In Windows/MSVS/C++ I can get a function pointer by pointing to its name like this:
void foo()
{
auto fooPtr = &foo;
}
But can I do the same thing without knowing the name of the function?
void foo()
{
auto fnPtr = &thisFunction; //no
}
Use case: I want to define a macro I can put at the top of many functions which will declare a pointer to the function. Ex:
#define defFnPtr auto fnPtr = &thisFunction
void foo()
{
defFnPtr;
}
void bar()
{
defFnPtr;
}
No, there is no way in standard C++ to get a pointer to the "current" function.
Best that you could do is perhaps to use meta programming: Write a program that generates the line auto fnPtr = &foo; into the source.
That said, I don't think that the goal is worth the effort.
Related
Is the following possible in c++?
int do_stuff(int one, int two)
{
...
return ...;
}
int main()
{
void *ptr = &do_stuff(6, 7);//DON'T CALL THE FUNCTION, just store a pointer
cout << *ptr;//call the function from the pointer without having to pass the arguments again
}
I know this can be done with classes, but is it possible the way I am trying to do it?
Well with c++11 and a little magic from std::function and std::bind you can.
std::function<int()> f = std::bind(&do_stuff,6,7);
std::cout << f();
No, not like that. There is nothing in the code
void *ptr = &do_stuff(6, 7);
that makes it parse like you want. I'm unsure if it would parse at all, if you can take the address of the return value. Taking the address of a function is basically a no-op, but function pointers don't convert to void * so it has problems anyway.
You need more magic, like C++11's lambda closures.
I'm no C++11 programmer, but I guess your code would look like:
int main(void)
{
auto func = [] () { do_stuff(6, 7); };
func();
return 0;
}
I will like to know, how to make a function where I define a function. And then I can call the defined function. Let me try with a example.
void funcToCall() {
std::cout<<"Hello World"<<std::endl;
}
void setFuncToCall(void func) {
//Define some variable with func
}
void testFuncCall() {
//Call function that has been defined in the setFuncToCall
}
setFuncToCall(funcToCall()); //Set function to call
testFuncCall(); //Call the function that has been defined
I Hope that you understand what I am trying to do here. But I don't know how to bring It down to the right code :-)
You need a function pointer. It's easier to work with function pointers if you typedef them first.
typedef void (*FuncToCallType)();
FuncToCallType globalFunction; // a variable that points to a function
void setFuncToCall(FuncToCallType func) {
globalFunction = func;
}
void testFuncCall() {
globalFunction();
}
setFuncToCall( funcToCall ); //Set function to call,NOTE: no parentheses - just the name
testFuncCall(); //Call the function that has been defined
As other answers have suggested you can use objects like functions, too. But this requires operator overloading (even if it remains hidden from you) and is typically used with templates.
It gives more flexibility (you can set some state to the object before passing it to the function and the object's operator() can use that state) but in your case function pointers may be just as good.
What you want to use is a callback, and is has been answered her : Callback functions in c++
I would advise you to use std::tr1::function ( generalized callback )
C syntax for function pointers is a bit weird, but here we go:
// this is a global variable to hold a function pointer of type: void (*)(void)
static void (*funcp)(void);
// you can typedef it like this:
//typedef void (*func_t)(void);
// - now `func_t` is a type of a pointer to a function void -> void
// here `func` is the name of the argument of type `func_t`
void setFuncToCall(void (*func)(void)) {
// or just: void setFuncToCall(func_t func) {
//Define some variable with func
...
funcp = func;
}
void testFuncCall(void) {
//Call function that has been defined in the setFuncToCall
funcp();
}
setFuncToCall(funcToCall); // without () !
testFuncCall();
I need to pass something like a pointer that takes anything as a function parameter. You know, something without any predefined type or a type that can take anything like this:
void MyFunc( *pointer );
And then use it like:
char * x = "YAY!";
MyFunc(x);
int y = 10;
MyFunc(&y);
MyObj *b = new MyObj();
MyFunc(b);
And I don't want to use templates because I am mostly using C in my project.
Is there anything that can be used here except a function macro?
