You don't convert the value before passing it to the addition operator, so any conversion that occurs is implicit.

You can argue it that way, but what's the difference between:

$z = add( $x->to_int(), $y->to_int() );

... and:

$z = $x + $y

... given that the + operator always performs numeric addition? If it's the presence of explicit type name hints, your argument gets awkward in the presence of type inference.

... conversion implies loss of the original value.

I agree; that's why I try to prefer the term "coercion".

In reply to Re^9: Strong typing and Type Safety.A multilanguage approach (implicit) by chromatic
in thread Strong typing and Type Safety.A multilanguage approach by nikosv

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