Extract data from heterogeneous list - c++

I was looking at this answer about the advantages of auto in template parameter.
Please consider the following heterogeneous list:
template <auto ... vs> struct HeterogenousValueList {};
using MyList1 = HeterogenousValueList<42, 'X', 13u>;
Now I've declared a type named MyList1. How can I extract the stored data from this type (i.e., 42, 'x' or 13u)?

To extract data from a template parameter pack, we usually do pattern matching in template.
Firstly, we create a class template At but without contents. Its template parameters are supposed to be an index, and an instance of HeterogenousValueList. This class template will be used like a function to access information in the list.
template <int Index, class ValueList>
struct At;
Next, we create a specialization of At. This is where pattern matching is used. Through pattern matching, the first element of the list will become u. The rest of the list will be vs. If the index is 0, u can be accessed through the static member value. Note that vs can be an empty list, so this also covers the case that u being the last of the list.
template <auto u, auto... vs>
struct At<0, HeterogenousValueList<u, vs...>>
{
static constexpr auto value = u;
};
What if the index is not 0? We shift the list and decrement the index by 1, and pass them into At again. In other words, this is a template recursion.
template <int Index, auto u, auto... vs>
struct At<Index, HeterogenousValueList<u, vs...>>
{
static constexpr auto value = At<Index - 1, HeterogenousValueList<vs...>>::value;
};
Now, we can try to use it: https://godbolt.org/g/51dpH8
int main()
{
volatile auto value0 = At<0, MyList1>::value;
volatile auto value1 = At<1, MyList1>::value;
volatile auto value2 = At<2, MyList1>::value;
// volatile auto value3 = At<-1, MyList1>::value;
// volatile auto value4 = At<3, MyList1>::value;
}
I use volatile variable so that the compiler does not optimize the effect away and you can see the effect in the assembly listing.
And one more great thing: the compiler checks the bound! We usually don't have bound check for run-time array for run-time efficiency reason. But this is a compile-time list. The compiler can do it for us!
Actually, there is a simpler implementation. This time, we use constexpr-if in a function template. But the idea of pattern matching and template recursion remain the same.
template <int Index, auto u, auto... vs>
auto at(HeterogenousValueList<u, vs...>)
{
if constexpr (Index == 0)
return u;
else
return at<Index - 1>(HeterogenousValueList<vs...>{});
}
And this time, when we use it, we need to instantiate MyList1 into an object.
https://godbolt.org/g/CA3VHj
int main()
{
volatile auto value0 = at<0>(MyList1{});
volatile auto value1 = at<1>(MyList1{});
volatile auto value2 = at<2>(MyList1{});
// volatile auto value3 = at<-1, MyList1>::value;
// volatile auto value4 = at<3, MyList1>::value;
}

As you mention "playing" and "learning it by manipulation", you may be interested in several alternatives.
First solution: convert to std::tuple and then std::get
The std::tuple, which is a container for heterogeneous values, can be utilized to access the elements of your heterogeneous value list. This is a simple two-step process:
convert HeterogenousValueList<42, 'X', 13u>{} to std::tuple<int, char, unsigned>{42, 'X', 13u}
access the tuple's values at the desired positions by std::get
Complete C++17 example:
#include <type_traits>
#include <tuple>
template<auto... vs>
struct HeterogenousValueList {};
template<int i, auto... vs>
constexpr auto get(HeterogenousValueList<vs...>) {
constexpr std::tuple tuple{vs...};// class-template argument deduction
static_assert(std::is_same<
decltype(tuple), const std::tuple<int, char, unsigned>
>{});
return std::get<i>(tuple);
}
int main() {
using MyList1 = HeterogenousValueList<42, 'X', 13u>;
constexpr auto at1 = get<1>(MyList1{});
static_assert(at1 == 'X');
static_assert(std::is_same<decltype(at1), const char>{});
}
Second solution: wrap in types for std::tuple_element_t
Another useful idiom is to wrap non-type template parameters in a single empty type. In your example, this allows to make use of std::tuple_element_t, which can yield the nth type of a variadic pack:
#include <type_traits>
#include <tuple>
template<auto... vs>
struct HeterogenousValueList {};
template<auto v_>
struct SingleValue {// similar to std::integral_constant, but uses auto
static constexpr auto v = v_;
};
template<int i, auto... vs>
constexpr auto get(HeterogenousValueList<vs...>) {
using TupleOfSingleValues = std::tuple<SingleValue<vs>...>;
using SingleValueAtPosition = std::tuple_element_t<i, TupleOfSingleValues>;
return SingleValueAtPosition::v;
// return std::tuple_element_t<i, std::tuple<SingleValue<vs>...>>::v;// same
}
// same `main` as first solution
Third solution: Implement your own logic
There are several ways to implement your own version for "getting the nth type/element". One aspect in this business is the compile-time performance: particularly the recursive strategies are said to cause long compilation times.
