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Comments:
<0> thanks
<0> now it works, ill stop to disturb you
<0> really thanks
<1> void f(int param);
<1> #define and *
<1> #define on -1
<1> f(3 and on);
<2> yes, it is
<1> or 'beyond' instead of 'on'? :)
<2> uh?
<1> Hey, I don't know english
<1> I started with an enum, with members like:
<1> param3 = 3;
<1> param3onwards = -3;
<1> But well... hmpf
<0> hmpf :p
<1> Maybe like this: f(3); or f(3, and_onwards);
<1> heheh
<1> At least I don't need macros then :p
<0> but Run.. suppose I do template
<0> damn, template <typename T> T foo() { return ...; }
<0> I need typename like: typename T foo() ?
<3> No, because the compiler already knows that T is a type.
<3> However, it doesn't know that T::whatever is a type unless you say so.
<0> ok
<1> or typename Foo<T>::whatever
<0> you type faster than me :p
<3> You already said "typename T" in the template declaration.
<1> Hmm
<1> Actually, it might know what when Foo<> is known.
<1> But I guess it doesn't keep track of specializations, so you still have to specify it there.
<0> but why the compiler doesn't know that T::whatever is a type..
<1> cl*** A {
<0> it is still declared template <typename T>
<3> CocaCola^: How would it?
<1> typedef int whatever; };
<1> cl*** B {
<3> CocaCola^: Maybe "whatever" is a static member.
<1> int whatever; };
<1> heh - static int :)
<0> hmpf :p
<0> template <typename T, typename C = list<T> > void print(C<T>& he) { }
<1> static boost::shared_ptr<Matcher> create(Identity* identity, int param, match0_type mt, bool single_param = true);
<0> is it possible to do that? : p
<2> CocaCola^ why would you need that?
<1> list<T><T> is nonsense
<0> I shoudl use typename C = list
<0> only ?
<1> template <typename T, typename C = std::list<T> > void print(C& he) { }
<1> hmpf
<1> template<typename T, typename C = std::list<T> > void print(C& he) { }
<0> it tells me: default template arguments may not be used in function templates
<1> I don't think it will be able to figure out what T is by itself then though.
<1> If C is an STL container, you'd either do:
<1> template<typename C = std::list<T> > void print(C& he) { } // use typename C::value_type to get T.
<1> or
<1> template<typename T> void print(std::list<T>& he) { }
<0> how can you use T if you don't define it
<1> Don't use it
<1> use typename C::value_type
<0> <1> template<typename C = std::list<T> >
<1> Oh.. hm
<1> You're right
<1> I never wrote a template that can take any type of container :/
<0> oh cool
<0> well, im trying all kind of stuff really
<1> It's possible to p*** a template (ie, just 'list') but I don't know how.
<0> but I just hate to follow tutorial and when examples don't work
<1> This is from AC++ ?
<0> nop
<0> im just talking in general
<4> i like the GUI of Winamp
<4> player
<4> lot better than windows mediay player
<0> agree
<4> i wonder what GUI Builder they're usin
<1> CocaCola^: I looked it up, you can do:
<1> template<template<cl*** U> cl*** V> cl*** C { ...
<0> oh
<0> ok
<0> where did you look
<1> So, maybe you want: template<template<typename T> C = std::list> void print(C<T>& he);
<1> however, I doubt it will work for functions.
<1> This example is for another cl***, not a template function.
<1> In the standard - 14.3.3
<2> why would you need to write a print function templated to a container of T?
<0> it becomes complex
<2> just copy the elements to output
<0> what do you mean
<0> I want to use print(d); d can be list<char> vector<bleh> deque <int>
<2> or you can just do things normally
<2> until we get concepts
<2> d can also be int
<2> well, maybe not int,
<2> but the whole design of the std library is to use algorithms for that stuff
<2> er
<2> iterators
<2> sigh, tired and frustrated
<0> ok but how can u get iterator
<0> in print definition
<0> void print(......::iterator b, ...::iterator e) { }
<0> if you don't know which type of iterator
<2> you don't need to write a print() at all
<0> lol :p
<2> std::copy( c.begin(), c.end(), std::ostream_iterator<element_type>(cout, " ") );
<2> see, all done
<0> cool
<0> hehe :p
<5> cl*** a { int x(); }; cl*** b { int x(int y); }; cl*** c : public a, public b { };
<5> shouldn't c inherit both functions?
<5> they're all public
<3> sk8ing: They are? You didn't write them that way.
<5> public: in both a and b
<3> sk8ing: Where?
<5> now they are :)
<5> come on.. I know how to write a cl***. cl*** a { public: int x(); };
<5> same with b
<3> Don't know, then.
<2> sk8ing how about describing your problem
<1> I don't know either... it could be that it first finds the scope of x, and then tries to resolve overloading...
<5> in the same scope?
<1> The scope if different.
<1> is
<3> sk8ing: Are you qualifying x() when referencing it as a member of c?
<5> well I'm trying to do something like c x; x->x(10); and it's not working
<3> sk8ing: Of course not. That's ambiguous.
<5> no it's not
<3> In c, it is.
<1> x.x(10); you mean
<5> yeah... x.x
<1> x.b::x(10); should work
<1> but you can put both x explicitely in c
<6> hi all. can anyone explain me how to use the lib with dll in my code when I don't have a header file? (using dev-cpp5)
<1> with a using, i think
<1> cl*** c .... { using a::x; using b::x;
<2> struct C: A, B{ using A::x; using B::x' };
<2> ;
<5> interesting, thanks
<3> danilov: What library is it, and why aren't there headers for it?
<6> it is bpmDetect... wait a sec...I'll find the link....and it has a header file, but it's not working, so it's the same as if there was no .h file....
<1> Ah, you're still in the stage "It doesn't work, so I'll delete it".
<3> You should probably figure out why it doesn't work.
<1> Hmm, I never got into that stage in the first place, actually.
<3> You might save yourself a great deal of effort... and probably program errors, besides.
<6> well....hm...it just declares functions....when I try using them the linker says the thing about undefined reference...
<3> danilov: Did you remember to link the library?
<6> yes
<3> Did you remember to link its dependencies?
<3> Did you do the same for any other libraries you might be linking?
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