如何在C中创建一个带有未知类型参数的函数？

Question

Say I have the following code:

struct test* t1;
t1 = get_t(1);

... where get_t is:

struct test* get_t(int);

How can I refactor the above code and put it in a function? Something like the following:

void r1(?* t, ?* (fn*)(int)) {
    t = fn(1);
}

/* ... */

struct test* t1;
r1(t1, &get_t);

Answer 1

使用void *param ，指向任何东西的指针......通常在glib中用作gpointer

Answer 2

I have two ideas:

[1] Pass void pointer to the variable/object, and type cast it in the function.

[2] Make a union of all datatypes along with an integer datatype which will identify which datatype variable in the union does hold the actual data. Pass this union as value or as void *

struct _unknown {
   union {
      int a;
      float b;
      char c;
      double d;
   } data;
   int type;
} unknown;
.
.
.
if (unknown.type == FLOAT)
{
  /* Process variable b */ 
}
else if (unknown.type == INT)
{
 /* Process variable a */
}
.
. 
.

Something like this.

You can hash define the FLOAT and INT and others as unique values.

Or simply

struct _unknown {
  void *data;
  int type;
} unknown;
.
.
.
if (unknown == FLOAT)
{
  /* process (float *) data */
}
else if (unknown == INT)
{
  /* process (int *) data */
}
else if (unknown == MY_DATA_TYPE)
{
  /* process (my_data_type *) data */
  /* where my_data_type could be a typedef, or struct */
}

Answer 3

If you have gcc, you can use this more typesafe version:

#define r1(varp,func) ({ \
    typeof(**varp)* (*_func_)(int); \
    typeof(**varp)* _varp_ = (varp); \
    _func_ = (func); \
    r1_((void**)(_varp_),(void*(*)(int))_func_); \
    })

void r1_(void** varp,void*(*func)(int))
    {
    *varp = func(1);
    }

Call as:

struct test* get_t(int);
struct test* t1;
r1(&t,get_t);

(You dont need to use & on functions, they decay to pointers automatically, like arrays). This checks that t is a pointer, and that get_t is a function returning that type of pointer. _varp_ is technically unneeded, but keeps the argument evaluation in the right order.

Edit:

If you don't have gcc, you can still do this, but you have to provide the type explicitly:

#define r1(T,varp,func) do { \
    T*(*_func_)(int); \
    T* _varp_ = (varp); \
    _func_ = (func); \
    r1_((void**)(_varp_),(void*(*)(int))_func_); \
    } while(0)

void r1_(void** varp,void*(*func)(int))
    {
    *varp = func(1);
    }

Call as:

struct test* get_t(int);
struct test* t1;
r1(struct test*,&t,get_t);

not quite as safe, and more redundant, but still fairly good.