My favorite non-recursive strategy is the one which is also used in Boost.Hana. If you prefer a video explanation you can watch two minutes of the talk "Metaprogramming for the brave" by Louis Dionne (Boost.Hana author) starting at 01 h 12 min. The idea is to use multiple inheritance. In your example, HeterogenousList<42, 'X', 13u> can have the base classes IndexedValue<0, 42>, IndexedValue<1, 'X'>, and IndexedValue<2, 13u>. Then you can pass the HeterogenousList<42, 'X', 13u> to a templated get function that takes a, say, const IndexedValue<1, [[DEDUCED]]>& as argument:
#include <type_traits>
#include <utility>
template<std::size_t i, auto v_>
struct IndexedValue {
static constexpr auto v = v_;
};
template<class Is, auto... vs>
struct IndexedHeterogenousList;
template<std::size_t... is, auto... vs>
struct IndexedHeterogenousList<
std::index_sequence<is...>,// partial specialization to extract the `is`
vs...
> : IndexedValue<is, vs>...// multiple "variadic" inheritance
{};
template<auto... vs>
struct HeterogenousValueList
: IndexedHeterogenousList<std::make_index_sequence<sizeof...(vs)>, vs...>
{};
template<std::size_t i, auto v>// `i` must be given; `v` is deduced
constexpr auto get(const IndexedValue<i, v>& iv) {// one base matches
return v;
}
// same `main` as first solution

Related

C++20 How to get the last element of a parameter pack of std::size_t

I've seen many answers online such as this one, but they do not seem to work when the parameter pack is of std::size_t.
template <typename ...Ts>
struct select_last
{
using type = typename decltype((std::type_identity<Ts>{}, ...))::type;
};
template<std::size_t... N>
class Example {
private:
using type = select_last<int, double>::type; // works
using size_t_type = select_last<size_t... N>::type; // doesn't work
};
How can I get the last element of a parameter pack of type std::size_t?
The template<std::size_t... N> is based on a non-type template parameter, so you cannot extract the type (or more precisely, there is no sense in trying to extract the type - I can just tell you it is std::size_t!), you may however extract the value, into a static constexpr.
Here is the proposed code:
template<std::size_t... N>
struct Last {
static constexpr std::size_t val = (N, ...); // take the last
};
int main() {
std::cout << Last<1, 2, 3, 99>::val; // 99
}
If you want, you can actually have it work for any type:
template<auto... N>
struct Last {
static constexpr auto val = (N, ...); // take the last
};

Convert between C++20 NTTP and type

I'm playing around with C++20 NTTPs (non-type template parameters), and I was wondering, is there a way to convert between the elements of an NTTP std::array and types in the form of T<int>?
Consider this example:
template<int X>
struct Func;
template<auto ints>
struct Foo{
// This doesn't work because within the std::apply the ints isn't a constant expression.
constexpr auto funcs() const {
return std::apply([](auto... xs){ return std::tuple{Func<xs>{}...};}, ints);
}
};
constexpr Foo<std::array{1,2,3}> f;
constexpr auto funcs = f.funcs();
It easily works if the elements are values in a variadic NTTP pack, but having them in a std::array would allow to use the full power of the standard library algorithms rather than having to operate on parameter packs through TMP (template metaprogramming).
template<int... ints>
struct Foo2{
// This here works, of course.
constexpr auto funcs() const {
return std::tuple{Func<ints>{}...};
}
};
constexpr Foo2<1,2,3> f2;
constexpr auto funcs = f2.funcs();
Full code here.
You can get a parameter pack with indices to the ints array as template parameter pack using the std::index_sequence method:
return []<std::size_t... I>(std::index_sequence<I...>){
return std::tuple{Func<ints[I]>{}...};
}(std::make_index_sequence<std::size(ints)>{});
Then ints[I] can be used exactly as you intend xs to be used in your example.

Function to generate a tuple given a size N and a type T

While trying to reply to this question, I found my self in the need of creating a bunch of parameters for a variadic function on the fly where:
the number of the parameters is not given
the types are all the same, but unknown (even if they must be default constructible)
At runtime, the standard containers and a for loop can be used to do that.
Anyway, I'd like to generate a set of parameters at compile time, so as to be able to forward them to a variadic function.
Because of that, a std::tuple seemed the obvious solution.
Here arose the question: given a size N and a default constructible type T at compile time, how can I write a function to generate a tuple of the given size?
I'm looking for something like this:
auto tup = gen<MyType, N>();
On SO is a notable example of a recursive generator based structs but I was struggling with a function based solution and I've not been able to find it anywhere.
A correctly written forwarding function (a la std::apply) should work with std::array<T, N> and anything else that implements the std::tuple_size/std::get interface. That said,
template<size_t, class T>
using T_ = T;
template<class T, size_t... Is>
auto gen(std::index_sequence<Is...>) { return std::tuple<T_<Is, T>...>{}; }
template<class T, size_t N>
auto gen() { return gen<T>(std::make_index_sequence<N>{}); }
Here is a possible implementation of such a function:
#include<utility>
#include<tuple>
template<typename T>
constexpr auto
params(std::index_sequence<0>) {
return std::tuple<T>{};
}
template<typename T, std::size_t I, std::size_t... O>
constexpr auto
params(std::index_sequence<I, O...>) {
auto tup = std::tuple<T>{ T{} };
auto seq = std::make_index_sequence<sizeof...(O)>{};
return std::tuple_cat(tup, params<T>(seq));
}
template<typename T, std::size_t N>
constexpr auto
gen(std::integral_constant<std::size_t, N>) {
return params<T>(std::make_index_sequence<N>{});
}
int main() {
auto tup = gen<int>(std::integral_constant<std::size_t, 3>{});
static_assert(std::tuple_size<decltype(tup)>::value == 3, "!");
}
For the sake of simplicity, I've used int as a type.