In C++, Boost.Any will let you do this in a type-safe way:
void func(boost::any const &x)
{
// any_cast a reference and it
// will throw if x is not an int.
int i = any_cast<int>(x);
// any_cast a pointer and it will
// return a null pointer if x is not an int.
int const *p = any_cast<int>(&x);
}
// pass in whatever you want.
func(123);
func("123");
In C, you would use a void pointer:
void func(void const *x)
{
// it's up to you to ensure x points to an int. if
// it's not, it might crash or it might silently appear
// to work. nothing is checked for you!
int i = *(int const*)x;
}
// pass in whatever you want.
int i = 123;
func(&i);
func("123");
You seem adverse to it but I'll recommend it anyway: if you're using C++, embrace it. Don't be afraid of templates. Things like Boost.Any and void pointers have a place in C++, but it is very small.
Update:
Well , I am making a small signals - slots - connections library to be
used with my gui toolkit. So that I can get rid of the Ugly WNDPROC. I
need these pointers for the connections.
If you need multi-target signals, Boost.Signals already provides a full and tested signals/slots implementation. You can use Boost.Bind (or std::bind, if you've got a C++0x compiler) to connect member functions:
struct button
{
boost::signal<void(button&)> on_click;
}
struct my_window
{
button b;
my_window()
{
b.on_click.connect(std::bind(&my_window::handle_click,
this, std::placeholders::_1));
}
void handle_click(button &b)
{
}
void simulate_click()
{
b.on_click(b);
}
};
If you only want a simple callback, Boost.Function (or std::function if you've got a C++0x compiler) will work well:
struct button
{
std::function<void(button&)> on_click;
}
struct my_window
{
button b;
my_window()
{
b.on_click = std::bind(&my_window::handle_click,
this, std::placeholders::_1);
}
void handle_click(button &b)
{
}
void simulate_click()
{
b.on_click(b);
}
};
You can use a function that takes a void*, but you must be aware of the pointer types that are not compatible with void*:
pointers to functions:
void MyFunc(void*);
MyFunc(&MyFunc); // WRONG
pointers to members:
void MyFunc(void*);
struct A { int x; };
MyFunc(&A::x); // WRONG
While these pointers are not compatible with void* (even with casting, on some compilers), they are themselves data. So you can pass a pointer to the pointer:
void MyFunc(void*);
void (*pfn)(void*) = &MyFunc;
MyFunc(&pfn); // ok
struct A { int x; };
int A::*px = &A::x;
MyFunc(&px); // ok
You can define the method as taking one void * argument. Of course, at that point, it's up to you to figure out what to do with the data (as far as accessing it or casting it.)
void MyFunc(void * ptr);
You could use:
void MyFunc( void* p){}
int g = 10;
MyFunc( (void*)&g );
void * is the way to do it. You can assign any pointer type to and from a void *. But to use the pointer in the called function, you'll have to know the type so you can create an appropriate local pointer or cast appropriately. You can encode a limited set of types as enum symbols, and perhaps use a switch to select type-specific behavior. But without a specific purpose or use-case, you might end up chasing your tail in a quest for generality for which C was never intended.
Another way would be to make a union to contain all the various types you know are needed.
typedef union {
int i;
char c;
float f;
} vartype;
Then if the value can carry around its own type-identifier, it becomes a tag-union or variant-record.
typedef struct {
enum type { INT, CHAR, FLOAT } type;
vartype var;
} varrec;
I have a class Test with a peculiar data structure.
A member of class Test is a std::map where the key is a std::string and the mapped value is a struct defined as follows:
typedef struct {
void (Test::*f) (void) const;
} pmf_t;
Initialization of the map is OK. The problem is when I am trying to call the function pointed. I made up a toy example reproducing the problem. Here it is:
#include <iostream>
#include <map>
using namespace std;
class Test;
typedef void (Test::*F) (void) const;
typedef struct {
F f;
} pmf_t;
class Test
{
public:
Test () {
pmf_t pmf = {
&Test::Func
};
m["key"] = pmf;
}
void Func (void) const {
cout << "test" << endl;
}
void CallFunc (void) {
std::map<std::string, pmf_t>::iterator it = m.begin ();
((*it).second.*f) (); // offending line
}
std::map<std::string, pmf_t> m;
};
int main ()
{
Test t;
t.CallFunc ();
return 0;
}
Thanks in advance,
Jir
The name of the pmf_t type is f, so the first change is to remove the * to get second.f. That gives you a pointer-to-member value. To use a pointer-to-member, you need an instance. The only one you have available of the correct type is this, so use it with the ->* operator:
(this->*it->second.f)();
You need parentheses around the whole thing, or else the compiler thinks you're trying to call it->second.f() (which isn't allowed) and then applying the result to ->*.
