从 C/C++ 中的函数返回函数指针

在 C/C++ 中，像普通数据指针（int *、char * 等）一样，可以有指向函数的指针。在程序中创建的每个函数都会在内存中获得一个地址，因为指针可以在 C/C++ 中使用，因此也可以创建指向函数的指针。

句法：

return type (*function_pointer_name) (argument_type_1,  argument_type_2, ……, argument_type_n) = &function_name;

OR

return type (*function_pointer_name) (argument_type_1, argument_type_2, ……, argument_type_n) = function_name;

注意：函数指针的参数类型和返回类型应与程序中存在的实际函数匹配。

方案一：

C

// C program for the above approach
#include 
 
// Function to add the value 10 to
// the variable a
void demo(int* a) { *a += 10; }
 
// Driver Code
int main()
{
    int num = 20;
 
    // ptr contains address of demo
    // function or void
    void (*ptr)(int*) = &demo;
 
    // or (*ptr)(&num);
    ptr(&num);
 
    printf("%d", num);
 
    return 0;
}

C++

// C++ program for the above approach
#include 
using namespace std;
 
void demo(int& a)
{
    a += 10;
}
 
// Driver Code
int main()
{
    int num = 20;
 
    // Now ptr contains address of demo
    // function or void
    void (*ptr)(int*) = &demo;
 
    // or (*ptr)(num);
    ptr(num);
 
    cout << num << endl;
 
    return 0;
}

输出

30

返回函数指针从函数：从一个函数返回一个函数指针，函数的返回类型应该是一个指向另一个函数。但是编译器不接受函数的这种返回类型，因此我们需要定义一个表示该特定函数指针的类型。

句法 ：

typedef return type (*function_pointer_name) (argument_type_1, argument_type_2, ……, argument_type_n);

This creates a type which represents a pointer for a particular function.

方案二：

C

// C program for the above approach
#include 
typedef int (*ptr)(int*);
typedef ptr (*pm)();
 
int fun1(int* y)
{
    printf("geeks!!\n");
    return *y + 10;
}
 
// Function that return type ptr
ptr fun()
{
    printf("Hello ");
 
    // or return fun1;
    /* or
     int(*pt)(int*)=fun1;
     return pt
  */
    return &fun1;
}
 
// Driver Code
int main()
{
    int a = 10;
 
    pm u = fun;
 
    printf("%d", (*u())(&a));
 
    return 0;
}

C++

// C++ program for the above approach
#include 
using namespace std;
typedef int (*ptr)(int*);
typedef ptr (*pm)();
 
int fun1(int* y)
{
    cout << "geeks!!" << endl;
    return *y + 10;
}
 
// Function that returns the type ptr
ptr fun()
{
    cout << "Hello ";
    return &fun1;
}
 
// Driver Code
int main()
{
    int a = 10;
    pm u = fun;
    cout << (*u())(&a) << endl;
 
    return 0;
}

输出

Hello geeks!!
20

宣称具有两个函数指针作为其元素的，并且这些函数指针阵列，反过来，返回其它函数指针，其指向其它功能。驱动代码main()函数的逻辑可以在上面的程序中改成：

方案三：

C

// C program for the above approach
#include 
 
// This defines a type for
// function prototypes
typedef int (*ptr)(int*);
typedef ptr (*pm)();
 
int fun1(int* y)
{
    printf("geeks!!\n");
    return *y + 10;
}
 
// fun() is a function with
// return type ptr
ptr fun()
{
    printf("Hello ");
 
    // or return fun1;
    /* or
     int(*pt)(int*)=fun1;
     return pt
  */
    return &fun1;
}
 
// Driver code
int main()
{
    int a = 10;
    pm u = fun;
 
    /*
    Above line assigns 'u' which is
    of type 'pm' to an array of size
    1 which has function pointers as
    its elements and these function
    pointers in turn return other
    function pointer which points to
    other functions.
 
    Now this 'p' array contains a function
    pointer 'u' which points to fun() and
    this fun() returns another function
    pointer which points to fun1().
  */
    int (*(*p[1])())(int*) = { u };
 
    printf("%d", (*p[0]())(&a));
}

C++

// C++ program for the above approach
#include 
using namespace std;
 
// This defines a type for
// function prototypes
typedef void (*ptr)(int&);
typedef ptr (*pm)();
 
void fun1(int& z)
{
    printf("geeks!!\n");
    cout << z + 10 << endl;
}
 
// Function that returns type ptr
ptr fun()
{
    printf("Hello ");
 
    // or return fun1;
    /* or
     int(*pt)(int*)=fun1;
     return pt
  */
    return &fun1;
}
 
// Driver Code
int main()
{
    int a = 10;
    pm u = fun;
 
    /*
    Above line assigns 'u' which is
    of type 'pm' to an array of size
    1 which has function pointers as its
    elements and these function pointers
    in turn return other function pointer
    which points to other functions.
 
    Now this 'p' array contains a function
    pointer 'u' which points to fun() and
    this fun() returns another function
    pointer which points to fun1() and
    this fun1() returns void.
  */
    void (*(*p[1])())(int&) = { u };
 
    (*p[0]())(a);
}

输出

Hello geeks!!
20

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