function-pointers

If I have a lambda which captures all automatic variables by reference ( [&] {} ), why can't it be converted to a function pointer? A regular function can modify variables just like a lambda that captures everything by reference can, so why is it not the same? I guess in other words, what is the functional difference between a lambda with a & capture list and a regular function such that the lambda is not convertible to a function pointer? So let's take the example of a trivial lambda: Object o; auto foo = [&]{ return o; }; What does the type of foo look like? It might look something like this

Is there any difference between pointer to const and usual pointer for functions? When it is suitable to use const qualifier for stand alone functions? I wrote short sample to illustrate my question: #include <iostream> using namespace std; int sum( int x, int y ) { return x + y; } typedef int sum_func( int, int ); int main() { const sum_func* sum_func_cptr = ∑ // const function sum_func* sum_func_ptr = ∑ // non-const function ? // What is the difference between sum_func_cptr and sum_func_ptr int x = sum_func_cptr( 2, 2 ); cout << x << endl; int y = sum_func_ptr( 2, 2 ); cout << y << endl; sum

I am using VBA behind MS Access Say I have global methods foo1 and foo2 that gets the same arguments but do different things. I know that in C++ I can assign a function an alias. Something like: instead of: If (term) then foo1 arg1, arg2, arg3 else foo2 arg1, arg2, arg3 End If I want to write: Var new_func = Iff (term, foo1,foo2) new_func arg1, arg2, arg3 Can I do this on vba? Would Run suit? new_func = IIf(term, "foo1", "foo2") ''new_func arg1, arg2, arg3 res = Run(new_func, "a", "b", 1) More information: http://msdn.microsoft.com/en-us/library/aa199108(office.10).aspx Alex K. If the 2

EDIT: MOTIVATION Suppose I define a Handler class as class Handler { public: class Message { /*...*/ }; typedef int (*Callback)(Message *msg); void registerCallback(int msgclass, Callback f); }; A client can do int f1(Handler::Message *msg) { /* handle message */ } int f2(Handler::Message *msg) { /* handle message */ } int main(){ Handler h; h.registerCallback(1, f1); h.registerCallback(2, f2); // .... } The compiler will indeed check that f1 and f2 are appropriate as parameters to registerCallback , however, it's up to the client to define f1 and f2 correctly. Since I've allready typedef ed

So here's a snippet of my code. void RoutingProtocolImpl::removeAllInfinity() { dv.erase(std::remove_if(dv.begin(), dv.end(), hasInfCost), dv.end()); } bool RoutingProtocolImpl::hasInfCost(RoutingProtocolImpl::dv_entry *entry) { if (entry->link_cost == INFINITY_COST) { free(entry); return true; } else { return false; } } I'm getting the following error when compiling: RoutingProtocolImpl.cc:368: error: argument of type bool (RoutingProtocolImpl::)(RoutingProtocolImpl::dv_entry*)' does not match bool (RoutingProtocolImpl:: )(RoutingProtocolImpl::dv_entry )' The problem is that this: bool

I'm looking at example abo3.c from Insecure Programming and I'm not grokking the casting in the example below. Could someone enlighten me? int main(int argv,char **argc) { extern system,puts; void (*fn)(char*)=(void(*)(char*))&system; char buf[256]; fn=(void(*)(char*))&puts; strcpy(buf,argc[1]); fn(argc[2]); exit(1); } So - what's with the casting for system and puts? They both return an int so why cast it to void? I'd really appreciate an explanation of the whole program to put it in perspective. [EDIT] Thank you both for your input! Jonathan Leffler , there is actually a reason for the code

I'm learning to program and I am using Python to start. In there, I see that I can do something like this: >>>> def myFunction(): return 1 >>>> test = myFunction >>>> test() 1 However, if I try and do the same with print it fails: >>>> test2 = print File "<stdin>", line 1 test2 = print ^ SyntaxError: invalid syntax Why is print different than a function I create? This is using Python v2.7.5. BrenBarn print is a statement , not a function. This was changed in Python 3 partly to allow you to do things like this. In Python 2.7 you can get print as a function by doing from __future__ import print