Re: How to invoke the methods of an object selecting them from a table
On 2008-02-05, Stefano Sabatini <stefano.sabatini@caos.org> wrote:
On 2008-02-05, Sam <sam@email-scan.com> wrote:
Stefano Sabatini writes:
[...]
The problem with this solution is that I have to create for each
method a corresponding static function, can you contrive a solution
which doesn't require this?
Many thanks in advance for any advice.
You want to use class member pointers:
// Declare:
class Foo {
public:
void x();
void y();
};
// Don't use arrays, when you really want a map:
std::map<std::string, void (Foo::*)() > methods;
// Initialize:
methods["x"]= &Foo:x;
methods["y"]= &Foo:y;
// Invoke some method on an object.
void callFoo(Foo *obj, std::string methodname)
{
std::map<std::string, void (Foo::*)() >::iterator p=
methods.find(methodname);
if (p == methods.end())
throw "You are a bozo";
(obj->*(p->second))();
}
Hi Ondra and Sam, yes I finally got in touch with the
member pointers feature of C++, thank you so much for your hints and
explanations!! :-)
I already posted a solution using the memeber pointers, well now I see
I should rather use std::map rather than a plain array.
And here it is a solution using map:
#include <string>
#include <iostream>
#include <map>
using namespace std;
class DumbThing {
public:
void sayFoo() { cout << "Foo!" << endl; }
void sayBar() { cout << "Bar!" << endl; }
};
int main(void) {
// declare a map
map<string, void (DumbThing::*)() > methods;
// initialize it
methods["sayFoo"] = &DumbThing::sayFoo;
methods["sayBar"] = &DumbThing::sayBar;
map<string, void (DumbThing::*)() >::iterator it;
DumbThing t;
for (it=methods.begin() ; it != methods.end(); it++ ) {
cout << "Invoking method: " << (*it).first << endl;
(t.*(it->second))();
}
return 0;
}
Best regards.
--
Stefano Sabatini
Linux user number 337176 (see http://counter.li.org)
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