Re: [Boost-users] [proto] defining function call operators on expressions

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.