A function pointer is like any other pointer, but it points to the address of a function instead of the address of data (on heap or stack). Like any pointer, it needs to be typed correctly. Functions are defined by their return value and the types of parameters they accept. So in order to fully describe a function, you must include its return value and the type of each parameter is accepts. When you typedef such a definition, you give it a 'friendly name' which makes it easier to create and reference pointers using that definition.

例如，假设你有一个函数:

float doMultiplication (float num1, float num2 ) {
    return num1 * num2; }

然后是下面的类型定义:

typedef float(*pt2Func)(float, float);

可以用来指向这个doMulitplication函数。它只是定义一个指向函数的指针，该函数返回一个浮点数，并接受两个参数，每个参数都是float类型。这个定义有一个友好的名字pt2Func。请注意，pt2Func可以指向返回一个浮点数并接受2个浮点数的任何函数。

所以你可以创建一个指针指向doMultiplication函数，如下所示:

pt2Func *myFnPtr = &doMultiplication;

你可以使用这个指针调用函数，如下所示:

float result = (*myFnPtr)(2.0, 5.1);

这是一个很好的阅读:http://www.newty.de/fpt/index.html

int add(int a, int b)
{
  return (a+b);
}
int minus(int a, int b)
{
  return (a-b);
}

typedef int (*math_func)(int, int); //declaration of function pointer

int main()
{
  math_func addition = add;  //typedef assigns a new variable i.e. "addition" to original function "add"
  math_func substract = minus; //typedef assigns a new variable i.e. "substract" to original function "minus"

  int c = addition(11, 11);   //calling function via new variable
  printf("%d\n",c);
  c = substract(11, 5);   //calling function via new variable
  printf("%d",c);
  return 0;
}

它的输出是:

22

6

注意，声明这两个函数时使用了相同的math_func定义器。

与typedef相同的方法可以用于extern struct。(在其他文件中使用struct)

A function pointer is like any other pointer, but it points to the address of a function instead of the address of data (on heap or stack). Like any pointer, it needs to be typed correctly. Functions are defined by their return value and the types of parameters they accept. So in order to fully describe a function, you must include its return value and the type of each parameter is accepts. When you typedef such a definition, you give it a 'friendly name' which makes it easier to create and reference pointers using that definition.

例如，假设你有一个函数:

float doMultiplication (float num1, float num2 ) {
    return num1 * num2; }

然后是下面的类型定义:

typedef float(*pt2Func)(float, float);

可以用来指向这个doMulitplication函数。它只是定义一个指向函数的指针，该函数返回一个浮点数，并接受两个参数，每个参数都是float类型。这个定义有一个友好的名字pt2Func。请注意，pt2Func可以指向返回一个浮点数并接受2个浮点数的任何函数。

所以你可以创建一个指针指向doMultiplication函数，如下所示:

pt2Func *myFnPtr = &doMultiplication;

你可以使用这个指针调用函数，如下所示:

float result = (*myFnPtr)(2.0, 5.1);

这是一个很好的阅读:http://www.newty.de/fpt/index.html

一个非常简单的方法来理解函数指针的typedef:

int add(int a, int b)
{
    return (a+b);
}

typedef int (*add_integer)(int, int); //declaration of function pointer

int main()
{
    add_integer addition = add; //typedef assigns a new variable i.e. "addition" to original function "add"
    int c = addition(11, 11);   //calling function via new variable
    printf("%d",c);
    return 0;
}

这是我作为练习写的关于函数指针和函数指针数组的最简单的例子。

    typedef double (*pf)(double x);  /*this defines a type pf */

    double f1(double x) { return(x+x);}
    double f2(double x) { return(x*x);}

    pf pa[] = {f1, f2};


    main()
    {
        pf p;

        p = pa[0];
        printf("%f\n", p(3.0));
        p = pa[1];
        printf("%f\n", p(3.0));
    }

宏呢?我已经看到，使用类型定义，我不能将一些东西强制转换为函数指针。所以我为它做了这个小宏，它能够做我到目前为止需要的所有事情(不知道它不能用于什么-我认为它基本上是一个类型定义，但有一个可选参数，使它在类型转换中可用):

#define funcptr_t(sym_name) void (*sym_name)(void)

// Declare a function pointer-returning function, and declare a function pointer variable.
funcptr_t (randomFunction(funcptr_t (func_ptr_variable)));
// Cast a variable to a function pointer
(funcptr_t()) some_variable;

PS:我做这个通用用途，不确定它是否可以修改为特定的功能使用。