Re: [Boost-users] [enable_if] using enable_if to extend std::numeric_limits?

7 Mar 2007

      From: "Ovanes Markarian" <om_boost@keywallet.com>
...
Sorry,
I ment enable_if< is_decimal_type<T> >::type. The ::type will be expanded to
either void or a derivable type.
Maybe I'm misunderstanding, but how can I derive std::numeric_limits from enable_if, since std::numeric_limits is already defined? I can't do this:

namespace std {
template <typename T>
class numeric_limits :
    private boost::enable_if<boost::decimal::is_decimal_type<T> >::type
{
public:
    static T min() { return boost::decimal::decimal_limits<T>::min(); }
};
}

and I can't do this:

namespace std {
template <typename T>
class numeric_limits<T> :
    private boost::enable_if<boost::decimal::is_decimal_type<T> >::type
{
public:
    static T min() { return boost::decimal::decimal_limits<T>::min(); }
};
}

because this is not a specialization.

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Re: [Boost-users] [enable_if] using enable_if to extend std::numeric_limits?

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