Fold Expression in C++17

Question

I'm reading "C++17 - The Complete Guide" book and I came across the example on page 107 and 108 regarding fold expression in C++17:

template<typename First, typename... Args>
void print(First first, const Args&... args)
{
    std::cout << first;
    auto outWithSpace = [](const auto& arg)
    {
        std::cout << " " << arg;
    };
    (..., outWithSpace(args));
    std::cout << "\n";
}

Is there any reason that the author couldn't do this as follows (without separating the first argument from the rest and apart from an extra printed space!):

template<typename... Types>
void print(Types const&... args)
{
    ([](const auto& x){ std::cout << x << " "; }(args), ...);
    std::cout << "\n";
}

I think you answered your own question? Yours prints an extra space. — Barry, Dec 16 '19 at 22:33
@Barry, just wanted to make sure I'm not missing the point of the book example :) — Ali, Dec 16 '19 at 22:35

Michael Kenzel · Accepted Answer · 2019-12-16T23:06:35.067

As you have already figured out, the author could not have done as you suggest because that would leave an additional space…

What the author could have done though, would be something like

template<typename First, typename... Args>
void print(First first, const Args&... args)
{
    ((std::cout << first), ..., (std::cout << ' ' << args)) << '\n';
}

or, rather

template <typename Arg, typename... Args>
std::ostream& print(Arg&& arg, Args&&... args)
{
    return ((std::cout << std::forward<Arg>(arg)), ..., (std::cout << ' ' << std::forward<Args>(args))) << '\n';
}

live example here

score 0 · Answer 2 · answered Dec 17 '19 at 03:24

0

Plainly the readable way to write this is

template<typename... Types>
void print(Types const&... args)
{
    std::size_t n = 0;
    (std::cout << " " + !n++ << args), ...);
    std::cout << '\n';
}

(with std::forward to taste). The temptation to do this sort of nonsense (which could use bool first if C++17 hadn’t killed that wonderful feature) is the motivation for the template for feature planned for C++23.

answered Dec 17 '19 at 03:24

Davis Herring

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@Barry i think he means arithmetic on `bool` – Richard Hodges Dec 17 '19 at 17:37
@RichardHodges: With tongue in cheek (like the answer), yes: for weird cases like this (and code golfing), post-increment (“has this been set already?”) and decrement (toggle) on `bool` were useful shorthands but far too obscure. – Davis Herring Dec 18 '19 at 02:07

score 0 · Answer 3 · answered Dec 17 '19 at 18:19

0

Since we're having fun times with friends

template<typename First, typename... Args>`
void print(First&& first, Args&&... args)
{
    auto emit = [](auto&&...x) { ((std::cout << x), ...); };

    (emit(first), ..., emit(' ', args)), emit('\n');
}

:)

answered Dec 17 '19 at 18:19

Richard Hodges

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score -1 · Answer 4 · answered Dec 16 '19 at 22:46

-1

No, it's just a question of aesthetics. No more reason.

answered Dec 16 '19 at 22:46

Zig Razor

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Fold Expression in C++17

4 Answers4