RE: [Boost-users] Defining less on member variables/functions

From: Chris Jurney (cjurney_at_[hidden])
Date: 2004-03-29 14:19:49

In this case, I could have used pair's operator<, but if I have a class
by value as the second item of the pair that doesn't have a meaningful
operator<, I want to ignore it in the sort.  Thanks for all the help,
guys.

Chris

-----Original Message-----
From: boost-users-bounces_at_[hidden]
[mailto:boost-users-bounces_at_[hidden]] On Behalf Of Victor A.
Wagner Jr.
Sent: Friday, March 26, 2004 8:40 AM
To: boost-users_at_[hidden]
Subject: Re: [Boost-users] Defining less on member variables/functions

why do you want to sort ONLY on the 1st element of the pair?

At Thursday 2004-03-25 11:34, you wrote:
>content-class: urn:content-classes:message
>Content-Type: multipart/alternative;
>         boundary="----_=_NextPart_001_01C41297.D800D40A"
>
>I've had this same pattern come up 4 or 5 times now, and each time I've

>been unable to find a stl + boost solution that allowed me to do what I

>want without writing my own less function.  Basically, I want something

>that will let me do a sort, or define a map on structs or pairs without

>writing a less that compares the member variable or ..first.  My last
attempt:
>
>vector<pair<float, float>> myVector
>std::sort(myVector.begin(), myVector.end(), _1.first < _2.first);
>
>No such luck.  It looks like I can do operator * and & on lambda
objects, 
>but not access members through them.  Is there anything in boost, stl,
or 
>elsewhere that lets me not write this:
>
>class myLess
>{
>     bool operator() (pair<float, float> a, pair<float, float>b)
>         {return a.first < b.first;}
>}
>
>Thanks for any help,
>Chris
>
>
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Victor A. Wagner Jr.      http://rudbek.com
The five most dangerous words in the English language:
               "There oughta be a law" 

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