Re: Unlimited integer: conversion to built-in int?

From: helmut.zeisel_at_[hidden]
Date: 2001-08-24 14:35:12

--- In boost_at_y..., scleary_at_j... wrote:
> > > Having said that, I must also say that I personally think
conversion
> > > functions are evil; they always allow users to hurt themselves.
> Providing a
> > > conversion to bool and hiding the other conversions will
probably make
> code
> > > such as this ambiguous (it will if your "BigInt" has an implicit
> conversion
> > > from int's):
> > >   BigInt a;
> > >   int b;
> > >   a + b; // operator+(a, BigInt(b)) or bool(a) + b ?
> > 
> > Fortunately this is no problem.
> > Thanks to operator.hpp (again)
> > I have an explicit overload for operator+(const BigInt&, int)
> > and all other related operators.
> 
> That would work on a bounded set of types.  But how would it react
to
> __int64?  Or __int128 on next year's compilers?
> 

Hopefully the set of built-in types will
always be finite.
OTOH, you are right,
it is not very elegant to be forced
to overload the operators for all built-in types.

Helmut