I have a function declared as int __stdcall MyFunction(int param, int param); and I need to get a type of the pointer to the function in a macro or a template when the name is passed as a parameter.
Is this possible in C++ or do I have to rewrite the function signature myself?
You can make a type alias for a function pointer (to which it decays, or which is returned by the address-of operator) the following way:
#include <iostream>
int MyFunction(int param, int param1)
{
std::cout << param << " " << param1 << std::endl;
return 0;
}
using Myfunction_ptr_t = std::decay<decltype(MyFunction)>::type;
// or using Myfunction_ptr_t = decltype(&MyFunction);
// or using Myfunction_ptr_t = decltype(MyFunction) *;
int main()
{
Myfunction_ptr_t Myfunction_ptr = MyFunction;
Myfunction_ptr(1, 2);
}
This example should rather use the auto specifier (since C++ 11):
auto Myfunction_ptr = MyFunction;
but that won't help in non-deducible context.
Function pointer behaves like every other pointer. It is a memory address which points to some entity (function code in this case). You can save these pointers somewhere and then fetch them in any convenient for your situation way.
For example you can create static map of std::string vs fun. pointer pairs:
static std::map<std::string, PTR_TYPE> funMap;
After that save pointers to this map and retreive them when needed.
If you don't have pointer yet, probably you speak about exported functions from libraries. In this case look at ldd for *nix-based or somethig simialar for other platform. You will need to search for runtime linker information and find fnction by it's name.
Actually I don't understand why you say you can't link against those functions. Do it in standard way: include header file with such declarations. Use declaration. Linker will do the work for you. Just provide it a path to library with those functions. It will link dynamically to required functions. That's all what you should do. If you don't have such library you need to search for function at runtime manualy with the help of (again) linker at runtime.
You could define a typedef like below:
typedef int (__stdcall *MYFUN)(int, int);
Test case:
int main() {
MYFUN f = MyFunction;
f(2, 3);
}
Alternatively, you could use a std::function object as bellow:
std::function<decltype(MyFunction)> myfun;
Test case:
std::function<decltype(MyFunction)> myfun = MyFunction;
myfun(4, 5);
Live Demo
Got it. decltype(FunctionName)* pointer; will do the job.
You could use decltype or auto, like the following
decltype(&FunctionName) pointer = &FunctionName;
or
auto pointer = &FunctionName;
I avoid manually translating function pointers to types myself in C++11 and C++14 :)
Related
Consider the following code:
file_1.hpp:
typedef void (*func_ptr)(void);
func_ptr file1_get_function(void);
file1.cpp:
// file_1.cpp
#include "file_1.hpp"
static void some_func(void)
{
do_stuff();
}
func_ptr file1_get_function(void)
{
return some_func;
}
file2.cpp
#include "file1.hpp"
void file2_func(void)
{
func_ptr function_pointer_to_file1 = file1_get_function();
function_pointer_to_file1();
}
While I believe the above example is technically possible - to call a function with internal linkage only via a function pointer, is it bad practice to do so? Could there be some funky compiler optimizations that take place (auto inline, for instance) that would make this situation problematic?
There's no problem, this is fine. In fact , IMHO, it is a good practice which lets your function be called without polluting the space of externally visible symbols.
It would also be appropriate to use this technique in the context of a function lookup table, e.g. a calculator which passes in a string representing an operator name, and expects back a function pointer to the function for doing that operation.
The compiler/linker isn't allowed to make optimizations which break correct code and this is correct code.
Historical note: back in C89, externally visible symbols had to be unique on the first 6 characters; this was relaxed in C99 and also commonly by compiler extension.
In order for this to work, you have to expose some portion of it as external and that's the clue most compilers will need.
Is there a chance that there's a broken compiler out there that will make mincemeat of this strange practice because they didn't foresee someone doing it? I can't answer that.
I can only think of false reasons to want to do this though: Finger print hiding, which fails because you have to expose it in the function pointer decl, unless you are planning to cast your way around things, in which case the question is "how badly is this going to hurt".
The other reason would be facading callbacks - you have some super-sensitive static local function in module m and you now want to expose the functionality in another module for callback purposes, but you want to audit that so you want a facade:
static void voodoo_function() {
}
fnptr get_voodoo_function(const char* file, int line) {
// you tagged the question as C++, so C++ io it is.
std::cout << "requested voodoo function from " << file << ":" << line << "\n";
return voodoo_function;
}
...