With a small effort, user defined types can be used and the constraint of having them default constructible can be relaxed.

Is it possible to define a std::array without size? [duplicate]

(Note: This question is about not having to specify the number of elements and still allow nested types to be directly initialized.)
This question discusses the uses left for a C array like int arr[20];. On his answer, #James Kanze shows one of the last strongholds of C arrays, it's unique initialization characteristics:
int arr[] = { 1, 3, 3, 7, 0, 4, 2, 0, 3, 1, 4, 1, 5, 9 };
We don't have to specify the number of elements, hooray! Now iterate over it with the C++11 functions std::begin and std::end from <iterator> (or your own variants) and you never need to even think of its size.
Now, are there any (possibly TMP) ways to achieve the same with std::array? Use of macros allowed to make it look nicer. :)
??? std_array = { "here", "be", "elements" };
Edit: Intermediate version, compiled from various answers, looks like this:
#include <array>
#include <utility>
template<class T, class... Tail, class Elem = typename std::decay<T>::type>
std::array<Elem,1+sizeof...(Tail)> make_array(T&& head, Tail&&... values)
{
return { std::forward<T>(head), std::forward<Tail>(values)... };
}
// in code
auto std_array = make_array(1,2,3,4,5);
And employs all kind of cool C++11 stuff:
Variadic Templates
sizeof...
rvalue references
perfect forwarding
std::array, of course
uniform initialization
omitting the return type with uniform initialization
type inference (auto)
And an example can be found here.
However, as #Johannes points out in the comment on #Xaade's answer, you can't initialize nested types with such a function. Example:
struct A{ int a; int b; };
// C syntax
A arr[] = { {1,2}, {3,4} };
// using std::array
??? std_array = { {1,2}, {3,4} };
Also, the number of initializers is limited to the number of function and template arguments supported by the implementation.
Best I can think of is:
template<class T, class... Tail>
auto make_array(T head, Tail... tail) -> std::array<T, 1 + sizeof...(Tail)>
{
std::array<T, 1 + sizeof...(Tail)> a = { head, tail ... };
return a;
}
auto a = make_array(1, 2, 3);
However, this requires the compiler to do NRVO, and then also skip the copy of returned value (which is also legal but not required). In practice, I would expect any C++ compiler to be able to optimize that such that it's as fast as direct initialization.
I'd expect a simple make_array.
template<typename ret, typename... T> std::array<ret, sizeof...(T)> make_array(T&&... refs) {
// return std::array<ret, sizeof...(T)>{ { std::forward<T>(refs)... } };
return { std::forward<T>(refs)... };
}
Combining a few ideas from previous posts, here's a solution that works even for nested constructions (tested in GCC4.6):
template <typename T, typename ...Args>
std::array<T, sizeof...(Args) + 1> make_array(T && t, Args &&... args)
{
static_assert(all_same<T, Args...>::value, "make_array() requires all arguments to be of the same type."); // edited in
return std::array<T, sizeof...(Args) + 1>{ std::forward<T>(t), std::forward<Args>(args)...};
}
Strangely, can cannot make the return value an rvalue reference, that would not work for nested constructions. Anyway, here's a test:
auto q = make_array(make_array(make_array(std::string("Cat1"), std::string("Dog1")), make_array(std::string("Mouse1"), std::string("Rat1"))),
make_array(make_array(std::string("Cat2"), std::string("Dog2")), make_array(std::string("Mouse2"), std::string("Rat2"))),
make_array(make_array(std::string("Cat3"), std::string("Dog3")), make_array(std::string("Mouse3"), std::string("Rat3"))),
make_array(make_array(std::string("Cat4"), std::string("Dog4")), make_array(std::string("Mouse4"), std::string("Rat4")))
);
std::cout << q << std::endl;
// produces: [[[Cat1, Dog1], [Mouse1, Rat1]], [[Cat2, Dog2], [Mouse2, Rat2]], [[Cat3, Dog3], [Mouse3, Rat3]], [[Cat4, Dog4], [Mouse4, Rat4]]]
(For the last output I'm using my pretty-printer.)
Actually, let us improve the type safety of this construction. We definitely need all types to be the same. One way is to add a static assertion, which I've edited in above. The other way is to only enable make_array when the types are the same, like so:
template <typename T, typename ...Args>
typename std::enable_if<all_same<T, Args...>::value, std::array<T, sizeof...(Args) + 1>>::type
make_array(T && t, Args &&... args)
{
return std::array<T, sizeof...(Args) + 1> { std::forward<T>(t), std::forward<Args>(args)...};
}
Either way, you will need the variadic all_same<Args...> type trait. Here it is, generalizing from std::is_same<S, T> (note that decaying is important to allow mixing of T, T&, T const & etc.):
template <typename ...Args> struct all_same { static const bool value = false; };
template <typename S, typename T, typename ...Args> struct all_same<S, T, Args...>
{
static const bool value = std::is_same<typename std::decay<S>::type, typename std::decay<T>::type>::value && all_same<T, Args...>::value;
};
template <typename S, typename T> struct all_same<S, T>
{
static const bool value = std::is_same<typename std::decay<S>::type, typename std::decay<T>::type>::value;
};
template <typename T> struct all_same<T> { static const bool value = true; };
Note that make_array() returns by copy-of-temporary, which the compiler (with sufficient optimisation flags!) is allowed to treat as an rvalue or otherwise optimize away, and std::array is an aggregate type, so the compiler is free to pick the best possible construction method.
