传递带有变量参数的函数作为void指针，并调用它

Question

I have a function which I want to pass through various functions

int func1(char *ptr)
{
    printf("%s",ptr);
    return 0;
}

and other function in which I want to call the func1

int func2(void *i)
{
    //call i
    //here i point to function func1
    //and do something with return value of i
}

So, how should I call it in main()?

int main()
{
    void *d;
    //SOMETHING wrong in next four line
    d=&func1("abcd");
    func2(d);
    d=&func1("xyz");
    func2(d);
    return 0;
}

Answer 1

You can simply create a function ( func2 ) that takes a function pointer to your desired function that you want called and another paramter that takes a pointer to your string. Then in func2 call the passed in function pointer with the str argument.

#include <stdio.h>

void func1(const char *str)
{
    puts(str);
}

void func2(void (*fp)(const char *), const char *str)
{
    fp(str);
}

int main(void)
{
    const char *str = "Hello world.";

    func2(func1, str);

    return 0;
}

Answer 2

You should first of all use Function pointers to do the job you desire. Using void pointer is a method where you before hand know what you will cast it to for dereferencing. Since you are passing the address of a function, you should cast it to a function pointer anyways. So avoid all this and declare func pointer anyways. Use void * only if you know that you can handle all the possiblities. eg. memcpy uses void * this is acceptable as you just want a bunch of data in soure location to be copied to a destination location.

#include <stdio.h>
void func1(char * str)
{
   printf("%s",str);
}
void func2(void * g)
{
   void (*p) (char *);   //function pointer
   p = (void (*)(char *))g;//casting void * to pi.e func pointer reqiuired eitherways.
   p("Func pointer caller");
}
void main()
{
   void * d = func1; // name of the function is itselt its address
   //printf("func1 = %d\nd = %d",func1,d);  
   func2(d);
}

all this can be avoided by a simple function pointer.

Answer 3

First of all: A function pointer refers to a function only. Parameters come into the game when the function is called, either directly or via a function pointer.

So if you want to achieve what you have in mind, that is binding a (set of) parameters to a specific function varaible you need to use a wrapper function:

int func1(char * pc)
{
  int i;
  /* assigne somehting to "i" and use "pc" */
  return i;
}

int func1_abcd(void)
{
  return func1("abcd")
}

int func1_xyz(void)
{
  return func1("xyz")
}

typedef int (*pfunc1_wrapper_t)(void);

int func2(pfunc1_wrapper_t pfunc1_wrapper)
{
  return pfunc1_wrapper();
}

int main(int argc, char ** argv)
{
  pfunc1_wrapper_t pfunc1_wrapper = NULL;
  int i = 0;

  pfunc1_wrapper = func1_abcd;
  i = func2(pfunc1_wrapper);

  pfunc1_wrapper = func1_xyz;
  i = func2(pfunc1_wrapper);

  ...

  return 0;
}

Answer 4

请参阅此函数指针并阅读Wiki - Function Pointers ，您将了解如何修复代码。

Answer 5

Jatin already posted a general link to function pointers, very good explaining function pointers.

As you seam to want to pass not a classic function pointer, but a function pointer WITH it's argument as one "function pointer" to an other function, I do not think that this will work.

But why passing a function with its argument as one parameter? This would mean, that the function you are passing with parameters, will return only one result, that is already known outside func2 (as func2 will call func1 with this already outside defined parameter set). So why not just pass the result of the func1 call to func2?