I'm currently studying C++ lvalue and rvalue these days. I got a little confuse in below code.
class Test {
public:
Test(){}
explicit Test(int _i): i(_i) {}
void doSomething()& {
cout << "L value ref" << endl;
}
void doSomething()&& {
cout << "R value ref" << endl;
}
private:
int i;
};
void callDoSomething(Test& t) {
t.doSomething();
}
void callDoSomething(Test&& t) {
t.doSomething();
}
int main() {
auto a = Test();
callDoSomething(a);
callDoSomething(Test());
return 0;
}
In above code, I will get this result:
L value ref
L value ref
I have check above code via debugger and pretty sure that in this code snippet: callDoSomething(Test()); It will go to the rvalue reference one which is callDoSomething(Test&& t). But why did it still call the lvalue member function?
I also tried template and still got same result.
template <typename T>
void callDoSomething(T &&t) {
t.doSomething();
}
I have already read this post and knows that in template version. The T &&t is actually a universal reference. But after check via debugger. The type of t is still Test&& and we can get Test&& &&t. By reference collapsing's definition, we should still get Test&& t.
Can anyone explain why this happen? Thanks a lot!