Finally, note that you cannot avoid copy/move construction when make_array sets up the initializer. So std::array<Foo,2> x{Foo(1), Foo(2)}; has no copy/move, but auto x = make_array(Foo(1), Foo(2)); has two copy/moves as the arguments are forwarded to make_array. I don't think you can improve on that, because you can't pass a variadic initializer list lexically to the helper and deduce type and size -- if the preprocessor had a sizeof... function for variadic arguments, perhaps that could be done, but not within the core language.
Using trailing return syntax make_array can be further simplified
#include <array>
#include <type_traits>
#include <utility>
template <typename... T>
auto make_array(T&&... t)
-> std::array<std::common_type_t<T...>, sizeof...(t)>
{
return {std::forward<T>(t)...};
}
int main()
{
auto arr = make_array(1, 2, 3, 4, 5);
return 0;
}
Unfortunatelly for aggregate classes it requires explicit type specification
/*
struct Foo
{
int a, b;
}; */
auto arr = make_array(Foo{1, 2}, Foo{3, 4}, Foo{5, 6});
EDIT No longer relevant:
In fact this make_array implementation is listed in sizeof... operator
The code below introduces undefined behavior as per [namespace.std]/4.4
4.4 The behavior of a C++ program is undefined if it declares a deduction guide for any standard library class template.
# c++17 version
Thanks to template argument deduction for class templates proposal we can use deduction guides to get rid of make_array helper
#include <array>
namespace std
{
template <typename... T> array(T... t)
-> array<std::common_type_t<T...>, sizeof...(t)>;
}
int main()
{
std::array a{1, 2, 3, 4};
return 0;
}
Compiled with -std=c++1z flag under x86-64 gcc 7.0
I know it's been quite some time since this question was asked, but I feel the existing answers still have some shortcomings, so I'd like to propose my slightly modified version. Following are the points that I think some existing answers are missing.
1. No need to rely on RVO
Some answers mention that we need to rely on RVO to return the constructed array. That is not true; we can make use of copy-list-initialization to guarantee there will never be temporaries created. So instead of:
return std::array<Type, …>{values};
we should do:
return {{values}};
2. Make make_array a constexpr function
This allow us to create compile-time constant arrays.
3. No need to check that all arguments are of the same type
First off, if they are not, the compiler will issue a warning or error anyway because list-initialization doesn't allow narrowing. Secondly, even if we really decide to do our own static_assert thing (perhaps to provide better error message), we should still probably compare the arguments' decayed types rather than raw types. For example,
volatile int a = 0;
const int& b = 1;
int&& c = 2;
auto arr = make_array<int>(a, b, c); // Will this work?
If we are simply static_asserting that a, b, and c have the same type, then this check will fail, but that probably isn't what we'd expect. Instead, we should compare their std::decay_t<T> types (which are all ints)).
4. Deduce the array value type by decaying the forwarded arguments
This is similar to point 3. Using the same code snippet, but don't specify the value type explicitly this time:
volatile int a = 0;
const int& b = 1;
int&& c = 2;
auto arr = make_array(a, b, c); // Will this work?
We probably want to make an array<int, 3>, but the implementations in the existing answers probably all fail to do that. What we can do is, instead of returning a std::array<T, …>, return a std::array<std::decay_t<T>, …>.
There is one disadvantage about this approach: we can't return an array of cv-qualified value type any more. But most of the time, instead of something like an array<const int, …>, we would use a const array<int, …> anyway. There is a trade-off, but I think a reasonable one. The C++17 std::make_optional also takes this approach:
template< class T >
constexpr std::optional<std::decay_t<T>> make_optional( T&& value );
Taking the above points into account, a full working implementation of make_array in C++14 looks like this:
#include <array>
#include <type_traits>
#include <utility>
template<typename T, typename... Ts>
constexpr std::array<std::decay_t<T>, 1 + sizeof... (Ts)>
make_array(T&& t, Ts&&... ts)
noexcept(noexcept(std::is_nothrow_constructible<
std::array<std::decay_t<T>, 1 + sizeof... (Ts)>, T&&, Ts&&...
>::value))
{
return {{std::forward<T>(t), std::forward<Ts>(ts)...}};
}
template<typename T>
constexpr std::array<std::decay_t<T>, 0> make_array() noexcept
{
return {};
}
Usage:
constexpr auto arr = make_array(make_array(1, 2),
make_array(3, 4));
static_assert(arr[1][1] == 4, "!");
C++11 will support this manner of initialization for (most?) std containers.
(Solution by #dyp)
Note: requires C++14 (std::index_sequence). Although one could implement std::index_sequence in C++11.