// question tagged as c++, so I'm using c++ syntax
auto* fn = get_voodoo_function(__FILE__, __LINE__);
but that's not really helping much, you really want a wrapper around execution of the function.
At the end of the day, there is a much simpler way to expose a function pointer. Provide an accessor function.
static void voodoo_function() {}
void do_voodoo_function() {
// provide external access to voodoo
voodoo_function();
}
Because here you provide the compiler with an optimization opportunity - when you link, if you specify whole program optimization, it can detect that this is a facade that it can eliminate, because you let it worry about function pointers.
But is there a really compelling reason not just to remove the static from infront of voodoo_function other than not exposing the internal name for it? And if so, why is the internal name so precious that you would go to these lengths to hide that?
static void ban_account_if_user_is_ugly() {
...;
}
fnptr do_that_thing() {
ban_account_if_user_is_ugly();
}
vs
void do_that_thing() { // ban account if user is ugly
...
}
--- EDIT ---
Conversion. Your function pointer is int(*)(int) but your static function is unsigned int(*)(unsigned int) and you don't want to have to cast it.
Again: Just providing a facade function would solve the problem, and it will transform into a function pointer later. Converting it to a function pointer by hand can only be a stumbling block for the compiler's whole program optimization.
But if you're casting, lets consider this:
// v1
fnptr get_fn_ptr() {
// brute force cast because otherwise it's 'hassle'
return (fnptr)(static_fn);
}
int facade_fn(int i) {
auto ui = static_cast<unsigned int>(i);
auto result = static_fn(ui);
return static_cast<int>(result);
}
Ok unsigned to signed, not a big deal. And then someone comes along and changes what fnptr needs to be to void(int, float);. One of the above becomes a weird runtime crash and one becomes a compile error.
My purpose is to call some C function from my C++ code and pass some C++ objects.
In fact I am using a integration routine from the GSL libray(written in C), see this link,
My code snippet:
// main.cpp
#include <stdio.h>
#include <gsl/gsl_integration.h>
#include <myclass.h>
/* my test function. */
double testfunction ( double x , void *param ) {
myclass *bar=static_cast<myclass*>(param);
/*** do something with x and bar***/
return val;
}
int main ( int argc , char *argv[] ) {
gsl_function F; // defined in GSL: double (* function) (double x, void * params)
/* initialize.*/
gsl_integration_cquad_workspace *ws =
gsl_integration_cquad_workspace_alloc( 200 ) ;
/* Prepare test function. */
myclass foo{}; // call myclass constructor
F.function = &testfunction;
F.params = &foo;
/* Call the routine. */
gsl_integration_cquad( &F, 0.0,1.0,1.0e-10,1.0e-10,ws, &res,&abserr,&neval);
/* Free the workspace. */
gsl_integration_cquad_workspace_free( ws );
return 0;
}
In my case, direct calling gsl_integration_cquad seems OK, provided the header includes sth like "ifdef __cplusplus", my concern is about the callback F,originally defined in C, am I allowed to pass the testfunction and also the C++ foo object in this way ? .
or is there any better way to do this kind of stuff, maybe overloading and use a functor?
P.S. Am I allowed to do exeption handling within the callback function? (use try catch inside "testfunction"). It works in my case but not sure if it's legal.
I'm not familiar with the library in question, but in general,
when passing a pointer to a callback and a void* to
a C routine, which will call the callback back with the void*,
there are two things you need to do to make it safe:
The function whose address you pass must be declared extern "C".
You'll get away with not doing this with a lot of compilers, but
it isn't legal, and a good compiler will complain.
The type you convert to the void* must be exactly the same
type as the type you cast it back to in the callback. The
classic error is to pass something like new Derived to the
C function, and cast it back to Base* in the callback. The
round trip Derived*→void*→Base* is undefined
behavior. It will work some of the time, but at other times, it
may crash, or cause any number of other problems.
And as cdhowie pointed out in a comment, you don't want to
allow exceptions to propagate accross the C code. Again, it
might work. But it might not.
For the exact example you posted, the only thing you need to do
is to declare testfunction as extern "C", and you're all
right. If you later start working with polymorphic objects,
however, beware of the second point.
You can use
myclass *bar=static_cast<myclass*>(param);
with void*.
If you meant something like transporting a c++ class pointer through a c callback's void* pointer, yes it's safe to do a static_cast<>.
