How do I convert a member function pointer to a static function?
Here is my code:
class ClassA
{
public:
int n;
void func();
};
void ClassA::func()
{
n = 89;
}
class ClassB
{
public:
float f1;
float f2;
void func(float f);
};
void ClassB::func( float f )
{
f1 += f;
f2 += f;
}
int main (int argc, char *argv[])
{
ClassA* a = new ClassA;
ClassB* b = new ClassB;
//PROBLEM IS HERE
void (* pf_func1)(void*) = ClassA.func;
void (* pf_func2)(void*, float) = ClassB.func;
pf_func1(a);
pf_func2(b, 10);
}
You could std::bind
it to an instance of the relevant class:
auto func1 = std::bind(&ClassA::func, a);
func1();
This binds the member function Class::func
to a
. And similarly:
auto func2 = std::bind(&ClassB::func, b, std::placeholders::_1);
func2(10.0f);
Alternatively, you can use std::mem_fn
to allow you to easily change the object that it is called on, providing the syntax that you asked for:
auto func1 = std::mem_fn(&ClassA::func);
func1(a);
auto func2 = std::mem_fn(&ClassB::func);
func2(b, 10.0f);
Not that in both cases func1
and func2
aren't actually function pointers, but they do behave like them. The types returned from std::bind
and std::mem_fn
are implementation defined. They are both, however, convertable to a std::function
.
void(ClassA::*pf1)() = &ClassA::func;
void(ClassB::*pf2)(float) = &ClassB::func;
void (__thiscall * pf_func1)(void*) = (void (__thiscall *)(void*)) ((void*&)pf1);
void (__thiscall * pf_func2)(void*, float) = (void (__thiscall *)(void*, float)) ((void*&)pf2);
SOLVED
:)
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