CODEWITHSUNDEEP
c++templatesvariadic-templatesvariadic-functionsperfect-forwarding
Drake Johnson
Drake Johnson

Reputation: 664

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!

Upvotes: 3

Views: 1099

Answers (1)

Remy Lebeau
Remy Lebeau

Reputation: 596377

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))

Upvotes: 6

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