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#ActualCornstalks

Posted 18 December 2012 - 11:02 AM

If you're using C++11, there's std::tie and you can just do:
return std::tie(a.dir, a.startx, a.starty) < std::tie(b.dir, b.startx, b.starty);

#2Cornstalks

Posted 18 December 2012 - 11:01 AM

If you're using C++11, there's std::tie and you can just do:

return std::tie(a.dir, a.startx, a.starty) < std::tie(b.dir, b.startx, b.starty);

#1Cornstalks

Posted 18 December 2012 - 11:00 AM

If you're using C++11, there's std::tie and you can just do:


return std::tie(a.dir, a.startx, a.starty) < std::tie(b.dir, b.startx, b.starty);

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