Do rvalues decay silently?

Question

std::vector foo( std::vector && rval )
{
    return std::move( rval );
}

If a function expects an rvalue reference but gets something else - e.g. a const reference or a temporary or whatever different from std::move(vec), will it silently make a copy instead of throwing an error or even a warning?

Answer 1

Try it yourself:

#include <iostream>

struct S {
    S() { }
    S(const S& other) { std::cout << "copy ctor" << std::endl; }
    S(S&& other) { std::cout << "move ctor" << std::endl; }
}; 

int foo( S && rval )
{
    return 1;
}

int main()
{
    S s1;
    foo (s1);
}

Copying S'es is not silent. So, what happens when you try to compile this?

This happens (Godbolt):

<source>: In function 'int main()':
<source>:17:10: error: cannot bind rvalue reference of type 'S&&' to lvalue of type 'S'
   17 |     foo (s1);
      |          ^~
<source>:9:15: note:   initializing argument 1 of 'int foo(S&&)'
    9 | int foo( S && rval )
      |          ~~~~~^~~~

So, the answer to your question is "no".