There's no kind of losing c++ specific attributes of this class pointer, when passed through c code. Though passing a derived class pointer, and static casting back to the base class, won't work properly as #James Kanze pointed out.
The void* will likely just be passed trough by the C library without looking at the pointed-to data, so it's not a problem if this contains a C++ class. As log as you cast the void* to the correctly there shouldn't be any problems.
To make sure the callback function itself is compatible, you can declare it as extern "C". Additionally you should make sure that no exceptions are thrown from the callback function, since the C code calling the callback won't expect those.
All together I would split up the code into one function that does the real work and another function that is used as the callback and handles the interface with the C library, for example like this:
#include <math.h>
double testfunction ( double x ,myclass *param ) {
/*** do something with x and bar***/
return val;
}
extern "C" double testfunction_callback ( double x , void *param ) {
try {
myclass *bar=reinterpret_cast<myclass*>(param);
return testfunction(x, bar);
}
catch(...) {
std::cerr << "Noooo..." << std::endl;
return NAN;
}
}
Certain situations in my code, I end up invoking the function only if that function is defined, or else I should not. How can I achieve this?
like:
if (function 'sum' exists ) then invoke sum ()
Maybe the other way around to ask this question is how to determine if function is defined at runtime and if so, then invoke?
When you declare 'sum' you could declare it like:
#define SUM_EXISTS
int sum(std::vector<int>& addMeUp) {
...
}
Then when you come to use it you could go:
#ifdef SUM_EXISTS
int result = sum(x);
...
#endif
I'm guessing you're coming from a scripting language where things are all done at runtime. The main thing to remember with C++ is the two phases:
Compile time
Preprocessor runs
template code is turned into real source code
source code is turned in machine code
runtime
the machine code is run
So all the #define and things like that happen at compile time.
....
If you really wanted to do it all at runtime .. you might be interested in using some of the component architecture products out there.
Or maybe a plugin kind of architecture is what you're after.
Using GCC you can:
void func(int argc, char *argv[]) __attribute__((weak)); // weak declaration must always be present
// optional definition:
/*void func(int argc, char *argv[]) {
printf("FOUND THE FUNCTION\n");
for(int aa = 0; aa < argc; aa++){
printf("arg %d = %s \n", aa, argv[aa]);
}
}*/
int main(int argc, char *argv[]) {
if (func){
func(argc, argv);
} else {
printf("did not find the function\n");
}
}
If you uncomment func it will run it otherwise it will print "did not find the function\n".
While other replies are helpful advices (dlsym, function pointers, ...), you cannot compile C++ code referring to a function which does not exist. At minimum, the function has to be declared; if it is not, your code won't compile. If nothing (a compilation unit, some object file, some library) defines the function, the linker would complain (unless it is weak, see below).
But you should really explain why you are asking that. I can't guess, and there is some way to achieve your unstated goal.
Notice that dlsym often requires functions without name mangling, i.e. declared as extern "C".
If coding on Linux with GCC, you might also use the weak function attribute in declarations. The linker would then set undefined weak symbols to null.
addenda
If you are getting the function name from some input, you should be aware that only a subset of functions should be callable that way (if you call an arbitrary function without care, it will crash!) and you'll better explicitly construct that subset. You could then use a std::map, or dlsym (with each function in the subset declared extern "C"). Notice that dlopen with a NULL path gives a handle to the main program, which you should link with -rdynamic to have it work correctly.
You really want to call by their name only a suitably defined subset of functions. For instance, you probably don't want to call this way abort, exit, or fork.
NB. If you know dynamically the signature of the called function, you might want to use libffi to call it.
I suspect that the poster was actually looking for something more along the lines of SFINAE checking/dispatch. With C++ templates, can define to template functions, one which calls the desired function (if it exists) and one that does nothing (if the function does not exist). You can then make the first template depend on the desired function, such that the template is ill-formed when the function does not exist. This is valid because in C++ template substitution failure is not an error (SFINAE), so the compiler will just fall back to the second case (which for instance could do nothing).
See here for an excellent example: Is it possible to write a template to check for a function's existence?
use pointers to functions.
//initialize
typedef void (*PF)();
std::map<std::string, PF> defined_functions;
defined_functions["foo"]=&foo;
defined_functions["bar"]=&bar;
//if defined, invoke it
if(defined_functions.find("foo") != defined_functions.end())
{
defined_functions["foo"]();
}
If you know what library the function you'd like to call is in, then you can use dlsym() and dlerror() to find out whether or not it's there, and what the pointer to the function is.
