我应该如何使用函数指针数组用C？
我该如何初始化呢？
Answer 1:
你有一个很好的例子在这里（函数指针数组） ，用详细的语法 。
int sum(int a, int b);
int subtract(int a, int b);
int mul(int a, int b);
int div(int a, int b);
int (*p[4]) (int x, int y);
int main(void)
{
int result;
int i, j, op;
p[0] = sum; /* address of sum() */
p[1] = subtract; /* address of subtract() */
p[2] = mul; /* address of mul() */
p[3] = div; /* address of div() */
[...]
调用这些函数指针之一：
result = (*p[op]) (i, j); // op being the index of one of the four functions
Answer 2:
以上回答可以帮助你，但你可能也想知道如何使用函数指针的数组。
void fun1()
{
}
void fun2()
{
}
void fun3()
{
}
void (*func_ptr[3]) = {fun1, fun2, fun3};
main()
{
int option;
printf("\nEnter function number you want");
printf("\nYou should not enter other than 0 , 1, 2"); /* because we have only 3 functions */
scanf("%d",&option);
if((option>=0)&&(option<=2))
{
(*func_ptr[option])();
}
return 0;
}
您只能使用相同的返回类型和相同的参数类型，没有参数，以一个单一的函数指针数组指定的函数的地址。
还可以传递参数如下面如果所有的上述功能都具有相同数目的相同类型的参数。
(*func_ptr[option])(argu1);
注意：此处所述阵列中的函数指针的编号从0相同的一般阵列来开始。 因此，在上面的例子中fun1可以称为如果选项= 0， fun2可以称为如果选项= 1和fun3可以被称为如果选项= 2。
Answer 3:
这里是你如何使用它：
New_Fun.h
#ifndef NEW_FUN_H_
#define NEW_FUN_H_
#include <stdio.h>
typedef int speed;
speed fun(int x);
enum fp {
f1, f2, f3, f4, f5
};
void F1();
void F2();
void F3();
void F4();
void F5();
#endif
New_Fun.c
#include "New_Fun.h"
speed fun(int x)
{
int Vel;
Vel = x;
return Vel;
}
void F1()
{
printf("From F1\n");
}
void F2()
{
printf("From F2\n");
}
void F3()
{
printf("From F3\n");
}
void F4()
{
printf("From F4\n");
}
void F5()
{
printf("From F5\n");
}
MAIN.C
#include <stdio.h>
#include "New_Fun.h"
int main()
{
int (*F_P)(int y);
void (*F_A[5])() = { F1, F2, F3, F4, F5 }; // if it is int the pointer incompatible is bound to happen
int xyz, i;
printf("Hello Function Pointer!\n");
F_P = fun;
xyz = F_P(5);
printf("The Value is %d\n", xyz);
//(*F_A[5]) = { F1, F2, F3, F4, F5 };
for (i = 0; i < 5; i++)
{
F_A[i]();
}
printf("\n\n");
F_A[f1]();
F_A[f2]();
F_A[f3]();
F_A[f4]();
return 0;
}
我希望这有助于理解Function Pointer.
Answer 4:
这个“答案”更是一增编VonC的回答; 刚刚指出的是，语法可以经由一个typedef被简化，并且可以使用集合初始化：
typedef int FUNC(int, int);
FUNC sum, subtract, mul, div;
FUNC *p[4] = { sum, subtract, mul, div };
int main(void)
{
int result;
int i = 2, j = 3, op = 2; // 2: mul
result = p[op](i, j); // = 6
}
// maybe even in another file
int sum(int a, int b) { return a+b; }
int subtract(int a, int b) { return a-b; }
int mul(int a, int b) { return a*b; }
int div(int a, int b) { return a/b; }
Answer 5:
哦，还有吨的例子。 只要看看巧舌如簧或GTK内任何东西。 你可以看到函数指针在工作中的工作有一路。
例如这里的gtk_button东西初始化。
static void
gtk_button_class_init (GtkButtonClass *klass)
{
GObjectClass *gobject_class;
GtkObjectClass *object_class;
GtkWidgetClass *widget_class;
GtkContainerClass *container_class;
gobject_class = G_OBJECT_CLASS (klass);
object_class = (GtkObjectClass*) klass;
widget_class = (GtkWidgetClass*) klass;
container_class = (GtkContainerClass*) klass;
gobject_class->constructor = gtk_button_constructor;
gobject_class->set_property = gtk_button_set_property;
gobject_class->get_property = gtk_button_get_property;
而在gtkobject.h可以找到以下声明：
struct _GtkObjectClass
{
GInitiallyUnownedClass parent_class;
/* Non overridable class methods to set and get per class arguments */
void (*set_arg) (GtkObject *object,
GtkArg *arg,
guint arg_id);
void (*get_arg) (GtkObject *object,
GtkArg *arg,
guint arg_id);
/* Default signal handler for the ::destroy signal, which is
* invoked to request that references to the widget be dropped.
* If an object class overrides destroy() in order to perform class
* specific destruction then it must still invoke its superclass'
* implementation of the method after it is finished with its
* own cleanup. (See gtk_widget_real_destroy() for an example of
* how to do this).
