ChummyJakes If you're having trouble with one test case, I know I did. The solution was to notice that the Calculate function returned a long long. This made me think that one test case was really large. I changed the private variables to long instead of ints and it worked. Hope this helps!
Sid_S But then you double the amount of memory used for no good reason.
A cast at the right spot is a better solution.
cupins If only typecasting worked had to use three extra lines of code :/
troywweber typecasting worked for me. I succesfully used "return (long long) l*b*h;"
h754548560 Thanks, it`s quite helpful.
lpassos Thanks
quick_dudley Does that syntax cast the full expression or does it cast l and then rely on the fact that int * long long = long long?
thomastc Good observation, no idea why it got a downvote. It's the latter -- which is why it works in the first place.
windak_szymon Comments worth reading ;]
thanks m8
aid888 Thanks bro
callum_white I went for "return static_cast< long long >(l) * b * h;" to avoid the c-style cast.
zero000_ece07g I prefer this solution. Thanks.
Praveenchandel81 thanks bro
michael_prantl Not quite the way you should generally do C++ type casting...
viditharwani Thanks :)
harjotsinghjio11 [deleted]
iqbal2929 that help me, ty.
gauri27 Thank you
christosg8 long long volume = l; volume *= b; volume *= h; return volume;This should do it, while maintaining l, b and h ints.
UST007 can u tell me why it will work plz?
LittleFox The product of two int will be a int (which don't fit in all tests cases) but the product of a long long with a int will be a long long. He cast l into a long long and then all the products will be long long. An another way to do that is :
return 1LL * l * b * h;
(one more product but less code =P )cupins Ahhh that makes more sense I thought it was more like: return (ll)l*(ll)b*(ll)h;
zenmumbler You only need to cast the first term to long long for it to work luckily.
(long long)(l) * b * hwill work fine, as in my entry.
mohitgarg Thanx #LittleFox for this hint, I really forgot and write long long volume=l*b*h, but infact it was claculating an int value instead of a long long value
jettatore [deleted]raviteja0914 return volume*b*h;
huangjiasen100 The 1st assignment should be long long volume = (long)(l) instead.
tanzeelsiddiqui [deleted]24anubhavmishra long long CalculateVolume() {long long v=l;v*=(b*h); return v;} //this worked for me
UST007 hmm,its test case 3 .and thanks it worked.but why would they they input l,b,h out of range of int?? it was mentioned to use int. as sid said long long would only inc the amound of memory ....
christosg8 None of
l,bandhare out of bounds. They all are <= 105 which will fit inside anint(109.3) (you can download the testcase to make sure of it). Butvolumewill be equal to the product of those three, and in the most extreme case, it's gonna be equal to 105x105x105=1015 which doesn't fit into anint, so you have to use along int(1018.9). Hope this helps you.
manishbsn188 thankls mn
osese tyy
DisjointTech Nice work! Typecasting the return value to "long long" didn't work, but declaring the variables as "long" did work. Could you explain to us the mystery behind this trick?
mitchel0022 Its a little late but essentially when you do (long long) ( a * b * c) if a,b and c are integers and an overflow has already occured then casting it to a long long wont help. It will compile but the overflow has already happened. You need
(long long)a * (long long)b * (long long)c
divyaprakash_131 Thank you so much!
1337ninja I don't understand why
static_cast<long long>(l*b*h)
didn't work here! I had to do (long long) l*b*h. Any reason why?
LittleFox In
static_cast<long long>(l*b*h)
You cast the result which already overflowed at this state.
static_cast<long long>(l)*b*h
Should work like your second example where you actually only cast l.
the_koolkarni hey! it works. Thanks a lot @ChummyJakes
NiceBuddy can also cast them something like this:
const uint64 CalculateVolume()const { return (static_cast<uint64>(m_length) * static_cast<uint64>(m_breadth) * static_cast<uint64>(m_height)); }
where uint64 is
Edit:
typedef unsigned long long uint64;
raufsaadiqkhan killed it
[deleted] Yes it is working, thanks!
arunkumaryadu thanks
ravisachdeva Thanks it helped me :-)
suyashchirme Thanks!! It helped.
ksbalaji2022 thank u it worked!!
asauve Thank you for the tip. I also noticed the overflow but wondered why the compiler did not adjust the size of the operations to the function output.
So it also works by casting the volume multiplication :
return (long)l * (long)b * (long)h;avinat123 you can also multiply the whole thing by 1.0 at the start.
should look like this:
return 1.0 * l * b * h;
sagarikashrote +1 dudeee
Praveenchandel81 welcome
satviksingh600 Thank you very much!!. You helped me to score 30.0 points!
