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

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(const Expr &)>::type x, typename boost::result_of(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.