Is it correct to capture the perfectly-forwarded mStuff variable with
the &mStuff syntax?
Yes, assuming that you don’t use this lambda outside doSomething. Your code captures mStuff per reference and will correctly forward it inside the lambda.
For mStuff being a parameter pack it suffices to use a simple-capture with a pack-expansion:
template <typename... T> void doSomething(T&&... mStuff)
{
auto lambda = [&mStuff...]{ doStuff(std::forward<T>(mStuff)...); };
}
The lambda captures every element of mStuff per reference. The closure-object saves an lvalue reference for to each argument, regardless of its value category. Perfect forwarding still works; In fact, there isn’t even a difference because named rvalue references would be lvalues anyway.