CODEWITHSUNDEEP
c++int
Andrew_Jacop
Andrew_Jacop

Reputation: 29

Is there a way to define an 'int' by 'greater/lesser than' instead of '='?

I was playing around in a challenge on some website, and I met this problem. i have an unidentified integer ... only I know that it is greater that x or\and lesser that y and so on ... is there a way to define a variable based on so? ... I mean as being greater\lesser than integer ..

some have noted that not_null would help but I couldn't understand how ..

here is some silly example :

int some_unknown_number > 8;
if [some_unknown_number<=1] 
    {cout << "wrong" << endl;}

so I expect the code to recognize that some_unknown_number can't be lesser than 1 since it's already larger than 8....

ps: I don't want the exact answer ... just tell me where to look if you know what I mean ....

Upvotes: 1

Views: 946

Answers (2)

vahancho
vahancho

Reputation: 21240

Interesting question, indeed. You can define your type. For example:

template<int Min, int Max>
struct Int
{
  static_assert(Max > Min, "Max should be greater than Min");

  bool operator<(int val) const
  {
    return val > Max;
  }

  bool operator>(int val) const
  {
    return Min > val;
  }
};

You can add more operators, if needed, to define necessary semantics, and use it like:

// Int<19, 1> wrongInt; <--- compile time error.

Int<1, 3> myInt;
if (myInt > 0)
  printf("Greater than 0\n");

if (myInt < 5)
  printf("Less than 5\n");

Upvotes: 2

Bathsheba
Bathsheba

Reputation: 234785

You could build a class representing this as

struct bounded
{
    std::optional<int> m_lower;
    std::optional<int> m_higher;
};

which models the lower and upper bound of an instance. If both are present and set to the same value, this clearly models a plain int.

You then build your < operator &c. in accordance with this model.

Upvotes: 4

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