#include <iostream>
// ---
#include <array>
#include <utility>
template <typename T>
using c_array = T[];
template<typename T, size_t N, size_t... Indices>
constexpr auto make_array(T (&&src)[N], std::index_sequence<Indices...>) {
return std::array<T, N>{{ std::move(src[Indices])... }};
}
template<typename T, size_t N>
constexpr auto make_array(T (&&src)[N]) {
return make_array(std::move(src), std::make_index_sequence<N>{});
}
// ---
struct Point { int x, y; };
std::ostream& operator<< (std::ostream& os, const Point& p) {
return os << "(" << p.x << "," << p.y << ")";
}
int main() {
auto xs = make_array(c_array<Point>{{1,2}, {3,4}, {5,6}, {7,8}});
for (auto&& x : xs) {
std::cout << x << std::endl;
}
return 0;
}
С++17 compact implementation.
template <typename... T>
constexpr auto array_of(T&&... t) {
return std::array{ static_cast<std::common_type_t<T...>>(t)... };
}
While this answer is directed more towards this question, that question was marked as a duplicate of this question. Hence, this answer is posted here.
A particular use that I feel hasn't been fully covered is a situation where you want to obtain a std::array of chars initialized with a rather long string literal but don't want to blow up the enclosing function. There are a couple of ways to go about this.
The following works but requires us to explicitly specify the size of the string literal. This is what we're trying to avoid:
auto const arr = std::array<char const, 12>{"some string"};
One might expect the following to produce the desired result:
auto const arr = std::array{"some string"};
No need to explicitly specify the size of the array during initialization due to template deduction. However, this wont work because arr is now of type std::array<const char*, 1>.
A neat way to go about this is to simply write a new deduction guide for std::array. But keep in mind that some other code could depend on the default behavior of the std::array deduction guide.
namespace std {
template<typename T, auto N>
array(T (&)[N]) -> array<T, N>;
}
With this deduction guide std::array{"some string"}; will be of type std::array<const char, 12>. It is now possible to initialize arr with a string literal that is defined somewhere else without having to specify its size:
namespace {
constexpr auto some_string = std::array{"some string"};
}
auto func() {
auto const arr = some_string;
// ...
}
Alright, but what if we need a modifiable buffer and we want to initialize it with a string literal without specifying its size?
A hacky solution would be to simply apply the std::remove_cv type trait to our new deduction guide. This is not recommended because this will lead to rather surprising results. String literals are of type const char[], so it's expected that our deduction guide attempts to match that.
It seems that a helper function is necessary in this case. With the use of the constexpr specifier, the following function can be executed at compile time:
#include <array>
#include <type_traits>
template<typename T, auto N>
constexpr auto make_buffer(T (&src)[N]) noexcept {
auto tmp = std::array<std::remove_cv_t<T>, N>{};
for (auto idx = decltype(N){}; idx < N; ++idx) {
tmp[idx] = src[idx];
}
return tmp;
}
Making it possible to initialize modifiable std::array-like buffers as such:
namespace {
constexpr auto some_string = make_buffer("some string");
}
auto func() {
auto buff = some_string;
// ...
}
And with C++20, the helper function can even be simplified:
#include <algorithm>
#include <array>
#include <type_traits>
template<typename T, auto N>
constexpr auto make_buffer(T (&src)[N]) noexcept {
std::array<std::remove_cv_t<T>, N> tmp;
std::copy(std::begin(src), std::end(src), std::begin(tmp));
return tmp;
}
C++20 UPDATE: Although there are some excellent answers that provide the desired functionality (such as Gabriel Garcia's answer that uses std::index_sequence), I am adding this answer because the simplest way to do this as of C++20 isn't mentioned: just use std::to_array(). Using the OP's last example of an array of structs:
struct A{ int a; int b; };
// C syntax
A arr[] = { {1,2}, {3,4} };
// using std::array
auto std_array = std::to_array<A>({ {1,2}, {3,4} });
If std::array is not a constraint and if you have Boost, then take a look at list_of(). This is not exactly like C type array initialization that you want. But close.
Create an array maker type.
It overloads operator, to generate an expression template chaining each element to the previous via references.
Add a finish free function that takes the array maker and generates an array directly from the chain of references.
The syntax should look something like this:
auto arr = finish( make_array<T>->* 1,2,3,4,5 );
It does not permit {} based construction, as only operator= does. If you are willing to use = we can get it to work:
auto arr = finish( make_array<T>= {1}={2}={3}={4}={5} );
or
auto arr = finish( make_array<T>[{1}][{2}[]{3}][{4}][{5}] );
None of these look like good solutions.
Using variardics limits you to your compiler-imposed limit on number of varargs and blocks recursive use of {} for substructures.
In the end, there really isn't a good solution.
What I do is I write my code so it consumes both T[] and std::array data agnostically -- it doesn't care which I feed it. Sometimes this means my forwarding code has to carefully turn [] arrays into std::arrays transparently.
None of the template approaches worked properly for me for arrays of structs, so I crafted this macro solution:
#define make_array(T, ...) \
(std::array<T,sizeof((T[]){ __VA_ARGS__ })/sizeof(T)> {{ __VA_ARGS__ }})
auto a = make_array(int, 1, 2, 3);
struct Foo { int x, y; };
auto b = make_array(Foo,
{ 1, 2 },
{ 3, 4 },
{ 5, 6 },
);
Note that although the macro expands its array arguments twice, the first time is inside sizeof, so any side effects in the expression will correctly happen only once.
