结构中的函数指针

Question

如何在Struct中使用函数指针？具体来说,给出以下示例,程序编译但运行时崩溃:

在头文件中

 #ifndef __FUNCTION_IN_STRUCT_H_
 #define __FUNCTION_IN_STRUCT_H_


struct functionDaemon {
     int id;
     //double (*funcp); // function pointer
     double  (*fp)(double);      // Function pointer
 };

 // #define NULL 'V'

 #endif /* _FUNCTION_IN_STRUCT_H_ */

在C文件中:

#include <math.h>
#include <stdio.h>

#include "function_in_struct.h"

extern struct functionDaemon *ftnAgent;

void do_compute_sum (void) {

     void* agent;
    // struct functionDaemon *ftnAgent = (struct functionDaemon *) agent;
    struct functionDaemon *ftnAgent;

    double  sum;

    // Use 'sin()' as the pointed-to function
    ftnAgent->fp = sin;
    sum = compute_sum(ftnAgent->fp, 0.0, 1.0);
    printf("sum(sin): %f\n", sum);

}

请建议我.

Answer 1

你快到了:

struct functionDaemon *ftnAgent;

double  sum;

// Use 'sin()' as the pointed-to function
ftnAgent->fp = sin;

你ftnAgent只是一个非初始化的指针.

struct functionDaemon ftnAgent;

double  sum;

// Use 'sin()' as the pointed-to function
ftnAgent.fp = sin;
sum = compute_sum(ftnAgent.fp, 0.0, 1.0);

这是一个工作示例:

#include <math.h>
#include <stdio.h>


struct functionDaemon {
     int id;
     //double (*funcp); // function pointer
     double  (*fp)(double);      // Function pointer
 };


int main()
{
        struct functionDaemon f;
        f.fp = sin;

        printf("%f\n", (f.fp)(10));

        return 0;
}

编辑

你有这个:

extern struct functionDaemon *ftnAgent;

我假设ftnAgent在其他地方实例化.在这种情况下,您不需要struct functionDaemon *ftnAgent;在内部do_compute_sum,因为它将隐藏已经声明的ftnAgent结构,因此您将访问错误的(未初始化的)变量.