How to define an implicit conversion from double?

Question

I have defined a class Complex which overloads the + operator:

Complex operator+(Complex const& x, Complex const& y);

I want to define an implicit conversion from double to Complex, such that, for example, if I write c + d, where c is a Complex and d a double, it will call my overloaded + that I defined above and return a Complex. How can I do this?

have a look at std::complex (it has the conversion, already) — , Jun 04 '15 at 16:29
@DieterLücking `Complex` was just an example. I wanted to understand how to write the implicit conversion in general. — a06e, Jun 04 '15 at 19:04

Cory Kramer · Answer 1 · 2015-06-04T15:44:47.217

You just need to define a constructor for it. This is referred to as a "converting constructor"

Complex::Complex(double x)
{
    // do conversion
}

This will allow for implicit conversion, as long as you don't use the explicit keyword, which will force you to have to use a cast to convert.

You can also define other versions of your operator+

Complex operator+(Complex const& x, double y);
Complex operator+(double x, Complex const& y);

score 4 · Answer 2 · answered Jun 04 '15 at 15:38

4

Simply add a non-explicit constructor for Complex taking a double :

Complex(double value);

The compiler will automatically use it to implicitly convert a double to Complex.

answered Jun 04 '15 at 15:38

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How to define an implicit conversion from double?

2 Answers2