Comment by fluoridation

>So, if you have f(T&&) and f(T), and call f(std::move(t)) then you'll get an ambiguous overload compilation error [...]

Okay, but that's not what I'm saying. I'm not talking about the program being invalid, I'm talking about the program being confusing to the programmer, which is what this article is about. Sure, in that case the program is ill-formed. What if you have f(const T &, double) and f(T, int)? If you call f(x, 0) you cause a copy, and if you call f(x, 0.0) you don't. A programmer who's not aware that the second overload exists will not realize this is happening, in the same way that he will not realize std::move() is not moving anything if they don't realize, for example, that the argument is const.