3
\$\begingroup\$

I decided to solve the Box It! challenge in Hackerrank to improve my C++ knowledge as I'm still new to the language.

Design a class named Box whose dimensions are integers and private to the class. The dimensions are labeled: length \$l\$ , breadth \$b\$ , and height \$h\$.

The default constructor of the class should initialise \$l\$ ,\$b\$ ,\$h\$ and to length, breadth and height .

The parameterized constructor Box(int length, int breadth, int height) should initialize Box's \$l\$ , \$b\$ and \$h\$ to length, breadth and height.

The copy constructor Box(Box B) should set \$l\$ , \$b\$ and \$h\$ to \$B\$ 's \$l\$ , \$b\$ and \$h\$ respectively.

Apart from the above, the class should have \$4\$ functions:

  • int getLength() - Return box's length
  • int getBreadth() - Return box's breadth
  • int getHeight() - Return box's height
  • long long CalculateVolume() - Return the volume of the box

Overload the operator for the class Box. Box \$A \leq \$ Box \$B\$ if:

  • \$ A.l \leq B.l\$
  • \$ A.b \leq B.b\$ and \$ A.l == B.l\$
  • \$ A.h \leq B.h\$ and \$ A.b == B.b\$ and \$ A.l == B.l\$

Overload operator \$<<\$ for the class Box().

If \$B\$ is an object of class Box:

\$cout << B\$ should print \$ B.l, B.b \$ and \$B.h\$ on a single line separated by spaces.

Constraints

\$0 \leq l,b,h \leq 10^5\$ Two boxes being compared using the \$<\$ operator will not have all three dimensions equal.

Here is my code

#include<bits/stdc++.h> using namespace std; class Box{ int length, breadth, height; public: Box (){ length =0; breadth = 0; height =0; } Box (int l, int b, int h) { length =l; breadth = b; height = h; } Box(const Box& b){ length = b.length; breadth = b.breadth; height = b.height; } int getLength(){ return length; } int getBreadth (){ return breadth; } int getHeight(){ return height; } long long CalculateVolume(){ return (long long) length * breadth * height; } bool operator<(Box& b){ if(this->length <= b.length || this->breadth <= b.breadth && this->length == b.length || this->height <= b.height && this->length == b.length && this->breadth == b.breadth) { return true; } return false; } friend ostream& operator<<(ostream &out, const Box &B); }; ostream& operator<<(ostream &out, const Box &B){ out<< B.length<<" "<< B.breadth<< " " << B.height; return out; } void check2() { int n; cin>>n; Box temp; for(int i=0;i<n;i++) { int type; cin>>type; if(type ==1) { cout<<temp<<endl; } if(type == 2) { int l,b,h; cin>>l>>b>>h; Box NewBox(l,b,h); temp=NewBox; cout<<temp<<endl; } if(type==3) { int l,b,h; cin>>l>>b>>h; Box NewBox(l,b,h); if(NewBox<temp) { cout<<"Lesser\n"; } else { cout<<"Greater\n"; } } if(type==4) { cout<<temp.CalculateVolume()<<endl; } if(type==5) { Box NewBox(temp); cout<<NewBox<<endl; } } } int main() { check2(); }

Final thoughts

  • I now I shouldn't be using this line of code using namespace std; as recommended in other reviews but Hackerrank had this pre-defined and I couldn't change it. The Check2() and main() were also predefined as well.

  • Was the use of friend method an overkill?

  • Did I override the operators << and < correctly?
  • How can I improve this code?
\$\endgroup\$
5
  • \$\begingroup\$ Just an fyi, you can change any of the code that Hackerrank supplies. As long as the code will compile, it will accept it. \$\endgroup\$ Commented Nov 10, 2016 at 21:30
  • \$\begingroup\$ You need parenthesis on your operator< otherwise the comparisons are made serially \$\endgroup\$ Commented Nov 10, 2016 at 21:30
  • \$\begingroup\$ @tinstaafl the code is in a locked editor and I'm only meant to write a class that will be called by the locked editor \$\endgroup\$ Commented Nov 10, 2016 at 21:32
  • \$\begingroup\$ OK. I also noticed the the < operator code only tests < but the description uses <=. \$\endgroup\$ Commented Nov 10, 2016 at 21:34
  • \$\begingroup\$ I must have changed it whilst copying the code to an editor . Good Spot on . I will make amends \$\endgroup\$ Commented Nov 10, 2016 at 21:35

2 Answers 2

Reset to default
2
\$\begingroup\$

  • The problem statement defines the less than condition differently.

  • I recommend to use tuple<int, int, int> for box dimensions. The default operator< for tuples does exactly what the problem statement asks for.

  • The

    if (condition) { return true; } return false;

    is a long way to say

    return condition;

  • this-> is just a noise. You may safely refer to this->length as length.

  • An indentation for operator<< is confusing, but I assume it a copy-paste error.

  • #include <bits/stdc++.h> is always wrong.

\$\endgroup\$
1
\$\begingroup\$

It's strange to use a signed type for length, but that appears to be imposed on you by the problem statement, so blame a poor specification for that.

Despite the letters std, <bits/stdc++.h> is not a standard header, so you have a portability bug. Even where it exists, it brings in far more than you need, so get used to including just the headers you need.

Avoid using namespace std; - that deprives you of the benefits of namespacing. If you really can't type std:: where needed, then import just the names you need into the smallest reasonable scope:

void check2() { using std::cin; using std::cout; using std::endl; //... }

Prefer to use initialisers to set members - a good compiler (e.g. g++ -Weffc++) can warn when you forget one:

Box() : Box{0, 0, 0} { } Box(int length, int breadth, int height) : length{length}, breadth{breadth}, height{height} { }

We can omit the Box(const Box&) constructor, as the compiler-generated one is identical.

Many methods are missing const qualifiers:

int getLength() const; int getBreadth () const; int getHeight() const; long long CalculateVolume() const; bool operator<(const Box& b) const;

check2 is a completely meaningless name; choose a name that demonstrates its purpose. Similarly, what does temp mean to the reader?

When reading input, always check the state of the stream afterwards:

if (!(std::cin >> l >> b >> h)) { // deal with input error here }
\$\endgroup\$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.