Why does a perfect forwarding function have to be templated?

Question

Why is the following code valid:

template
void foo(T1 &&arg) { bar(std::forward(arg)); }

std::string str = \"Hello Worl         


        

Answer 1

First of all, read this to get a full idea of forwarding. (Yes, I'm delegating most of this answer elsewhere.)

To summarize, forwarding means that lvalues stay lvalues and rvalues stay rvalues. You can't do that with a single type, so you need two. So for each forwarded argument, you need two versions for that argument, which requires 2^N combinations total for the function. You could code all the combinations of the function, but if you use templates then those various combinations are generated for you as needed.

---

If you're trying to optimize copies and moves, such as in:

struct foo
{
    foo(const T& pX, const U& pY, const V& pZ) :
    x(pX),
    y(pY),
    z(pZ)
    {}

    foo(T&& pX, const U& pY, const V& pZ) :
    x(std::move(pX)),
    y(pY),
    z(pZ)
    {}

    // etc.? :(

    T x;
    U y;
    V z;
};

Then you should stop and do it this way:

struct foo
{
    // these are either copy-constructed or move-constructed,
    // but after that they're all yours to move to wherever
    // (that is, either: copy->move, or move->move)
    foo(T pX, U pY, V pZ) :
    x(std::move(pX)),
    y(std::move(pY)),
    z(std::move(pZ))
    {}

    T x;
    U y;
    V z;
};

You only need one constructor. Guideline: if you need your own copy of the data, make that copy in the parameter list; this enables the decision to copy or move up to the caller and compiler.