The offending line is trying to call a member function without any object to call it on. If the intention is to call it for the this object, I believe the call should look like
( this->* ((*it).second.f) )();
Where this->* is the syntax for dereferencing a pointer-to-member for the current object. ((*it).second.f) is the pointer retrieved from the map, and () is the call operator for actually calling the function.
This is perhaps good as an exercise, but otherwise of limited use.
I think you might want to check out the C++ FAQ on this one. The syntax is apparently pretty tricky to get right (they actually recommend using a macro).
It might be too late for this question but, the seemingly complex synatax can be break down to two simple lines so it looks pretty clear:
void CallFunc (void)
{
pmf_t t = m["key"]; //1>get the data from key
(this->*t.f)(); //2>standard procedure to call pointer to member function
}
try this:
(this->*((*it).second.f)) ();
Is there a way, I can switch between 2 similar function sets (C/C++) in an effective way?
To explain better what I mean, lets say I have 2 sets of global functions like:
void a_someCoolFunction();
void a_anotherCoolFunction(int withParameters);
…
void b_someCoolFunction();
void b_anotherCoolFunction(int withParameters);
…
And I want to able to "switch" in my program at runtime which one is used. BUT: I dont want to have one if condition at every function, like:
void inline someCoolFunction(){
if(someState = A_STATE){
a_someCoolFunction();
}else{
b_someCoolFunction();
}
}
Because, I expect that every function is called a lot in my mainloop - so It would be preferable if I could do something like this (at start of my mainloop or when someState is changed):
if(someState = A_STATE){
useFunctionsOfType = a;
}else{
useFunctionsOfType = b;
}
and then simply call
useFunctionsOfType _someCoolFunction();
I hope its understandable what I mean… My Background: Im writing an App, that should be able to handle OpenGL ES 1.1 and OpenGL ES 2.0 both properly - but I dont want to write every render Method 2 times (like: renderOpenGL1() and renderOpenGL2() I would rather to write only render()). I already have similiar Methods like: glLoadIdentity(); myLoadIdentity(); … But need a way to switch between these two somehow.
Is there any way to accomplish this in an efficent way?
Several options, including (but not limited to):
Use function pointers.
Wrap them in classes, and use polymorphism.
Have two separate copies of the loop.
But please profile to ensure this is actually a problem, before you make any large changes to your code.
As the question seems to be interested in a C++ solution and no-one has spelt out the polymorphic solution (too obvious?), here goes.
Define an abstract base class with the API you require, and then implement a derived class for each supported implementation:
class OpenGLAbstract
{
public:
virtual ~OpenGLAbstract() {}
virtual void loadIdentity() = 0;
virtual void someFunction() = 0;
};
class OpenGLEs11 : public OpenGLAbstract
{
public:
virtual void loadIdentity()
{
// Call 1.1 API
}
virtual void someFunction()
{
// Call 1.1 API
}
};
class OpenGLEs20 : public OpenGLAbstract
{
public:
virtual void loadIdentity()
{
// Call 2.0 API
}
virtual void someFunction()
{
// Call 2.0 API
}
};
int main()
{
// Select the API to use:
bool want11 = true;
OpenGLAbstract* gl = 0;
if (want11)
gl = new OpenGLEs11;
else
gl = new OpenGLEs20;
// In the main loop.
gl->loadIdentity();
delete gl;
}
Note that this is exactly the sort of thing that C++ was intended for, so if can use C++ here, this is the simplest way to go.
