Initialize C-style array with non-zero value without loops

Question

We know that C++ allows initialization of C-style arrays with zeros:

int a[5] = {0};
// or
int a[5] = {};

The same works for std::array

std::array a<int, 5> = {};

However, this won't work:

int a[5] = {33}; // in memory( 33, 0, 0, 0, 0 )
std::array<int, 5> = {33}; // in memory( 33, 0, 0, 0, 0 )

Is there any way to initialize the whole array with a non-zero value without using vector or algorhtm?

Maybe constexpr could help? What would be the best solution?

P.S.:

GCC offers this syntax

int a[5] = {[0 ... 4] = 33};

but I am not sure if it is valid for other compilers.

Maybe you need to tighten the requirements because int a[5] = {33, 33, 33, 33, 33}; — juanchopanza
– juanchopanza, Commented Apr 23, 2016 at 12:28
Well, Googling your exact title gives: 'About 1,470,000 results'. Maybe there is something in there that may be of use. — Martin James
– Martin James, Commented Apr 23, 2016 at 12:31
@LightnessRacesinOrbit Personally I prefer such a qualifier, it makes discussion clearer IMO. — M.M
– M.M, Commented Apr 23, 2016 at 12:35
@LightnessRacesinOrbit people, even programmers, rarely speak nor type namespaces in English. It is very common to discuss about vector and map and set, even though in C++, such things only exists in the std namespace. People discuss about arrays too, and depending on context, they may refer to either form of array or even a vector which is also an array of objects. Therefore there is a need for a way to disambiguate the term. — eerorika
– eerorika, Commented Apr 23, 2016 at 12:41

Barry · Accepted Answer · 2016-04-23 14:06:30Z

6

What do you have against <algorithm>? I think this is pretty clean:

int a[5];                                  // not initialized here yet
std::fill(std::begin(a), std::end(a), 33); // everything initialized to 33

answered Apr 23, 2016 at 14:06

Barry

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1 Comment

paceholder Over a year ago

Nothing personal, it was more a theoretical question. We have a certain syntax which works for zeros, is there anything similar which might work for non-zeros.

simpel01 · Accepted Answer · 2016-04-23 13:15:12Z

3

I have some code around to achieve compile-time initialization of an std::array using template meta-programming (of course).

namespace impl {

    template <class SeqTy, size_t N, SeqTy FillVal, SeqTy... Seq>
    struct make_fill {
        using type = typename make_fill<SeqTy, N-1, FillVal, FillVal, Seq...>::type;
    };

    template <class SeqTy, SeqTy FillVal, SeqTy... Seq>
    struct make_fill<SeqTy, 0, FillVal, Seq...> {
        using type = std::integer_sequence<SeqTy, Seq...>;
    };

    template <class T>
    struct make_array;

    template <class SeqTy, SeqTy... Seq>
    struct make_array<std::integer_sequence<SeqTy, Seq...>> {
        static constexpr std::array<SeqTy, sizeof...(Seq)> value() { 
            return std::array<SeqTy, sizeof...(Seq)>{ {Seq...} };
        }
    };

} // end impl namespace

template <class SeqTy, size_t N, SeqTy FillVal = 0ul>
constexpr std::array<SeqTy, N> fill() {
    return impl::make_array<typename impl::make_fill<SeqTy, N, FillVal>::type>::value();
};

which you can use as follows:

std::array<size_t, N> ones = fill<size_t,N,1ul>();

I think you can easily adapt it if you don't want to use std::array

answered Apr 23, 2016 at 13:15

simpel01

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1 Comment

paceholder Over a year ago

I must admit that my question was valid for std::array as well; the same initialization rules are applied there. Your answer is pretty much what I was looking for.

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