I want to create a static function pointer array, so I can jump to a certain function regarding a received index. Like an index jumper.
So imagine a class like this:
Class A
{
private:
static void 1stFunction();
static void 2ndFunction();
static void(*functionPointer[20])(void);
};
Then I would like that functionPointer to get the value of the 1stFunction and 2ndFunction, and maybe even more.
So, how do I initialize it?
As far as I know, when a static member is declared, you can use it even before an instance is created. So I though, lets initialize that function pointer, so later I can call it like this
functionPointer[receivedIndex]();
So i tried to initilize it like this, in the same .h file
void (*A::functionPointer[])(void) =
{
A::1stFunction,
A::2ndFunction,
};
But the compiler gives me redifinition, it says it's already created.
So, pretty sure I'm missing something. I don't know though, if it is syntax or simply it is not possible to do it this way.
I know that function pointers to class's member functions are different than normal function pointers... But this is a static function, so I believe it doesn't belong to an instance and therefore it should work with normal function pointers.
Any help would be appreciated.
Thanks
The following would be a working example that probably achieves what you need.
You need C++11 for the initializer list.
It is a good practice to initialize the static member in the cpp file, as you don't want to have a definition of the static member everytime the header is included (this can lead to linking issues).
You can call callf with the desired index and have the corresponding function called, based on the initialization of the function pointer array.
The output of the program would be:
I am 2ndFunction
Header file
class A
{
private:
static void Function1();
static void Function2();
static void(*functionPointer[20])();
public:
static void callf(int index);
};
Implementation
#include <iostream>
#include "ex.h"
void(*A::functionPointer[20])() {
A::Function1,
A::Function2
};
void A::Function1() {
std::cout << "I am 1stFunction" << std::endl;
}
void A::Function2() {
std::cout << "I am 2ndFunction" << std::endl;
}
void A::callf(int index) {
A::functionPointer[index]();
}
int main(int argc, char const *argv[]) {
A::callf(1);
return 0;
}
Here you have a more modern C++ approach (C++14 needed)
I would advise you to explore lambda functions if you are not restricted to C++03.
#include <iostream>
#include <functional>
#include <vector>
class A {
public:
using f_type = std::function<void(void)>;
f_type f1 = []() { std::cout << "f0" << std::endl;};
f_type f2 = []() { std::cout << "f1" << std::endl;};
static void f3() { std::cout << "f3" << std::endl; }
std::vector<f_type> functions{f1, f2, f3};
};
int main() {
A a;
a.functions[0]();
a.functions[1]();
//adding custom lambda
a.functions.emplace_back([](){ std::cout << "custom f" << std::endl;});
a.functions[2]();
return 0;
}
you can add both functions and lambdas to your container.
Related
Can I send an instance to a function not by using the . operator?
For example:
// header file
class A
{
public:
void foo() {std::cout << "Hello" << std::endl;}
};
// main file
A instance = new A;
instance.foo();
// instead do something like this
A::foo(instance);
Can I do something like that?
Yes, you can indirectly via std::invoke:
#include <functional>
#include <iostream>
struct A {
void foo() {
std::cerr << "hi\n";
}
};
int main() {
A a;
std::invoke(&A::foo,a);
}
But std::invoke's implementation will internally probably just apply the .* operator.
You're more than welcome to use the pointer to member syntax.
A instance;
auto fn = &A::foo;
(instance.*fn)();
.* is a different operator than .. Whether this is more readable is left as an exercise to the reader (hint: it's not)
I want to programmatically retrieve the identifier of a C++ class instance at runtime. I'm aware that C++ doesn't support reflection yet, but is there any alternative solution out there?
For instance, given the following example:
class Foo {
Foo() {
auto name = reflect::getIdentifierName(this);
std::cout << name << std::endl;
}
};
void main() {
Foo my_obj;
}
Executing this program should print out "my_obj".
I'm looking for any utility library that I could use to implement this basic reflection function.
I'm particularly wondering if libclang can be used to extract such information - if so, any hint for how to build the reflect function to do this.
Yes, but this is implementation defined. Proceed at your own risk.
Yunnosch's suggestion sounds much more reasonable without more context.
