How to call a c function from name , which is stored in a char* pointer?

Question

I wanted to dynamically call a function by its name , e.g , suppose have the following function and string:

void do_fork()
{
   printf ("Fork called.\n");

}

char *pFunc = "do_fork";

Now i need to call do_fork() just by *pFunc. So is that possible ?

Either C/C++ code is welcomed , many thanks !

Answer 1

Neither C nor C++ have enough reflection to do this out of the box, so you will have to implement your own scheme.

In C++, the more or less canonical way to do that is using a map of strings to function pointers. Something like this:

typedef void (*func_t)();
typedef std::map<std::string,func_t> func_map_t;

// fill the map
func_map_t func_map;
func_map["do_fork"] = &do_fork;
func_map["frgl"] = &frgl;

// search a function in the map
func_map_t::const_iterator it = func_map.find("do_fork");
if( it == func_map.end() ) throw "You need error handling here!"
(*it->second)();

Of course, this is limited to functions with exactly the same signature. However, this limitation can be somewhat lifted (to encompass reasonably compatible signatures) by using std::function and std::bind instead of a plain function pointer.

Answer 2

Not entirely sure if it's what you're looking for but you could easily use dlopen and dlsym.

void *dlsym(void *restrict handle, const char *restrict name);

The dlsym() function shall obtain the address of a symbol defined within an object made accessible through a dlopen() call. The handle argument is the value returned from a call to dlopen() (and which has not since been released via a call to dlclose()), and name is the symbol's name as a character string.

That said, it usually doesn't come to this in C so you likely don't actually need it.