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    You are wrong. You can call a constructor of the same class. It will be determined which constructor to call using its argument list. Doing B(int x, inty) : B(x) will first call the constructor with signature B(int x). Commented Nov 29, 2019 at 9:54
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    Yes. But I was correct in November of 2008, before C++11 was published. Commented Dec 3, 2019 at 20:42