Re: Question on implicit type conversion

"Alf P. Steinbach" <>
Wed, 07 Mar 2012 20:32:45 +0100
On 07.03.2012 20:06, Ruby Stevenson wrote:

I am reading Effective C++, item 24 on implicit type conversion. The
example given is that:

class Rational {
   Rational (int numerator=0, int denominator=1);
   const Rational operator*(const Rational&rhs) const;

Rational oneHalf(1,2);

result = oneHalf * 2; // fine
result = 2 * oneHalf; //error

The book then says "It turns out that parameters are eligible for
implicit type conversion only if they are listed in the parameter
list" ...

I am confused as to what parameter list it is talking about??

In the first case '2' is implicitly converted to 'Rational', because the
left hand side operand of '*' is of type 'Rational' and that type has a
member function 'operator*' that requires the right hand side operand to
be of type 'Rational const&'.

In the second case the left hand side operand is '2', of type 'int'. The
only '*' operator associated with that type is the built-in '*'. The
built-in arithmetic operators like '*' do force some implicit
conversions, called "promotions", essentially bringing both operands up
to the same common supertype, but type `Rational` does not offer any
conversion to any of the built-in arithmetic types.

What if it did, though?

#include <iostream>

struct Rational
     operator double() const { return 7; }

int main()
     using namespace std;

     cout << 6*Rational() << endl;

That works perfectly fine (according to the C++ standard, not just
according to a specific compiler), so the book is a bit misleading.

Cheers & hth.,

- Alf

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