c++ - Right reference on lvalue -

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when rvalue become right reference on lvalue?

example following, , should solved std::forward:

void g(int && v1, int & v2) {     std::cout << v1 << " " << ++v2 << std::endl; }  template <typename f, typename t1, typename t2> void flip(f f,t1 && t1, t2  && t2){     g(t2, t1); }

error occurs on call this:

flip(g, j, 2);

my problem is: in flip(g, j 2), third argument 2 rvalue.
in function flip, argument t2 right reference on 2. when g called, seems t2 right reference lvalue.
in point did rvalue, received reference (without copying?), became lvalue?

here example: https://ideone.com/cjvfcg

there's "if has name" rule saying within code this, if variable has name (in case, t2) @ point becomes lvalue. (see scott meyers effective modern c++.)

to preserve original intent of caller, indeed use perfect forwarding:

g(std::forward<t2>(t2), ...)

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