4 ms·

It's matching std::vector's implicit constructor which takes two iterators, which can be initialized with any pointer. Note that it won't compile with one, thre

It's matching std::vector's implicit constructor which takes two iterators, which can be initialized with any pointer. Note that it won't compile with one, three, or more string-literal char*s, but exactly two.

metadat 2y ago
Why only exactly two? The whole "UB but it still compiles" thing, is pretty gross.
- comex 2y ago
  Because the constructor takes two arguments, a "begin" iterator and an "end" iterator. And pointers count as valid iterators. Normally you wouldn't use the "= {}" syntax to invoke a constructor this way. Instead of `std::vector<int> v = {begin, end}` you would usually write `std::vector<int> v(begin, end)`. But for some reason C++11 decided to make those two things mostly equivalent.
  - gpderetta 2y ago
    List initialization is quite nice, but the interaction with brace initialization introduced at the same time is a giant foot-gun.
  - BobbyJo 2y ago
    > But for some reason C++XX decided to.... Why I gave up C++. That and because somehow package management is still a nightmare.
- UncleMeat 2y ago
  This sort of UB detection requires bespoke work. The compiler probably can figure out that the begin and end iterators here are referencing different objects but if you add just a bit more complexity to the code then the compiler won't be able to prove that.
  - mwkaufma 2y ago
    That's consistent with STL's move to use "explicit" and other narrowing annotations, as described in the article.