#include <iostream>
#include <typeinfo>
class Foo {
public:
Foo() {
const char * const name = typeid(this).name();
std::cout << name << std::endl;
}
};
int main()
{
Foo my_obj;
}
why does this code work? with c++14
// Example program
#include <iostream>
#include <string>
using namespace std;
auto fun()
{
struct a
{
int num = 10;
a()
{
cout << "a made\n";
}
~a()
{
cout << "a destroyed\n";
}
};
static a a_obj;
return a_obj;
}
int main()
{
auto x = fun();
cout << x.num << endl;
}
how is the type a visible in main? if i change auto x= to a x= it obviously doesn't compile, but how does main know about the type a?
The static declaration is there since I was trying to test for something else but then I stumbled upon this behavior.
Running it here: https://wandbox.org/permlink/rEZipLVpcZt7zm4j
This is all surprising until you realize this: name visibility doesn't hide the type. It just hides the name of the type. Once you understand this it all makes sense.
I can show you this without auto, with just plain old templates:
auto fun()
{
struct Hidden { int a; };
return Hidden{24};
}
template <class T> auto fun2(T param)
{
cout << param.a << endl; // OK
}
auto test()
{
fun2(fun()); // OK
}
If you look closely you will see this is the same situation as yours:
you have a struct Hidden which is local to fun. Then you use an object of type Hidden inside test: you call fun which returns a Hidden obj and then you pass this object to the fun2 which in turn has no problem at all to use the object Hidden in all it's glory.
as #Barry suggested the same thing happens when you return an instance of a private type from a class. So we have this behavior since C++03. You can try it yourself.
C++14 is made to be more and more tolerant with auto. Your question is not clear, because you're not stating what the problem is.
Now let's tackle your question differently: Why does it not work with a x = ...?
The reason is that the struct definition is not in the scope of the main. Now this would work:
// Example program
#include <iostream>
#include <string>
using namespace std;
struct a
{
int num = 10;
};
auto fun()
{
static a a_obj;
return a_obj;
}
int main()
{
a x = fun();
cout << x.num << endl;
}
Now here it doesn't matter whether you use a or auto, because a is visible for main(). Now auto is a different story. The compiler asks: Do I have enough information to deduce (unambiguously) what the type of x is? And the answer is yes, becasue there's no alternative to a.
What do I declare with the following definition:
void (*bar)(A*){ }; //1
My first thought was that I declare and define function pointer and a function the pointer point to. But it's wrong, because any call to the bar() leads to a segmentation fault:
#include <iostream>
#include <vector>
#include <memory>
struct A{ };
void foo(A*){ std:cout << "foo" << std::endl; }
void (*bar)(){ };
int main(){
bar();
}
Moreover, I can't imbed any statement into the "definition":
void (*bar)(A*){ std::cout << "foo" << std::endl };
yeilds compile-time error.
So, what does the declaration //1 mean?
This statement:
void (*bar)(A*){ };
declares a variable named bar of type void(*)(A*), ie "pointer to function taking pointer to A and returning void", and zero-initializes it. Thus, it's equivalent to this:
void (*bar)(A*) = nullptr;
Obviously, when calling this bar, a segfault should be no surprise.
It's not possible to declare a function and a pointer to that function in a single declaration.
When you say
void (*bar)(A*){ }; //1
it means "bar" is a function pointer which can point to some function which takes "A*" as parameter.
In your case, it is not pointing to any function yet.
to make it working use,
void (*bar)(A*) = foo;
This means you have declared a function pointer that points to nothing at the moment. You should able to validate that using a debugger.
void (*bar)(A*){ }; //1
You could make the pointer point to a function like this:
void foo(A*){ std::cout << "foo" << std::endl };
bar = &foo;
And call it like this now:
A a;
bar(&a);
Full snippet:
#include <iostream>
class A {};
void (*bar)(A*){};
void foo(A*) { std::cout << " foo " << std::endl;}
int main() {
A a;
bar = &foo;
bar(&a);
}
Your code should be changed to the following code.
#include <iostream>
#include <vector>
#include <memory>
struct A{ };
void foo(A*){ std::cout << "foo" << std::endl; }
void (*bar)(A*);
int main(){
A a;
bar = &foo;
bar(&a);
}
To declare an actual function, get rid of the (*) portion around the function name:
void bar(A*){ std::cout << "foo" << std::endl };
https://ideone.com/UPIYxg
So, what does the declaration //1 mean?