Praveenchandel81 welcome
saahil96 Thanks! It works!
zaishachavada1 thanks dude , it helped me alot.
sratnakumar99 Thanks mate :)
hernandezurbina it helped! thanks!
rachitjain1626 thanks @ChummyJakes really helpful...
[deleted] All test case passed!
class Box{ private: int l, b, h; public: Box(){ l = 0; b = 0; h = 0; } Box(int length, int breadth, int height){ l = length; b = breadth; h = height; } Box(const Box& B){ l = B.l; b = B.b; h = B.h; } int getLenght(){ return l; } int getBreadth(){ return b; } int getHeight(){ return h; } long long CalculateVolume(){ return (long long)l*b*h; } friend bool operator < ( Box&A,Box& B){ if( (A.l < B.l) || ((A.b < B.b) && (A.l == B.l)) || ((A.h < B.h) && (A.l == B.l) && (A.b == B.b)) ){ return true; }else{ return false; } }; friend ostream& operator<< (ostream& output, const Box& B){ output << B.l << " " << B.b << " " << B.h; return output; } };
jeetjpavani [deleted][deleted] A friend function is defined outside the scope of the class but has access to the class private and protected data which is needed here for operator overloading.
zeems1234 why there is ; sign at the end of < operator function
raghuvendra867 without it can work too ..may be it is by mistake!!
JimNova [deleted]arka_cool1996 Very nicely written code. I have been trying and failing to understand the concept of operator overloading. I get that it is a way to define what an operator does but the syntax seems very confusing. Can someone explain it to me like i am a 5 year old
buffcoder operator overloads are just like other function just instead of a function name you have to use the specified operator symbol.
hustler1 thank you. that was really helpful
raj_rajet Why is 'friend' used while defining the operators ?
akankshyam A friend function is a function that can access the private members of a class as though it were a member of that class. This is needed here because we're overloading operators, and we're accessing outside the class scope.
johnmanderson78 but you can use getters which are public.
Shu7bh Yes, you can use getters, but sometimes if you have a lot of variables, it becomes tedious and long. And if you want to make sure that you won't by-mistake change it's value, you can declare the object as a const
Here, you don't need to have operator < as a friend, you can just make it as a member of the class and remove the 1st parameter as the operator will be called from the operand which is on the left hand side of the operator.
But operators << and >> are usually declared as friends because if you don't, you'll have to make the overload in the ostream class which isn't encouraged.
manasnaik96 Why is long long used for return instead of long?
maurya159dheeraj friend ostream& operator<< (ostream& output ,const Box &B) { output<
please can you explain bold part :(
Shu7bh It is a bit difficult to explain so bare with me
ostream is the class which handles the output for us. So for overloading the << operator, you need to have a parameter as ostream, we take it by referece since its a class type and also since the object value will change.
We are returning ostream& since if we are cascading we need the ostream object again to print the other stuff.
Laraib07 why is ampersand(&) used ?
akshyvrma8 to get a reference of ofstream class
benaffleks Your < operator doesn't require a friend method, it should work without it as both Box's are part of the same class and there isn't anything else acccessing those two objects.
Also to refactor, instead of passing Box A, you can just use a this pointer to access the current Box's elements like this-> l < B.l etc.
shubhampcvn Why is it necessary to define operation overloading as friend function? it isn't working without it.
benaffleks It depends on what you are overloading
Since an ostream does not have access to private variables of Box, it makes sense to enable access via friend
However for a greater / less than comparison, since we are just dealing with two Box's and already have access to their private variables, we don't need to friend that method
I may be COMPLETELY wrong here but mine works without using a friend method for less than or equal comparison overloads
shubhampcvn Thanks Ben
Can you post your code
benaffleks [deleted]benaffleks class Box{
private: int l, b, h; public: Box(){ this->l = 0; this->b = 0; this->h = 0; } Box(int x1, int x2, int x3){ this->l = x1; this->b = x2; this->h = x3; } Box(Box &b){ this->l = b.l; this->b = b.b; this->h = b.h; } int getLength(){ return l;} int getBreadth(){ return b; } int getHeight() { return h; } long long CalculateVolume(){ return (long long) l * b * h; } bool operator < (Box &b){ if(this->l < b.l){ return true; } else if(this->l == b.l && this->b < b.b){ return true; } else if(this->h < b.h && this->l == b.l && this->b == b.b){ return true; } return false; } friend ostream& operator <<(ostream& out, const Box& b){ out << b.l << " " << b.b << " " << b.h; return out; }};
remoteblue Thanks @Benaffleks. Best explain on why using friend for this problem ever.
