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c++templatesvariadic-templatesvariadic-functionsperfect-forwarding

Syntax of '...' with Parameter Pack Expansion using std::forward


In C++, it seems that a parameter may be normally expanded with ... directly after the parameter pack's name. For example,

template <class... Tys>
void function(Tys... params) {
    function(params...);
}

However, when using std::forward, the ... goes seemingly after the parameter pack's name. For example,

template <class... Tys>
void function(Tys... params) {
    function(std::forward<Tys>(params)...); // The '...' is outside the parenthesis of 'params'
}

My question is why is there a difference? Why is the following code incorrect?

template<class... Tys>
void function(Tys... params) {
    function(std::forward<Tys>(params...));
}

At what point in the parameter pack expanded? I probably am not fully understanding how std::forward works or how it forwards arguments perfectly.

I appreciate any answers!


Solution

  • The ... applies to the thing on its left.

    • function(params...)

      expands to

      function(param1, param2, ..., paramN).

    • function(std::forward<Tys>(params)...)

      expands to

      function(std::forward<Tys1>(param1), std::forward<Tys2>(param2), ..., std::forward<TysN>(paramN))

    • function(std::forward<Tys>(params...))

      expands to

      function(std::forward<Tys>(param1, param2, ..., paramN))