Have fun!

How to emulate C array initialization "int arr[] = { e1, e2, e3, ... }" behaviour with std::array?

(Note: This question is about not having to specify the number of elements and still allow nested types to be directly initialized.)
This question discusses the uses left for a C array like int arr[20];. On his answer, #James Kanze shows one of the last strongholds of C arrays, it's unique initialization characteristics:
int arr[] = { 1, 3, 3, 7, 0, 4, 2, 0, 3, 1, 4, 1, 5, 9 };
We don't have to specify the number of elements, hooray! Now iterate over it with the C++11 functions std::begin and std::end from <iterator> (or your own variants) and you never need to even think of its size.
Now, are there any (possibly TMP) ways to achieve the same with std::array? Use of macros allowed to make it look nicer. :)
??? std_array = { "here", "be", "elements" };
Edit: Intermediate version, compiled from various answers, looks like this:
#include <array>
#include <utility>
template<class T, class... Tail, class Elem = typename std::decay<T>::type>
std::array<Elem,1+sizeof...(Tail)> make_array(T&& head, Tail&&... values)
{
return { std::forward<T>(head), std::forward<Tail>(values)... };
}
// in code
auto std_array = make_array(1,2,3,4,5);
And employs all kind of cool C++11 stuff:
Variadic Templates
sizeof...
rvalue references
perfect forwarding
std::array, of course
uniform initialization
omitting the return type with uniform initialization
type inference (auto)
And an example can be found here.
However, as #Johannes points out in the comment on #Xaade's answer, you can't initialize nested types with such a function. Example:
struct A{ int a; int b; };
// C syntax
A arr[] = { {1,2}, {3,4} };
// using std::array
??? std_array = { {1,2}, {3,4} };
Also, the number of initializers is limited to the number of function and template arguments supported by the implementation.
Best I can think of is:
template<class T, class... Tail>
auto make_array(T head, Tail... tail) -> std::array<T, 1 + sizeof...(Tail)>
{
std::array<T, 1 + sizeof...(Tail)> a = { head, tail ... };
return a;
}
auto a = make_array(1, 2, 3);
However, this requires the compiler to do NRVO, and then also skip the copy of returned value (which is also legal but not required). In practice, I would expect any C++ compiler to be able to optimize that such that it's as fast as direct initialization.
I'd expect a simple make_array.
template<typename ret, typename... T> std::array<ret, sizeof...(T)> make_array(T&&... refs) {
// return std::array<ret, sizeof...(T)>{ { std::forward<T>(refs)... } };
return { std::forward<T>(refs)... };
}
Combining a few ideas from previous posts, here's a solution that works even for nested constructions (tested in GCC4.6):
template <typename T, typename ...Args>
std::array<T, sizeof...(Args) + 1> make_array(T && t, Args &&... args)
{
static_assert(all_same<T, Args...>::value, "make_array() requires all arguments to be of the same type."); // edited in
return std::array<T, sizeof...(Args) + 1>{ std::forward<T>(t), std::forward<Args>(args)...};
}
Strangely, can cannot make the return value an rvalue reference, that would not work for nested constructions. Anyway, here's a test:
auto q = make_array(make_array(make_array(std::string("Cat1"), std::string("Dog1")), make_array(std::string("Mouse1"), std::string("Rat1"))),
make_array(make_array(std::string("Cat2"), std::string("Dog2")), make_array(std::string("Mouse2"), std::string("Rat2"))),
make_array(make_array(std::string("Cat3"), std::string("Dog3")), make_array(std::string("Mouse3"), std::string("Rat3"))),
make_array(make_array(std::string("Cat4"), std::string("Dog4")), make_array(std::string("Mouse4"), std::string("Rat4")))
);
std::cout << q << std::endl;
// produces: [[[Cat1, Dog1], [Mouse1, Rat1]], [[Cat2, Dog2], [Mouse2, Rat2]], [[Cat3, Dog3], [Mouse3, Rat3]], [[Cat4, Dog4], [Mouse4, Rat4]]]
(For the last output I'm using my pretty-printer.)
Actually, let us improve the type safety of this construction. We definitely need all types to be the same. One way is to add a static assertion, which I've edited in above. The other way is to only enable make_array when the types are the same, like so:
template <typename T, typename ...Args>
typename std::enable_if<all_same<T, Args...>::value, std::array<T, sizeof...(Args) + 1>>::type
make_array(T && t, Args &&... args)
{
return std::array<T, sizeof...(Args) + 1> { std::forward<T>(t), std::forward<Args>(args)...};
}
Either way, you will need the variadic all_same<Args...> type trait. Here it is, generalizing from std::is_same<S, T> (note that decaying is important to allow mixing of T, T&, T const & etc.):
template <typename ...Args> struct all_same { static const bool value = false; };
template <typename S, typename T, typename ...Args> struct all_same<S, T, Args...>
{
static const bool value = std::is_same<typename std::decay<S>::type, typename std::decay<T>::type>::value && all_same<T, Args...>::value;
};
template <typename S, typename T> struct all_same<S, T>
{
static const bool value = std::is_same<typename std::decay<S>::type, typename std::decay<T>::type>::value;
};
template <typename T> struct all_same<T> { static const bool value = true; };
Note that make_array() returns by copy-of-temporary, which the compiler (with sufficient optimisation flags!) is allowed to treat as an rvalue or otherwise optimize away, and std::array is an aggregate type, so the compiler is free to pick the best possible construction method.
