如何在C中实现将名称与void函数关联的映射？

Question

After reading Structure and Interpretation of Computer Programs (SICP) I decided to find a way to implement some of these functional programming techniques using C. I tried to write a program that makes a pair whose first argument is a name of the function and second arg is any function that takes one arg and returns one arg. Using implementation below I was expecting to see an output like:

fact(7) = 5040
fib(7) = 13

but instead I am getting

fact(7) = 5040
fib(7) = 0

along with warnings

$ cc map.c
map.c: In function ‘main’:
map.c:41:17: warning: assignment from incompatible pointer type [enabled by default]
   maps[0].f_ptr = &fact;
                 ^
map.c:43:17: warning: assignment from incompatible pointer type [enabled by default]
   maps[1].f_ptr = &fib;
                 ^
map.c:47:7: warning: passing argument 1 of ‘maps[i].f_ptr’ makes pointer from integer without a cast [enabled by default]
       ans = (int) maps[i].f_ptr((int) num);
       ^
map.c:47:7: note: expected ‘void *’ but argument is of type ‘int’
map.c:47:13: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
       ans = (int) maps[i].f_ptr((int) num);
             ^
map.c:52:7: warning: passing argument 1 of ‘maps[i].f_ptr’ makes pointer from integer without a cast [enabled by default]
       ans2 = (int) maps[i].f_ptr((int) num);
       ^
map.c:52:7: note: expected ‘void *’ but argument is of type ‘int’
map.c:52:14: warning: cast from pointer to integer of different size [-Wpointer-to-int-cast]
       ans2 = (int) maps[i].f_ptr((int) num);

during compilation. Looking at the code I don't see the problem but then again I haven't used C in quite some time. Is there a better way to implement such a construct and why is fib(7) printing a 0 instead of 13?

Here's my code:

struct Map
{
  char* name;
  void* (*f_ptr)(void*);
};

int fact(int a) {
    if (a == 0)
      return 0;
    if (a == 1)
      return 1;
    return a * fact (a-1);
}

int fib(int a) {
  if (a == 0)
    return 0;
  if (a == 1)
    return 1;
  return fib(a-1) + fib(a-2);
}

int findFunc (char* str, struct Map map)
{
  if (map.name == str)
    return 1;
  return 0;
}

int main()
{
  int i = 0;
  int ans = 0;
  int ans2 = 0;
  int num = 7;

  struct Map maps[2];
  maps[0].name = "fact";
  maps[0].f_ptr = &fact;
  maps[1].name = "fib";
  maps[1].f_ptr = &fib;

  for (i; i < (sizeof(maps)/sizeof(maps[0])); i++) {
    if (findFunc("fact", maps[i]))
      ans = (int) maps[i].f_ptr((int) num);
  }

  for (i; i < (sizeof(maps)/sizeof(maps[0])); i++) {
    if (findFunc("fib", maps[i]))
      ans2 = (int) maps[i].f_ptr((int) num);
  }

  printf("fact(%d) = %d\n", num, ans);
  printf("fib(%d) = %d", num, ans2);
  return 0;
}

Answer 1

String comparisons

This is not how you do string comparison in C.

if (map.name == str)

This is how you do string comparison in C.

if (0 == strcmp(map.name, str))

Because strings in C are just pointers to characters, map.name == str checks if map.name and str are identical pointers (point to the same block of memory), not whether what they point to is the same.

for loops

Your code is probably reporting fib(7) = 0 because it's failing to find fib. One possible culprit is the string comparison issue I mentioned. However, your for loop syntax is also odd:

for (i; i < (sizeof(maps)/sizeof(maps[0])); i++) {

You don't set i to anything, so this means, "Starting from wherever i happens to be, do the following..."

To loop over all of maps, use this:

for (i = 0; i < (sizeof(maps)/sizeof(maps[0])); i++) {

type warnings

As @alk said in a comment, the reason you're getting all of those warnings is because you've declared a function type of void* (*f_ptr)(void*); , even though your functions are int (*)(int) . If you want to keep using void* to allow different types, and you're careful enough with your types to make this work, then you can add casts to silence the warnings.

maps[0].f_ptr = (void *(*)(void*)) &fact;
ans2 = (int) maps[i].f_ptr((void*) num);

Etc.

Better implementations?

A "real" implementation of mapping functions to names would use a hash table, instead of linearly searching for matching names. Implementing a hash table in C would add complexity and may not be worth it for this exercise.

Answer 2

but instead I am getting

[...]

 fib(7) = 0 

The code misses to initialise i to 0 for the 2nd for -loop.