2013-07-26 13:03, Murray Cumming skrev:
C+11 has std::function<> which is a bit like sigc::slot, though C++11
doesn't have anything like sigc::signal<>. I played with that here:
http://www.murrayc.com/permalink/2013/07/08/c-in-glom-shared_ptr-and-slotsfunctions/
I've noticed that std::function<> works with libsigc++, as in the code
below, but I wonder why it works. Can anyone explain?
#include <iostream>
#include <string>
#include <sigc++/sigc++.h>
#include <functional>
void on_print(const std::string& str)
{
std::cout << str;
}
int main()
{
sigc::signal<void, const std::string&> signal_print;
std::function<void(const std::string&)> slot = &on_print;
signal_print.connect(slot);
signal_print.emit("hello world\n");
return 0;
}
std::function<> is a functor (function object). Not much more than that
is required in order to convert it to a sigc::slot. What *is* required,
is that the compiler must be able to check that the return type is
correct. That's described at
https://developer.gnome.org/libsigc++/unstable/group__sigcfunctors.html
Your example works because the return type is void. That's some kind of
default assumption in libsigc++. std::function<>s with other return
types work only if you include /namespace sigc {
SIGC_FUNCTORS_DEDUCE_RESULT_TYPE_WITH_DECLTYPE }/ in your code.
sigc::trackable derived objects don't get any special treatment, if they
are included in std::function or a C++11 lambda expression. Therefore I
added sigc::track_obj(), but it has not yet made it into a libsigc++
release.
There has been some discussion about libsigc++ and C++11 lambda
expressions over the last two years or so. I think most of what has been
said about C++11 lambda expressions can be directly translated to
std::function.
Kjell
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