Evaluating a parameter pack

Question

Is this the only way to evaluate a parameter pack without using folding (since it requires the use of operators)?

#include <iostream>

template<int ...Is, typename Function>
void eval(Function&& f)
{
    // (f(Is)...);
    auto op = [&f](int i){f(i); return 0;};
    auto doNothing = [](auto...){};
    doNothing(op(Is)...);
}

int main()
{
    eval<0,1,2>([](int x){std::cout << x << "\n";});
}

Essentially I want to do (f(Is)...), but for some reason, this is disallowed in C++. Is there a more elegant way this can be achieved than by using the workaround presented above?

You just need a coma between `f(Is)` and `...`. `(f(Is)...)` is a parameter pack expansion that can appear as a function argument. `(f(Is),...)` is a fold expression on the comma operator. — Oliv, Oct 20 '19 at 18:15
`since it requires the use of operators`: coma is an operator! — ph3rin, Oct 20 '19 at 18:18

score 7 · Accepted Answer · answered Oct 20 '19 at 18:09

7

There is a simpler solution:

#include <iostream>

template<int ...Is, typename Function>
void eval(Function&& f)
{
    (f(Is),...);
}

int main()
{
    eval<0,1,2>([](int x){std::cout << x << "\n";});
}

answered Oct 20 '19 at 18:09

Oliv

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1

What happens if someone overloads `,` though? – lightxbulb Oct 20 '19 at 18:26
4

@lightxbulb: you fire them :) `((void)f(Is),...)` might prevent the overload, or something like that. – Mat Oct 20 '19 at 18:45

Evaluating a parameter pack

1 Answers1