From: musiphil@bawi.org   
      
   On 2011-12-08 09:21, Nevin ":-]" Liber wrote:   
   >   
   > Sure, but I can use other facilities to minimize the pain, as in:   
   >   
   >    bool operator<(complex const& a, complex const& b)   
   >    { return std::tie(a.re, a.im) < std::tie(b.re, b.im); }   
   >   
      
   That leads me to another question: should[n't] the library have   
   provided std::less for std::tie that doesn't just fall back to   
   operator< for std::tie, but that uses std::less for each member   
   in the same manner as operator< for std::tie does?   
      
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   Seungbeom Kim   
      
      
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