Now a more subtle issue you might face is if your 2.0 version requires the process to load a dynamic linked library at run time with the 2.0 platform implementation. In that case just supporting the API switch is not enough (whatever the solution). Instead put each OpenGL concrete class in its own linked library and in each provide a factory function to create that class:
OpenGlAbstract* create();
Then load the desired library at run time and call the create() method to access the API.
In C (since it seems you want both C and C++) this is done with pointer to functions.
// Globals. Default to the a_ functions
void(*theCoolFunction)() = a_someCoolFunction;
void(*theOtherCoolFunction)(int) = a_anotherCoolFunction;
// In the code ...
{
...
// use the other functions
theCoolFunction = b_someCoolFunction;
theOtherCoolFunction = b_anotherCoolFunction;
...
}
You might probably want to switch those functions in groups, so you better set a array of pointers to functions and pass that array around. If you decide to do so, you might probably want to also define some macro to ease the reading:
void (*functions_a[2])();
void (*functions_b[2])();
void (**functions)() = functions_a;
....
#define theCoolFunction() functions[0]()
#define theOtherCoolFunction(x) functions[1](x)
....
// switch grooup:
functions = functions_b;
but in this case you'll lose the static check on argument types (and you have to initialize the array, of course).
I guess in C++ you will have instatiate two different objects with the same parent class and different implementation for their methods (but I'm no C++ prograammer!)
You could use functions pointers. You can read a lot about them if you google it, but briefly a function pointer stores a pointer to a function's memory address.
Function pointers can be used the same way as a funcion, but can be assigned the address of different functions, making it a somehow "dynamic" function. As an example:
typedef int (*func_t)(int);
int divide(int x) {
return x / 2;
}
int multiply(int x) {
return x * 2;
}
int main() {
func_t f = ÷
f(2); //returns 1
f = &multiply;
f(2); //returns 4
}
Something like boost::function (std::function) would fit the bill. Using your example:
#include <iostream>
#include <boost/function.hpp> //requires boost installation
#include <functional> //c++0x header
void a_coolFunction() {
std::cout << "Calling a_coolFunction()" << std::endl;
}
void a_coolFunction(int param) {
std::cout << "Calling a_coolFunction(" << param << ")" << std::endl;
}
void b_coolFunction() {
std::cout << "Calling b_coolFunction()" << std::endl;
}
void b_coolFunction(int param) {
std::cout << "Calling b_coolFunction(" << param << ")" << std::endl;
}
float mul_ints(int x, int y) {return ((float)x)*y;}
int main() {
std::function<void()> f1; //included in c++0x
boost::function<void(int)> f2; //boost, works with current c++
boost::function<float(int,int)> f3;
//casts are necessary to resolve overloaded functions
//otherwise you don't need them
f1 = static_cast<void(*)()>(a_coolFunction);
f2 = static_cast<void(*)(int)>(a_coolFunction);
f1();
f2(5);
//switching
f1 = static_cast<void(*)()>(b_coolFunction);
f2 = static_cast<void(*)(int)>(b_coolFunction);
f1();
f2(7);
//example from boost::function documentation. No cast required.
f3 = mul_ints;
std::cout << f3(5,3) << std::endl;
}
Compiled with g++-4.4.4, this outputs:
Calling a_coolFunction()
Calling a_coolFunction(5)
Calling b_coolFunction()
Calling b_coolFunction(7)
15
The biggest limitation is that the types of f1,f2, etc cannot change, so any function you assign to them must have the same signature (i.e. void(int) in the case of f2).
The simple way could be storing pointers to functions, and change them od demand.
But the better way is to use something similar to abstract factory design pattern. The nice generic implementation can be found in Loki library.
In C you would typically do this with a struct containing function pointers:
struct functiontable {
void (*someCoolFunction)(void);
void (*anotherCoolFunction)(int);
};
const struct functiontable table_a = { &a_someCoolFunction, &a_anotherCoolFunction };
const struct functiontable table_b = { &b_someCoolFunction, &b_anotherCoolFunction };
const struct functiontable *ftable = NULL;
To switch the active function table, you'd use:
ftable = &table_a;
To call the functions, you'd use:
ftable->someCoolFunction();