Edit: I probably wouldn't actually use this approach - instead I would recommend Matiu's solution, as I think it's much better practice. However, dlsym() isn't very well known, so I thought I'd point it out.
You can use #pragma weak for the compilers that support it (see the weak symbol wikipedia entry).
This example and comment is from The Inside Story on Shared Libraries and Dynamic Loading:
#pragma weak debug
extern void debug(void);
void (*debugfunc)(void) = debug;
int main() {
printf(“Hello World\n”);
if (debugfunc) (*debugfunc)();
}
you can use the weak pragma to force the linker to ignore unresolved
symbols [..] the program compiles and links whether or not debug()
is actually defined in any object file. When the symbol remains
undefined, the linker usually replaces its value with 0. So, this
technique can be a useful way for a program to invoke optional code
that does not require recompiling the entire application.
So another way, if you're using c++11 would be to use functors:
You'll need to put this at the start of your file:
#include <functional>
The type of a functor is declared in this format:
std::function< return_type (param1_type, param2_type) >
You could add a variable that holds a functor for sum like this:
std::function<int(const std::vector<int>&)> sum;
To make things easy, let shorten the param type:
using Numbers = const std::vectorn<int>&;
Then you could fill in the functor var with any one of:
A lambda:
sum = [](Numbers x) { return std::accumulate(x.cbegin(), x.cend(), 0); } // std::accumulate comes from #include <numeric>
A function pointer:
int myFunc(Numbers nums) {
int result = 0;
for (int i : nums)
result += i;
return result;
}
sum = &myFunc;
Something that 'bind' has created:
struct Adder {
int startNumber = 6;
int doAdding(Numbers nums) {
int result = 0;
for (int i : nums)
result += i;
return result;
}
};
...
Adder myAdder{2}; // Make an adder that starts at two
sum = std::bind(&Adder::doAdding, myAdder);
Then finally to use it, it's a simple if statement:
if (sum)
return sum(x);
In summary, functors are the new pointer to a function, however they're more versatile. May actually be inlined if the compiler is sure enough, but generally are the same as a function pointer.
When combined with std::bind and lambda's they're quite superior to old style C function pointers.
But remember they work in c++11 and above environments. (Not in C or C++03).
In C++, a modified version of the trick for checking if a member exists should give you what you're looking for, at compile time instead of runtime:
#include <iostream>
#include <type_traits>
namespace
{
template <class T, template <class...> class Test>
struct exists
{
template<class U>
static std::true_type check(Test<U>*);
template<class U>
static std::false_type check(...);
static constexpr bool value = decltype(check<T>(0))::value;
};
template<class U, class = decltype(sum(std::declval<U>(), std::declval<U>()))>
struct sum_test{};
template <class T>
void validate_sum()
{
if constexpr (exists<T, sum_test>::value)
{
std::cout << "sum exists for type " << typeid(T).name() << '\n';
}
else
{
std::cout << "sum does not exist for type " << typeid(T).name() << '\n';
}
}
class A {};
class B {};
void sum(const A& l, const A& r); // we only need to declare the function, not define it
}
int main(int, const char**)
{
validate_sum<A>();
validate_sum<B>();
}
Here's the output using clang:
sum exists for type N12_GLOBAL__N_11AE
sum does not exist for type N12_GLOBAL__N_11BE
I should point out that weird things happened when I used an int instead of A (sum() has to be declared before sum_test for the exists to work, so maybe exists isn't the right name for this). Some kind of template expansion that didn't seem to cause problems when I used A. Gonna guess it's ADL-related.
This answer is for global functions, as a complement to the other answers on testing methods. This answer only applies to global functions.
First, provide a fallback dummy function in a separate namespace. Then determine the return type of the function-call, inside a template parameter. According to the return-type, determine if this is the fallback function or the wanted function.
If you are forbidden to add anything in the namespace of the function, such as the case for std::, then you should use ADL to find the right function in the test.
For example, std::reduce() is part of c++17, but early gcc compilers, which should support c++17, don't define std::reduce(). The following code can detect at compile-time whether or not std::reduce is declared. See it work correctly in both cases, in compile explorer.
