bugprone-move-forwarding-reference

Warns if std::move is called on a forwarding reference, for example:

template <typename T>
void foo(T&& t) {
  bar(std::move(t));
}

Forwarding references should typically be passed to std::forward instead of std::move, and this is the fix that will be suggested.

(A forwarding reference is an rvalue reference of a type that is a deduced function template argument.)

In this example, the suggested fix would be

bar(std::forward<T>(t));

Background

Code like the example above is sometimes written with the expectation that T&& will always end up being an rvalue reference, no matter what type is deduced for T, and that it is therefore not possible to pass an lvalue to foo(). However, this is not true. Consider this example:

std::string s = "Hello, world";
foo(s);

This code compiles and, after the call to foo(), s is left in an indeterminate state because it has been moved from. This may be surprising to the caller of foo() because no std::move was used when calling foo().

The reason for this behavior lies in the special rule for template argument deduction on function templates like foo() -- i.e. on function templates that take an rvalue reference argument of a type that is a deduced function template argument. (See section [temp.deduct.call]/3 in the C++11 standard.)

If foo() is called on an lvalue (as in the example above), then T is deduced to be an lvalue reference. In the example, T is deduced to be std::string &. The type of the argument t therefore becomes std::string& &&; by the reference collapsing rules, this collapses to std::string&.

This means that the foo(s) call passes s as an lvalue reference, and foo() ends up moving s and thereby placing it into an indeterminate state.