Finally, note that you cannot avoid copy/move construction when make_array sets up the initializer. So std::array<Foo,2> x{Foo(1), Foo(2)}; has no copy/move, but auto x = make_array(Foo(1), Foo(2)); has two copy/moves as the arguments are forwarded to make_array. I don't think you can improve on that, because you can't pass a variadic initializer list lexically to the helper and deduce type and size -- if the preprocessor had a sizeof... function for variadic arguments, perhaps that could be done, but not within the core language.
Using trailing return syntax make_array can be further simplified
#include <array>
#include <type_traits>
#include <utility>
template <typename... T>
auto make_array(T&&... t)
-> std::array<std::common_type_t<T...>, sizeof...(t)>
{
return {std::forward<T>(t)...};
}
int main()
{
auto arr = make_array(1, 2, 3, 4, 5);
return 0;
}
Unfortunatelly for aggregate classes it requires explicit type specification
/*
struct Foo
{
int a, b;
}; */
auto arr = make_array(Foo{1, 2}, Foo{3, 4}, Foo{5, 6});
EDIT No longer relevant:
In fact this make_array implementation is listed in sizeof... operator
The code below introduces undefined behavior as per [namespace.std]/4.4
4.4 The behavior of a C++ program is undefined if it declares a deduction guide for any standard library class template.
# c++17 version
Thanks to template argument deduction for class templates proposal we can use deduction guides to get rid of make_array helper
#include <array>
namespace std
{
template <typename... T> array(T... t)
-> array<std::common_type_t<T...>, sizeof...(t)>;
}
int main()
{
std::array a{1, 2, 3, 4};
return 0;
}
Compiled with -std=c++1z flag under x86-64 gcc 7.0
I know it's been quite some time since this question was asked, but I feel the existing answers still have some shortcomings, so I'd like to propose my slightly modified version. Following are the points that I think some existing answers are missing.
1. No need to rely on RVO
Some answers mention that we need to rely on RVO to return the constructed array. That is not true; we can make use of copy-list-initialization to guarantee there will never be temporaries created. So instead of:
return std::array<Type, …>{values};
we should do:
return {{values}};
2. Make make_array a constexpr function
This allow us to create compile-time constant arrays.
3. No need to check that all arguments are of the same type
First off, if they are not, the compiler will issue a warning or error anyway because list-initialization doesn't allow narrowing. Secondly, even if we really decide to do our own static_assert thing (perhaps to provide better error message), we should still probably compare the arguments' decayed types rather than raw types. For example,
volatile int a = 0;
const int& b = 1;
int&& c = 2;
auto arr = make_array<int>(a, b, c); // Will this work?
If we are simply static_asserting that a, b, and c have the same type, then this check will fail, but that probably isn't what we'd expect. Instead, we should compare their std::decay_t<T> types (which are all ints)).
4. Deduce the array value type by decaying the forwarded arguments
This is similar to point 3. Using the same code snippet, but don't specify the value type explicitly this time:
volatile int a = 0;
const int& b = 1;
int&& c = 2;
auto arr = make_array(a, b, c); // Will this work?
We probably want to make an array<int, 3>, but the implementations in the existing answers probably all fail to do that. What we can do is, instead of returning a std::array<T, …>, return a std::array<std::decay_t<T>, …>.
There is one disadvantage about this approach: we can't return an array of cv-qualified value type any more. But most of the time, instead of something like an array<const int, …>, we would use a const array<int, …> anyway. There is a trade-off, but I think a reasonable one. The C++17 std::make_optional also takes this approach:
template< class T >
constexpr std::optional<std::decay_t<T>> make_optional( T&& value );
Taking the above points into account, a full working implementation of make_array in C++14 looks like this:
#include <array>
#include <type_traits>
#include <utility>
template<typename T, typename... Ts>
constexpr std::array<std::decay_t<T>, 1 + sizeof... (Ts)>
make_array(T&& t, Ts&&... ts)
noexcept(noexcept(std::is_nothrow_constructible<
std::array<std::decay_t<T>, 1 + sizeof... (Ts)>, T&&, Ts&&...
>::value))
{
return {{std::forward<T>(t), std::forward<Ts>(ts)...}};
}
template<typename T>
constexpr std::array<std::decay_t<T>, 0> make_array() noexcept
{
return {};
}
Usage:
constexpr auto arr = make_array(make_array(1, 2),
make_array(3, 4));
static_assert(arr[1][1] == 4, "!");
C++11 will support this manner of initialization for (most?) std containers.
(Solution by #dyp)
Note: requires C++14 (std::index_sequence). Although one could implement std::index_sequence in C++11.