#include <numeric>
namespace fallback
{
// fallback
std::false_type reduce(...) { return {}; }
// Depending on
// std::recuce(Iter from, Iter to) -> decltype(*from)
// we know that a call to std::reduce(T*, T*) returns T
template <typename T, typename Ret = decltype(reduce(std::declval<T*>(), std::declval<T*>()))>
using return_of_reduce = Ret;
// Note that due to ADL, std::reduce is called although we don't explicitly call std::reduce().
// This is critical, since we are not allowed to define any of the above inside std::
}
using has_reduce = fallback::return_of_reduce<std::true_type>;
// using has_sum = std::conditional_t<std::is_same_v<fallback::return_of_sum<std::true_type>,
// std::false_type>,
// std::false_type,
// std::true_type>;
#include <iterator>
int main()
{
if constexpr (has_reduce::value)
{
// must have those, so that the compile will find the fallback
// function if the correct one is undefined (even if it never
// generates this code).
using namespace std;
using namespace fallback;
int values[] = {1,2,3};
return reduce(std::begin(values), std::end(values));
}
return -1;
}
In cases, unlike the above example, when you can't control the return-type, you can use other methods, such as std::is_same and std::contitional.
For example, assume you want to test if function int sum(int, int) is declared in the current compilation unit. Create, in a similar fashion, test_sum_ns::return_of_sum. If the function exists, it will be int and std::false_type otherwise (or any other special type you like).
using has_sum = std::conditional_t<std::is_same_v<test_sum_ns::return_of_sum,
std::false_type>,
std::false_type,
std::true_type>;
Then you can use that type:
if constexpr (has_sum::value)
{
int result;
{
using namespace fallback; // limit this only to the call, if possible.
result = sum(1,2);
}
std::cout << "sum(1,2) = " << result << '\n';
}
NOTE: You must have to have using namespace, otherwise the compiler will not find the fallback function inside the if constexpr and will complain. In general, you should avoid using namespace since future changes in the symbols inside the namespace may break your code. In this case there is no other way around it, so at least limit it to the smallest scope possible, as in the above example
/* Thanks to anyone looking at this who might attempt to answer it. I'm really not trying to waste anyone's time here, but I have beat my head on this for about three days. I realize it is probably very simple for someone who understands it. I have tried most every possible combination I can think of and still get compiler errors.
C:\random\RNDNUMTEST.cpp(41) : error C2102: '&' requires l-value
I am trying to pass a pointer as a parameter to a function makeRndmNumber() for the member function fstream.open(). I want to open the file in RNDNUMTEST.cpp and then pass it to makeRndmNumber() so that it can be modified in some way. I have looked online for help, including this website, but I feel like I am overlooking something important or simple or maybe I am just missing the concept altogether.
This isn't for homework, I'm not a college student. Although I did go to school for it, it has been over 10 years since I've done any programming and I never really understood this that well to begin with. Any suggestions would be appreciated.
// These are only excerpts from the actual files.
// RndmNum_Class.h file
typedef void(fstream::*fStream_MPT)(const char*); // fStream_MPT (Member Pointer Type)
class RandomNumber {
public:
RandomNumber();
~RandomNumber() {};
static void loadDigits(double, double, char array[]);
static int getLastNDigits(char array[], int);
static int makeRndmNumber(int, int, fStream_MPT);
};
//*************************************************************8
//RndmNum_Class.cpp file
int RandomNumber::makeRndmNumber(int seed, int _fileSize, fStream_MPT FILE) {
......
}
//**************************************************************/
// RNDNUMTEST.cpp file
#include "RndmNum_Class.h"
int main() {
const char* RNDM_FILE = "c:\\RandomFile.txt";
fstream FStream_Obj;
// FStream_Obj.open(RNDM_FILE);
fStream_MPT FileMembPtr = &FStream_Obj.open(RNDM_FILE);
//fStream_MPT FileMembPtr = &fstream::open;
int seed = 297814;
int size = 20000;
cout << RandomNumber::makeRndmNumber(seed, size, FileMembPtr);
return 0;
}
This: &FStream_Obj.open(RNDM_FILE) is not taking the address of the function, it's trying to take the address of the return value of a call to that function. But that function returns void, hence the error message.
First, change the function definition from typedef void(fstream::*fStream_MPT)(const char*); to typedef void(fstream::*fstream_MPT)(const char*,ios_base::openmode), there is a default parameter you are forgetting.