*/
void (*destroy) (GtkObject *object);
};
该（* set_arg）的东西是起作用的指针，这个例如可以在指定的一些派生类中另一种实现方式。
你经常看到这样的事情
struct function_table {
char *name;
void (*some_fun)(int arg1, double arg2);
};
void function1(int arg1, double arg2)....
struct function_table my_table [] = {
{"function1", function1},
...
所以，你可以通过名字达到了桌子上，称之为“相关”的功能。
或者，也许你使用，你把功能与“名”把它称为哈希表。
问候
弗里德里希
Answer 6:
这应该是一个短的简单的复制和粘贴片上面的响应的代码的例子。 希望这有助于。
#include <iostream>
using namespace std;
#define DBG_PRINT(x) do { std::printf("Line:%-4d" " %15s = %-10d\n", __LINE__, #x, x); } while(0);
void F0(){ printf("Print F%d\n", 0); }
void F1(){ printf("Print F%d\n", 1); }
void F2(){ printf("Print F%d\n", 2); }
void F3(){ printf("Print F%d\n", 3); }
void F4(){ printf("Print F%d\n", 4); }
void (*fArrVoid[N_FUNC])() = {F0, F1, F2, F3, F4};
int Sum(int a, int b){ return(a+b); }
int Sub(int a, int b){ return(a-b); }
int Mul(int a, int b){ return(a*b); }
int Div(int a, int b){ return(a/b); }
int (*fArrArgs[4])(int a, int b) = {Sum, Sub, Mul, Div};
int main(){
for(int i = 0; i < 5; i++) (*fArrVoid[i])();
printf("\n");
DBG_PRINT((*fArrArgs[0])(3,2))
DBG_PRINT((*fArrArgs[1])(3,2))
DBG_PRINT((*fArrArgs[2])(3,2))
DBG_PRINT((*fArrArgs[3])(3,2))
return(0);
}
Answer 7:
能在这样的方式来使用它：
//! Define:
#define F_NUM 3
int (*pFunctions[F_NUM])(void * arg);
//! Initialise:
int someFunction(void * arg) {
int a= *((int*)arg);
return a*a;
}
pFunctions[0]= someFunction;
//! Use:
int someMethod(int idx, void * arg, int * result) {
int done= 0;
if (idx < F_NUM && pFunctions[idx] != NULL) {
*result= pFunctions[idx](arg);
done= 1;
}
return done;
}
int x= 2;
int z= 0;
someMethod(0, (void*)&x, &z);
assert(z == 4);
Answer 8:
这个问题已经被回答非常好的例子。 这可能是唯一缺少的例子是在函数返回指针。 我写这个又如，并增加了很多的意见，如果有人发现了它有帮助：
#include <stdio.h>
char * func1(char *a) {
*a = 'b';
return a;
}
char * func2(char *a) {
*a = 'c';
return a;
}
int main() {
char a = 'a';
/* declare array of function pointers
* the function pointer types are char * name(char *)
* A pointer to this type of function would be just
* put * before name, and parenthesis around *name:
* char * (*name)(char *)
* An array of these pointers is the same with [x]
*/
char * (*functions[2])(char *) = {func1, func2};
printf("%c, ", a);
/* the functions return a pointer, so I need to deference pointer
* Thats why the * in front of the parenthesis (in case it confused you)
*/
printf("%c, ", *(*functions[0])(&a));
printf("%c\n", *(*functions[1])(&a));
a = 'a';
/* creating 'name' for a function pointer type
* funcp is equivalent to type char *(*funcname)(char *)
*/
typedef char *(*funcp)(char *);
/* Now the declaration of the array of function pointers
* becomes easier
*/
funcp functions2[2] = {func1, func2};
printf("%c, ", a);
printf("%c, ", *(*functions2[0])(&a));
printf("%c\n", *(*functions2[1])(&a));
return 0;
}
Answer 9:
用于与函数指针“多维数组这个简单的例子：
void one( int a, int b){ printf(" \n[ ONE ] a = %d b = %d",a,b);}
void two( int a, int b){ printf(" \n[ TWO ] a = %d b = %d",a,b);}
void three( int a, int b){ printf("\n [ THREE ] a = %d b = %d",a,b);}
void four( int a, int b){ printf(" \n[ FOUR ] a = %d b = %d",a,b);}
void five( int a, int b){ printf(" \n [ FIVE ] a = %d b = %d",a,b);}
void(*p[2][2])(int,int) ;
int main()
{
int i,j;
printf("multidimensional array with function pointers\n");
p[0][0] = one; p[0][1] = two; p[1][0] = three; p[1][1] = four;
for ( i = 1 ; i >=0; i--)
for ( j = 0 ; j <2; j++)
(*p[i][j])( (i, i*j);
return 0;
}
Answer 10:
最简单的办法是给你想要的最终载体的地址，并修改它的函数内部。
void calculation(double result[] ){ //do the calculation on result
result[0] = 10+5;
result[1] = 10 +6;
.....
}
int main(){
double result[10] = {0}; //this is the vector of the results
calculation(result); //this will modify result
}
文章来源: How can I use an array of function pointers?
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