#include <iostream>
// ---
#include <array>
#include <utility>
template <typename T>
using c_array = T[];
template<typename T, size_t N, size_t... Indices>
constexpr auto make_array(T (&&src)[N], std::index_sequence<Indices...>) {
return std::array<T, N>{{ std::move(src[Indices])... }};
}
template<typename T, size_t N>
constexpr auto make_array(T (&&src)[N]) {
return make_array(std::move(src), std::make_index_sequence<N>{});
}
// ---
struct Point { int x, y; };
std::ostream& operator<< (std::ostream& os, const Point& p) {
return os << "(" << p.x << "," << p.y << ")";
}
int main() {
auto xs = make_array(c_array<Point>{{1,2}, {3,4}, {5,6}, {7,8}});
for (auto&& x : xs) {
std::cout << x << std::endl;
}
return 0;
}
С++17 compact implementation.
template <typename... T>
constexpr auto array_of(T&&... t) {
return std::array{ static_cast<std::common_type_t<T...>>(t)... };
}
While this answer is directed more towards this question, that question was marked as a duplicate of this question. Hence, this answer is posted here.
A particular use that I feel hasn't been fully covered is a situation where you want to obtain a std::array of chars initialized with a rather long string literal but don't want to blow up the enclosing function. There are a couple of ways to go about this.
The following works but requires us to explicitly specify the size of the string literal. This is what we're trying to avoid:
auto const arr = std::array<char const, 12>{"some string"};
One might expect the following to produce the desired result:
auto const arr = std::array{"some string"};
No need to explicitly specify the size of the array during initialization due to template deduction. However, this wont work because arr is now of type std::array<const char*, 1>.
A neat way to go about this is to simply write a new deduction guide for std::array. But keep in mind that some other code could depend on the default behavior of the std::array deduction guide.
namespace std {
template<typename T, auto N>
array(T (&)[N]) -> array<T, N>;
}
With this deduction guide std::array{"some string"}; will be of type std::array<const char, 12>. It is now possible to initialize arr with a string literal that is defined somewhere else without having to specify its size:
namespace {
constexpr auto some_string = std::array{"some string"};
}
auto func() {
auto const arr = some_string;
// ...
}
Alright, but what if we need a modifiable buffer and we want to initialize it with a string literal without specifying its size?
A hacky solution would be to simply apply the std::remove_cv type trait to our new deduction guide. This is not recommended because this will lead to rather surprising results. String literals are of type const char[], so it's expected that our deduction guide attempts to match that.
It seems that a helper function is necessary in this case. With the use of the constexpr specifier, the following function can be executed at compile time:
#include <array>
#include <type_traits>
template<typename T, auto N>
constexpr auto make_buffer(T (&src)[N]) noexcept {
auto tmp = std::array<std::remove_cv_t<T>, N>{};
for (auto idx = decltype(N){}; idx < N; ++idx) {
tmp[idx] = src[idx];
}
return tmp;
}
Making it possible to initialize modifiable std::array-like buffers as such:
namespace {
constexpr auto some_string = make_buffer("some string");
}
auto func() {
auto buff = some_string;
// ...
}
And with C++20, the helper function can even be simplified:
#include <algorithm>
#include <array>
#include <type_traits>
template<typename T, auto N>
constexpr auto make_buffer(T (&src)[N]) noexcept {
std::array<std::remove_cv_t<T>, N> tmp;
std::copy(std::begin(src), std::end(src), std::begin(tmp));
return tmp;
}
C++20 UPDATE: Although there are some excellent answers that provide the desired functionality (such as Gabriel Garcia's answer that uses std::index_sequence), I am adding this answer because the simplest way to do this as of C++20 isn't mentioned: just use std::to_array(). Using the OP's last example of an array of structs:
struct A{ int a; int b; };
// C syntax
A arr[] = { {1,2}, {3,4} };
// using std::array
auto std_array = std::to_array<A>({ {1,2}, {3,4} });
If std::array is not a constraint and if you have Boost, then take a look at list_of(). This is not exactly like C type array initialization that you want. But close.
Create an array maker type.
It overloads operator, to generate an expression template chaining each element to the previous via references.
Add a finish free function that takes the array maker and generates an array directly from the chain of references.
The syntax should look something like this:
auto arr = finish( make_array<T>->* 1,2,3,4,5 );
It does not permit {} based construction, as only operator= does. If you are willing to use = we can get it to work:
auto arr = finish( make_array<T>= {1}={2}={3}={4}={5} );
or
auto arr = finish( make_array<T>[{1}][{2}[]{3}][{4}][{5}] );
None of these look like good solutions.
Using variardics limits you to your compiler-imposed limit on number of varargs and blocks recursive use of {} for substructures.
In the end, there really isn't a good solution.
What I do is I write my code so it consumes both T[] and std::array data agnostically -- it doesn't care which I feed it. Sometimes this means my forwarding code has to carefully turn [] arrays into std::arrays transparently.
None of the template approaches worked properly for me for arrays of structs, so I crafted this macro solution:
#define make_array(T, ...) \
(std::array<T,sizeof((T[]){ __VA_ARGS__ })/sizeof(T)> {{ __VA_ARGS__ }})
auto a = make_array(int, 1, 2, 3);
struct Foo { int x, y; };
auto b = make_array(Foo,
{ 1, 2 },
{ 3, 4 },
{ 5, 6 },
);
Note that although the macro expands its array arguments twice, the first time is inside sizeof, so any side effects in the expression will correctly happen only once.
Have fun!