If you need it more than once in a single function consider: auto &mymap = *mptr; mymap[a] = x; mymap[b] = y; for even less uglyness.
A recurring question for me: Given `std::unordered_map<Key, T> *mptr`, which is least ugly? A: (*mptr)[X] = Y; B: mptr->operator[](X) = Y; C: mptr->insert(std::make_pair(X, Y));
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I guess it heavily depends on who reads the code. I as c++ detester would favor 3 because it is most clear but I assume a) is readable enough to be fine for those whose souls were eaten by the stroustrup.
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Oh yeah. That's better.
End of conversation
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mptr->insert(std::pair(X,Y)); // C++17 If you use mptr more than once, consider auto &m = *mptr; // use m thereafter; so much nicer!
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Thanks. I think https://twitter.com/Alnurn/status/1025713301337063424 … is better, and it only requires C++11 afaict.
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the first. It's almost pythonic.
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This isn't just a syntactic choice. A & B both default-construct and assign; C can in-place construct, depending on the type of X & Y.
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