same effect as foo_plus(1, 2); above.

Answer 2

Callable objects mainly include function pointers, std::function and classes/structures that overload the meta bracket operator.