Re: Assign the value of a class to an unsigned integer (built-in) type

From:
Victor Bazarov <v.bazarov@comcast.invalid>
Newsgroups:
comp.lang.c++
Date:
Tue, 01 Jun 2010 13:16:21 -0400
Message-ID:
<hu3fb8$417$1@news.eternal-september.org>
On 6/1/2010 1:04 PM, Jonathan Lee wrote:

On Jun 1, 12:42 pm, Peter Olcott<NoS...@OCR4Screen.com> wrote:

I would like to be able to assign the value of this class to an unsigned
integer using operator=() syntax.


You can overload casting operators:

   class Dummy {
     unsigned long x;
   public:
     Dummy(unsigned long y): x(y) { }
     operator unsigned long() const { return x; }
   };

   #include<iostream>
   using namespace std;
   int main() {
     Dummy d(5);
     unsigned long t = d;
     std::cout<< t<< std::endl;
   }

I don't personally recommend it, though. It invites
conversions where you don't intend it, for one thing.


To the OP:

A better way, perhaps, would be to write a 'toXXX()' const member
function and use it *explicitly*. With a conversion function like 'op
long', a typo with an object of your type used in an arithmetic
expression can easily go unnoticed (and that's what Jonathan's warning
you about).

Compare:

    unsigned long a42(42);
    Dummy aa42(84);
    ...
    unsigned long ul_result = aa42; // If no conversion exists, error.
                                    // Actually, it was a typo with
                                    // the intended initializer 'a42'.

Otherwise

    unsigned long ul_result = s42.toULong(); // no doubt

V
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