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Subject: Re: [Boost-users] [proto] defining function call operators on expressions
From: Joel Falcou (joel.falcou_at_[hidden])
Date: 2010-03-02 03:58:00


Eric Niebler wrote:
> It's not actually quite that simple in this case. Manjunath has defined
> his function call overloads as follows:
>
> result_type operator()(
> typename
> boost::result_of<call_param_type<0>(const Expr &)>::type x,
> typename
> boost::result_of<call_param_type<1>(const Expr &)>::type y)
>
> This function is *not* a template, so the types mentioned in its
> signature are evaluated eagerly. The call_param_type transform extracts
> the Nth child from Expr. If Expr doesn't have enough child nodes, it
> violates the transform's preconditions. Compiler errors are sure to result.
>

Oh good catch, I missed that.


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