Change the fStream_MPT FileMembPtr = &FStream_Obj.open(RNDM_FILE); to fStream_MPT FileMembPtr = &fstream::open; as per your comment, and add an additional parameter to makeRndNumber, a pointer to an fstream to operate on.
int RandomNumber::makeRndmNumber(int seed, int _fileSize, fStream_MPT FILE, fstream *file)
{
((*file).*FILE)("ExampleText",ios_base::in | ios_base::out);
}
FILE = fstream::open;
EDIT
This could also be done a little cleaner with std::function objects.
First redefine your type.
typedef std::function<void(const char*)> fStream_MPT;
Then when you assign, be sure to bind your objects.
fStream_MPT FILE = std::bind(&fstream::open,&file,std::placeholders::_1, ios_base::in | ios_base::out);
Then in your function you simply call the function
int RandomNumber::makeRndmNumber(int seed, int _fileSize, fStream_MPT FILE)
{
FILE("Example text");
}
It doesn't make any sense: member function pointers is used so you can apply different member functions somewhere without knowing which exact function is called. It is like passing the function's name around (except that the name is resolved at compile-time). It doesn't seem that this is what you want to do!
Even if you would correctly obtain the function's address (rather than trying to get the address of the result of calling open()), it wouldn't work because std::fstream::open() takes two arguments: the second argument is for the open-mode and it is defaulted to std::ios_base::in | std::ios_base::out.
I'm not quite sure what you really want to d but it seems you want to pass the file stream around. The normal way to do this is to pass a reference to a std::iostream as argument to the function. Well, actually you probably want to use a std::ifstream initially and hence pass the argument as std::istream&.
Is it possible to store pointers to various heterogenous functions like:
In the header:
int functionA (int param1);
void functionB (void);
Basically this would the part I don't know how to write:
typedef ??boost::function<void(void)>?? functionPointer;
And afterwards:
map<char*,functionPointer> _myMap;
In the .cpp
void CreateFunctionMap()
{
_myMap["functionA"] = &functionA;
_myMap["functionB"] = &functionB;
...
}
And then reuse it like:
void execute(int argc, char* argv[])
{
if(argc>1){
int param = atoi(argv[1]);
int answer;
functionPointer mfp;
mfp = map[argv[0]];
answer = *mfp(param);
}
else{
*map[argv[0]];
}
}
etc.
Thanks
--EDIT--
Just to give more info:
The reason for this question is that I am implementing a drop-down "quake-style" console for an already existing application. This way I can provide runtime command line user input to access various already coded functions of various types i.e.:
/exec <functionName> <param1> <param2> ...
If you want to have "pointer to something, but I'm not going to define what, and it could be a variety of things anyway" you can use void *.
But you really shouldn't.
void * is purely a pointer. In order to do anything with it, you have to cast it to a more meaningful pointer, but at that point, you've lost all type safety. What's to stop someone from using the wrong function signature? Or using a pointer to a struct?
EDIT
To give you a more useful answer, there's no need to put this all into a single map. It's ok to use multiple maps. I.e.
typedef boost::function<void(void)> voidFunctionPointer;
typedef boost::function<int(int)> intFunctionPointer;
map<std::string, voidFunctionPointer> _myVoidMap;
map<std::string, intFunctionPointer > _myIntMap;
void CreateFunctionMap()
{
_myVoidMap["functionA"] = &functionA;
_myIntMap["functionB"] = &functionB;
...
}
void execute(int argc, char* argv[])
{
if(argc>1){
int param = atoi(argv[1]);
int answer;
// todo: check that argv[0] is actually in the map
intFunctionPointer mfp = _myIntMap[argv[0]];
answer = mfp(param);
}
else{
// todo: check that argv[0] is actually in the map
voidFunctionPointer mfp = _myVoidMap[argv[0]];
mfp();
}
}
You can use
boost::variant<
boost::function<void(void)>,
boost::function<void(int)> >
Why not just add functions of type int (*func)(int argc, char* argv[])? You could easily remove first arg from execute's params and call the relevant one.
Can you not use the command pattern to encapsulate the function calls. So you can store the functions in functors and call them after wards. For functor implementation you can have a look at Modern C++ Design by Andrei Alexandrescu.
Each of your functions has a different type, so you need some kind of type erasure. You could use the most generic of them: Boost.Any. You can have a map of boost::any, but you need to know the type of the function in order to get it back and call it.
Alternatively, if you know your arguments ahead of time you can bind them with the function call and have all functions in the map be nullary functions: function< void() >. Even if you don't, you may be able to get away with it by binding the argument to references, and then at call time fill the referred variables with the appropiate arguments.