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Comments:
<0> fstream too, right?
<1> shut up :P
<0> ha ha ha.. good question
<2> Ok, next question: When is it needed to add virtual base cl***es to an initializer list?
<0> give up
<1> if they dont have a default constructor -.-
<0> Run, how do we do that?
<0> is it like constructor chaining?
<2> Suppose we have: cl*** V; cl*** B1 : virtual V {}; cl*** B2 : virtual V {}; cl*** A : public B1, public B2 {}; And V doesn't have a default constructor. Then is it necessary to add V to the initializer list of A?
<0> i dont think, interviewer will be knowing about this... :)
<2> heheh
<0> thanks anyways... :)
<2> The answer is yes, because the compiler don't know how to initialize V: through path B1 or B2.
<3> e vreun roman in aer
<1> so my answer was right
<1> :P
<2> Ie, B1 could have V(3) and B2 V(5)
<3> why visual doesnt work fi
<2> Tamama: actually, I think even if it has a default constructor, you still need to specify one.
<2> But I never use virtual inheritance.. so I don't have a lot of experience with it.
<1> Run: ok, i should have been more specific: if a default constructor can not be automatically chosen
<1> double = cant choose, non existant = cant choose
<2> The term 'default constructor' is reserved for V(void)
<1> meh
<1> terms schmerms
<3> why
<3> #include<iostream.h>
<3> int a[10][10],i,j,n,m,st[67],c1,c2,x,y;
<3> void citire()
<3> { cin>>n>>m;
<3> for(i=1;i<=n;i++)
<3> a[i][i]=0;
<3> for(i=1;i<=m;i++)
<0> default can be V(int=0) too , right?
<3> {cin>>x>>y;
<3> a[x][y]=1;
<3> a[y][x]=1;
<3> }
<1> bah
<4> ;)
<1> i wasnt op :S
<0> :P
<1> I even typed a kick message
<5> wow
<0> default constructor can be V(int=0) too , right?
<5> that really is stretching it lol
<5> um
<5> int i = 0 ?
<2> int=0 is ok too
<2> I think
<5> cool
<5> how would you use that?
<3> i send earlyer aprog
<0> SB_, that is declaration part...
<1> Well I never found the compiler randomly choosing constructors with arguments that dont default
<0> in definition, we need to use a variable.
<2> It would all you to call V(), and get no compiler warning if you don't use 'i' inside that constructor
<1> luckely
<5> DoIt, but i mean, if you have f(int = 0) as opposed to f(int i = 0), how would you use 'int = 0' in the function
<1> previous statement stands then :)
<5> I guess I don't see the purpose of it.
<5> Run, f(void)?
<6> You don't need the ***ignment in a definition. Just leave out the name of the variable if you don't use it.
<0> SB_, in the function you dont need to specify the default argument
<2> SB_: suppose you call f like f() and f(x), but in this particular f you don't USE the parameter...
<5> agh, but why even have it
<6> int foobar(int, int); /* declaration */ int foobar(int thisIsUsed, int) { std::cout << thisIsUsed; }
<6> LIke that.
<6> No need for the default param value.
<0> same function will invoke default constructor as well as one argument constructor
<3> sb wake up
<4> the virtual base constructor is called based on the one invoked from the actual cl*** being instanciated
<6> That should prevent a warning.
<3> you are so boring
<3> prevent a warning.
<3> <3> you are so boring
<3> prevent a warning.
<3> <3> you are so boring
<3> prevent a warning.
<3> <3> you are so boring
<3> prevent a warning.
<3> <3> you are so boring
<4> the ***ist, heh
<5> I mean, all that to prevent a compiler warning?
<5> that's really all that is accomplishing?
<0> SB_, nope
<6> What I just typed does.
<6> I don't know what DoIt is doing, or Run.
<0> you can even have V(int=0,int=0,int=0)
<5> yeah
<6> All you have to do is not supply the parameter name.
<6> No need for the =0
<5> f(int) ?
<4> run is prepping DoIt for his interview, maybe
<6> I guess
<0> :)
<2> Solamente: You sure you can call it as f() then?
<5> Sorry for being stupid
<5> I was just curious. I mean I still don't fully understand it, but that's ok.
<6> Ah.... okay, we're conflating two different things.
<6> I see now.
<6> We're talking about two sides of one issue
<5> The limit of the issue as approached from the left and the right ;)
<6> You're talking about not having to supply a value on the call.
<2> The question was if the syntax f(int = 0) ever made sense.
<6> I'm talking about getting rid of the warning in the definition.
<4> mmm, i feel like ordering pizza
<5> Run, More like why do it lol
<6> So what I typed above gets rid of the warning if the parameter is not used.
<5> rdragon, lucky bastard, i don't have money for that.. been a while since i've had decent pizza
<6> What you're describing makes the parameter unnecessary at point-of-call.
<4> oh yeah, money...
<6> (And gets rid of the warning)
<5> rdragon, haha about that
<4> well, today is payday... or tomorrow
<6> So, yeah.
<0> SB_, if you decare a function like A{public: A(int=0);};A::A(int i) {} , this constructor will invoke for V() and V(9);
<6> I'm going to shut up now.
<5> rdragon, i need a job still
<2> SB_: I already said that, because it does TWO things: you don't have to (but can) supply a value on the call, AND it won't give a warning for not using the argument inside the function.
<6> Yeah, I see that.
<2> Ok, play time is over.
<4> I can't get "the cars - hello again" out of my head... heard it on a commercial and it's been stuck
<6> Man, I actually have to go to the office today. That ****s.
<7> have to go? awwww, I'm already there
<6> "All I Want" >>>> "Hello Again"
<5> DoIt, But why not just use A(void) for when you call A()...
<0> SB_, the same fucntion is getting called for both...
<0> you dont need to define both functions
<0> if all you need to do is same for all kind of constructors....
<5> DoIt, so i gets the default value of 0?
<5> DoIt, and it does the same thing as A(int i) ?
<7> why would one bother??
<2> DoIt: Where are you _allowed_ to use the typename keyword? :)
<4> DoIt a virtual base gets constructed exactly once, and it uses the form of the constructor invoked from the complete object - that is, the type of the object being constructed
<5> DoIt, what is it called, I'll stop wasting y'alls time and look it up :)
<4> irrgardless if base cl***es have syntax for invoking other constructors of the virtual base, they're ignored
<7> rdragon the _most_ derived, rdragon which makes it even messier
<0> Run, no idea....
<5> hrm
<4> most derived, yeah
<5> so what is it called?
<4> you're at the office already? that seemed quick
<5> Or is that just building on a basic principle of c++ i should know?
<4> Run don't prep him for his interview - either he knows what he's talking about or doesn't
<5> rdragon, haha
<2> DoIt: They won't ask that. But the answer is (apart from in a template argument list), inside a template definition for types that are depending on a template argument.
<4> he'll be worse off if he repeats something he heard here, and then can't actually explain what it means
<0> rdragon, i wont, unless i have the full concept
<4> you say that
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