No Suitable User Defined Conversion

Question

I am trying to write a C++ program which wraps numeric values, I am doing this by writing a super class which will handle two simple functions, and an operator overloading function. This is my code:

#include <iostream>
#include <string>
#include <sstream>
using namespace std;


template <class T>
class Number {
protected:
    T number;

public:
    Number(T num) {
        number = num;
    }

    string mytype() {
        return typeid(number).name();
    }

    string what_am_i() {
        ostringstream oss;
        oss << "I am " << Number<T>::mytype() << " and my nanana is " << number;
        return oss.str();
    }

    Number operator+ (Number an) {
        Number brandNew = NULL;
        brandNew.number = number + an.number;
        return brandNew;
    }
};

class MyInt : public Number<int> {
public:
    MyInt() : Number<int>(0){};
    MyInt(int num) : Number(num){
    }


};

In the Main function I would like to do something like:

 void main() {

    MyInt three = 3;
    MyInt two = 2;
    MyInt five = three + two;
    cout << five.what_am_i();

}

My problem is the addition between three and two, the compiler says:

no suitable user-defined conversion from "Number" to "MyInt" exists

I could solve this by implementing the overloading function in MyInt but since i want to support many classes like MyShort and MyFloat I would like to leave it in the Superclass. Is there any solution? Thanks!

Why do you need a wrapper for numeric types? – John Zwinck Apr 05 '14 at 11:59 — John Zwinck, Apr 05 '14 at 11:59
@JohnZwinck it is an exercise – Steinfeld Apr 05 '14 at 12:21 — Steinfeld, Apr 05 '14 at 12:21

score 13 · Accepted Answer · edited Mar 25 '20 at 13:27

13

The problem is when you inherit from a class templated the same as the current class. The inherited type will not replaced as you might expect. For example, Number<int> will not be replaced with MyInt for inherited operator +.

The return value and entry parameter of operator + is a Number<int>, not a MyInt. The inherited class must be able to construct a MyInt from a Number<int>. Put below line in MyInt class:

MyInt(const Number<int> &x) : Number<int>(x) {}

To avoid the extra work, it's better not to inherit from Number, but instead just put a typedef for int:

typedef Number<int> MyInt;

... and then everything else is OK.

edited Mar 25 '20 at 13:27

Joshua Swain

571
2
4
22

answered Apr 05 '14 at 12:11

masoud

55,379
16
141
208

@M.M, compiler still shows same error, is there a specific place I should be putting it in? – Steinfeld Apr 05 '14 at 12:24
1

Do as [this](http://coliru.stacked-crooked.com/a/bee979a14bc3c1c1) live code. It should be inside `MyInt` under your constructors. And also I recommend to use my second suggestion. – masoud Apr 05 '14 at 12:28
1

The second suggestion works well but I must use inheritance. Thanks for the live code it works great! – Steinfeld Apr 05 '14 at 12:34
Could you please give a short explanation to the specified line you gave? – Steinfeld Apr 05 '14 at 12:52
2

First one? It's a constructor that tries to construct a `MyInt` from `Number`. For example the result of `+` is a `Number` and it should be assigned to a `MyInt` as you've written `five = ...`. This constructor makes `MyInt` able to accept `Number`. – masoud Apr 05 '14 at 13:20

No Suitable User Defined Conversion

1 Answers1

Linked