Re: [boost] [unordered] Buffered functions?

Subject: Re: [boost] [unordered] Buffered functions?
From: Howard Hinnant (hinnant_at_[hidden])
Date: 2008-12-28 11:06:30

On Dec 28, 2008, at 1:25 AM, David Abrahams wrote:

> The conundrum here is that if we use that technique for tuple, then a
> tuple of empty types can never be empty itself.  I'm starting to get  
> ill
> just thinking about a generic framework for composing "logically  
> empty"
> types that undoes this "EBO erasure" effect when necessary.

Perhaps I'm misunderstanding.  Reverting from English to C++ in the  
hopes of clarification. :-)  The C++0X program and its output below  
look quite reasonable to me:

#include <tuple>
#include <type_traits>
#include <cstdio>

struct empty1 {};
struct empty2 {};
struct empty3 {};

int main()
{
     typedef std::tuple<empty1> T1;
     std::printf("is_empty<tuple<empty1>> = %d\n",  
std::is_empty<T1>::value);
     typedef std::tuple<empty1, empty2> T2;
     std::printf("is_empty<tuple<empty1, empty2>> = %d\n",  
std::is_empty<T2>::value);
     typedef std::tuple<T2, empty3> T3;
     std::printf("is_empty<tuple<tuple<empty1, empty2>, empty3>> = %d 
\n", std::is_empty<T3>::value);
     typedef std::tuple<int, T3> T4;
     std::printf("is_empty<tuple<int, tuple<tuple<empty1, empty2>,  
empty3>>> = %d\n", std::is_empty<T4>::value);
     std::printf("sizeof(int) = %zu, sizeof(tuple<int,  
tuple<tuple<empty1, empty2>, empty3>>) = %zu\n", sizeof(int),  
sizeof(T4));
}

is_empty<tuple<empty1>> = 1
is_empty<tuple<empty1, empty2>> = 1
is_empty<tuple<tuple<empty1, empty2>, empty3>> = 1
is_empty<tuple<int, tuple<tuple<empty1, empty2>, empty3>>> = 0
sizeof(int) = 4, sizeof(tuple<int, tuple<tuple<empty1, empty2>,  
empty3>>) = 4

I currently favor the "shallow hierarchy" design Doug outlined in c+ 
+std-lib-18989.  Though EMO isn't shown in that description, it is  
pretty simple to add to that design.

-Howard