vincent_jadraque when overloading the < operator, you could just return the conditions instead of using the if and else statements
return (l < B.l || ((b < B.b) && (l == B.l)) || ((h < B.h) && (l == B.l) && (b == B.b)));
daymianpac Box(const Box& B){ l = B.getLength(); b = B.getBreadth(); h = B.getHeight(); }
guptakeshav_rs Why is the operator overloading function declared as friend function?
elijahelrod this should fail because the function is called getLength() not getLenght()
aam3r77 somebody please explain this
friend bool operator < ( Box&A,Box& B){ if( (A.l < B.l) || ((A.b < B.b) && (A.l == B.l)) || ((A.h < B.h) && (A.l == B.l) && (A.b == B.b)) ){ return true; }else{ return false; } }; friend ostream& operator<< (ostream& output, const Box& B){ output << B.l << " " << B.b << " " << B.h; return output; }poulamorcoss i am not using const in the copy constructor and and it gives me erroe could you explain to me why it is a necessity to use const?
poulamorcoss why we overload smaller than and stream insertion operator we overload it as friend?
poulamorcoss how are ostream and bool are friend function and you defined them inside the class?
Ankur_trivedi Why do you put the overloaded function as a friend function ?
bhanu7416 Here, data members are not using outside class so, why the operator (<<) function has to be friend function?
sampritipatel but A.l and B.l are integers here aren't they. And '>' works fine on integers. Why then do we need to overload the operator ?
akshyvrma8 it is clearly mentioned in the problem to do (overload <<) so you can't edit the locked stub code .
bharatmundra119 why you are youing file handling in this problem cam you explain me is that is necessary
akshyvrma8 Noone used file handling my friend
imsinghofficial why we use friend here ??
akshyvrma8 friend function has access to private members of class. Here to overload << operator , you need an instance of stream (ostream) which dont have access to l , b and h of Class Box . Using friend function enables the ostream instance to access private data members.
pranaykumar31399 why used friend for 'bool operator <' even though it is inside the class ?
otc_simon Why do you not have to declare the type of Box function and I do not understand Box&A and Box& B ordeal, could somebody pls help me?
akkusahu559 Why you have taken parameter Box as a const in overloading << operator?
manishbsn188 why friend function is used?
linjunzh To overload operator <<, I use the following code
ostream& operator<<(ostream& out, const Box& B){ return out << B.l << " " << B.b << " " << B.h; }
But the system says "std::ostream& Box::operator<<(std::ostream&, const Box&)’ must take exactly one argument". Can anyone help me with this problem? Thanks in advance.
NicoleB I am having the exact same problem!
raulbojalil You need to remove "const Box& B". It's not needed, you already have the this context.
brynmurrell97 Wait, I have done this and now it says that "B" was not declared in this scope.
veronicamdoyle operator overloads declared in the class assume "this" instead of the Box argument.
What you could do is declre the operator overload function outside of the class, and then add the operator overload function as a "friend" to your class.
class Box(){...
friend ostream& operator<<(ostream& os, const Box& b);};
ostream& operator<<(ostream& os, const Box& b){ os << b.l << " " << b.b << " " << b.h; return os; }anushaec_ec16 thanks
Surabhi_97 Please elaborate the reason for using this function as a friend!
akshyvrma8 friend has access to private members of a class(which is friend)
vivek_nitd you need to use B.getLength() instead of B.l and similarly for other.
krithikaa_mahe even though i couldn't get the result
DJAvanci The problem is that youdefined the operator << overload inside of the class. You must do it outside class scope.
hara_toshiya Thanks!! I couldn't figure it out for the longest time.
rishabh_ags But why?
rahul_tulani The operators ‘<<' and '<<' are called like 'cout << ob1' and 'cin << ob1'. So if we want to make them a member method, then they must be made members of ostream and istream classes, which is not a good option most of the time. Therefore, these operators are overloaded as global functions with two parameters, cout and object of user defined class
krithikaa_mahe make it as friend function
vrundavyas01 [deleted]
lfliang Hi, does anyone know that this operator has to be defined outside the class? Thx.
NiceBuddy make it as a
friendof class like this: I used this manner and didn't make any of the getfunctions.
class Box { // class definition friend std::ostream& operator<< (std::ostream& out, const Box &obj); }; std::ostream& operator<< (std::ostream& out, const Box &obj) { return out<<obj.m_length<<" "<<obj.m_breadth<<" "<<obj.m_height; }
akshyvrma8 use friend in front of declaration
dwilches This is a bad practice: ostream& operator<<(ostream& out, Box B) The second parameter should be const Box& B. Otherwise, you are invoking the copy constructor in a place where it is not necessary.
This is the reason some people get 4 instead of 7 for the sample test case.
Also, all those getXX() functions are missing the const qualifier at the end.
RaviSwami009 I don't understand what this problem helped me learn. Bad code practice let you pass test cases.
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