Unlimited integer: conversion to built-in int?

From: helmut.zeisel_at_[hidden]
Date: 2001-08-23 01:47:13

In the big_int implementation I provided a constructor
for big_int from every built in integral type,
but no operator to implictely convert a big_int to a built-in
integral type; the conversion must be done explicitely:

int i=5;
unlimited u=i; // possible
i=convert<int>(u); // possible, might throw std::overflow_error
// i=u; - not possible!

A user of my big_int implementation asked me if
I could add implicit conversion to less precise integral types.
since this is offered also buy built-in integers.

This implicit conversion, however, is an unsafe operation,
so I am not sure whether it should be offered.

What do you think?

Helmut