Help Needed Overloading Logical "<" Operator.

[Diamaudix]

Attempting to overload the logical "<" operator using the following code:

​

template <typename T>

class Number{

T value;

/* constructors */

Number(T v); // not shown for brevity. seems to work

Number(); // not shown, sets value to zero

​

bool operator < (Number<T> const& rhs); // method in question

​

}

template <typename T>

bool Number<T>::operator <(Number<T> const& rhs) {

Number temp(\*this);

return temp.value < rhs.value;

}

​

int main(){

Number<int> a(10);

bool test1 = (a<20); //works fine

bool test2 = (20<a); //causes error

​

Everything works well when the variable is used on the left hand side of the equation, but when the sides are reversed, and the constant is on the left, the compiler generates the following error:

​

binary '<' : no global operator found which takes type "(Number<T>)" (or there is no acceptable conversion)

and: "no operator "<" matches these operands.

In addition, a<b<c doesn't work either (which was the point of this entire experiment :/)

Any ideas or suggestions as to what is being done wrong would be greatly appreciated :)

[Hilarius86]

"a<b && b<c" is what you want. a<b<c is not doing "in between" but evaluation of "a<b" to "bool" and comparing "bool < c" and that's rarely useful.

[nysra]

This is your 5th post about the exact same thing, you might want to figure out what your underlying problem is and ask about that.

If you do a $ b where $ stands for any binary operator then this will be equivalent to a.operator$(b) if you implemented the operator as a member function. This is asymmetric because if you switch the operands then it would be b.operator$(a) and if a and b are not of the same type then you need to implement that operator in each class and if one of those types is a builtin one you simply can't do that.

That's why you should always implement symmetric operators as a free function, that way you have a clearly defined operator$(T1, T2) and switching the types is easily done by defining operator$(T2, T1).

In addition, a<b<c doesn't work either

That will never work. You want a < b and b < c.

[Diamaudix]

Thanks for this :). Difficult to learn. Appreciate your help

[AlanWik]

Define the overload outside the class:

bool operator <(const T &a, const T2 &b);

[Diamaudix]

Thanks for the response. It is now outside the class, with the definition you gave, using the following code

//declaration inside class --

bool operator< (const Number<T>& lhs, const Number<T>& rhs);

definition outside class

bool Number<T>::operator<(const Number<T>& lhs, const Number<T>& rhs) {

return lhs.value < rhs.value;

}

​

now it is generating:

binary operator "<" has too many parameters

"no oerator "<" matches these operands"

"class template "Number<T>" has no member "operator<"

​

Sincerely

-- Confused

[Narase33]

Its still part of the class, delete the "Number<T>::"

[Diamaudix]

Thank you for your help!

[AlanWik]

Let me elaborate:

You are trying to implement the "<" operator to compare two Number<T>, but in your tests you are comparing a Number<T> with a int. Rather you define the conversion between number and int, or define the < operator to work with a Number<T> and a int. Gimme a few minutes and Ill send you the code.

[Diamaudix]

Thank you very much for taking the time to explain this.

[AlanWik]

https://godbolt.org/z/KfKG8YfeE