Re: [Boost-users] Generically comparing structs

Subject: Re: [Boost-users] Generically comparing structs
From: Christopher Jefferson (chris_at_[hidden])
Date: 2009-08-28 04:24:48

On 27 Aug 2009, at 19:42, Steven Watanabe wrote:

> AMDG
>
> Bill Buklis wrote:
>> Frequently I have to write comparison functions (usually of the  
>> "less"
>> variety) that involve multiple keys or members of a struct. For  
>> example:
>>
>> struct s
>> {
>>  A   a;
>>  B   b;
>>  C   c;
>>  D   d;
>>  // there may be other members not used in the comparison
>> };
>>
>> // operator< function or an equivalent function object
>> bool operator<( const s& lhs, const s& rhs )
>> {
>>  if( lhs.a < rhs.a )
>>     return(true);
>>  else if( lhs.a == rhs.a )
>>  {
>>     if( lhs.b < rhs.b )
>>        return(true);
>>     else if( lhs.b == rhs.b )
>> //and so on...
>>  }
>>  return(false);
>> }
>>
>> This gets quite repetitive, especially if I have to write alternate  
>> sort
>> orders (e.g. b,a,c,d or c,d,a,b)
>>
>> I'm thinking there has to an easier way to do this or a way to do a  
>> generic
>> algorithm for this. Is there anything in boost that would help?
>>
>> I had an idea that one might be able to fill a vector with binders  
>> for each
>> element and then call lexicographical_compare, but I'm not sure how  
>> to make
>> it work.
>>
>
> Boost.Fusion provides several ways to do this.
>
> #include <boost/fusion/include/less.hpp>
> #include <boost/fusion/include/adapt_struct.hpp>
>
> BOOST_FUSION_ADAPT_STRUCT(s, (A, a)(B, b)(C, c)(D, d))
>
> bool operator<( const s& lhs, const s& rhs )
> {
>   return boost::fusion::less(lhs, rhs);
> }
>
> If you need different sort orders, you can use tie:
>
> #include <boost/fusion/include/vector_tie.hpp>
>
> bool operator<( const s& lhs, const s& rhs )
> {
>   return boost::fusion::vector_tie(lhs.a, lhs.b, lhs.c, lhs.d) <
>       boost::fusion::vector_tie(rhs.a, rhs.b, rhs.c, rhs.d);
> }

If you don't want to pull in fusion, then tuple::tie will perform  
exactly the same task (this is what I usually use).

Chris