It is just a comment.
I have the following problem. I have a function from an external library (which cannot be modified) like this:
void externalFunction(int n, void udf(double*) );
I would like to pass as the udf function above a function member of an existing class. Please look at the following code:
// External function (tipically from an external library)
void externalFunction(int n, void udf(double*) )
{
// do something
}
// User Defined Function (UDF)
void myUDF(double* a)
{
// do something
}
// Class containing the User Defined Function (UDF)
class myClass
{
public:
void classUDF(double* a)
{
// do something...
};
};
int main()
{
int n=1;
// The UDF to be supplied is myUDF
externalFunction(n, myUDF);
// The UDF is the classUDF member function of a myClass object
myClass myClassObj;
externalFunction(n, myClassObj.classUDF); // ERROR!!
}
I cannot declare the classUDF member function as a static function, so the last line of the code above results in a compilation error!
This is impossible to do - in c++, you must use either a free function, or a static member function, or (in c++11) a lambda without capture to get a function pointer.
GCC allows you to create nested function which could do what you want, but only in C. It uses so-called trampolines to do that (basically small pieces of dynamically generated code). It would be possible to use this feature, but only if you split some of the code calling externalFunction to a separate C module.
Another possibility would be generating code at runtime eg. using libjit.
So if you're fine with non-reenrant function, create a global/static variable which will point to this and use it in your static function.
class myClass
{
public:
static myClass* callback_this;
static void classUDF(double* a)
{
callback_this.realUDF(a);
};
};
Its really horrible code, but I'm afraid you're out of luck with such a bad design as your externalFunction.
You can use Boost bind or TR1 bind (on recent compilers);;
externalFunction(n, boost::bind(&myClass::classUDF, boost::ref(myClassObj)));
Unfortunately, I lived in a pipe dream for the last 10 minutes. The only way forward is to call the target using some kind of a static wrapper function. The other answers have various neat (compiler-specific) tidbits on that, but here's the main trick:
void externalFunction(int n, void (*udf)(double*) )
{ double x; udf(&x); }
myClass myClassObj;
void wrapper(double* d) { myClassObj.classUDF(d); }
int main()
{
externalFunction(1, &wrapper);
}
std::function<>
Store a bound function in a variable like this:
std::function<void(double*)> stored = std::bind(&myClass::classUDF, boost::ref(myClassObj))
(assuming C++0x support in compiler now. I'm sure Boost has a boost::function<> somewhere)
Vanilla C++ pointers-to-member-function
Without magic like that, you'd need pointer-to-memberfunction syntax:
See also live on http://ideone.com/Ld7It
Edit to clarify to the commenters, obviously this only works iff you have control over the definition of externalFunction. This is in direct response to the /broken/ snippet int the OP.
struct myClass
{
void classUDF(double* a) { };
};
void externalFunction(int n, void (myClass::*udf)(double*) )
{
myClass myClassObj;
double x;
(myClassObj.*udf)(&x);
}
int main()
{
externalFunction(1, &myClass::classUDF);
}
C++98 idiomatic solution
// mem_fun_ref example
#include <iostream>
#include <functional>
#include <vector>
#include <algorithm>
#include <string>
int main ()
{
std::vector<std::string> numbers;
// populate vector:
numbers.push_back("one");
numbers.push_back("two");
numbers.push_back("three");
numbers.push_back("four");
numbers.push_back("five");
std::vector <int> lengths (numbers.size());
std::transform (numbers.begin(), numbers.end(), lengths.begin(),
std::mem_fun_ref(&std::string::length));
for (int i=0; i<5; i++) {
std::cout << numbers[i] << " has " << lengths[i] << " letters.\n";
}
return 0;
}
Here is how I do this, when MyClass is a singleton:
void externalFunction(int n, void udf(double) );
class MyClass
{
public:
static MyClass* m_this;
MyClass(){ m_this = this; }
static void mycallback(double* x){ m_this->myrealcallback(x); }
void myrealcallback(double* x);
}
int main()
{
MyClass myClass;
externalFunction(0, MyClass::mycallback);
}