Graznarak Graznarak - 1 year ago 114
C++ Question

How to declare array with auto

I have been playing with

and I noticed that for most cases you can replace a variable definition with
and then assign the type.

In the following code
are equivalent (default initialized
, but lets not get into potential copies). Is there a way to declare
such that it has the same type as

int w{};
auto x = int{};
int y[5];
auto z = int[5];

Answer Source


template<typename T, int N> using raw_array = T[N];

auto &&z = raw_array<int,5>{};

Your example of auto z = int[5]; isn't legal any more than auto z = int; is, simply because a type is not a valid initializer. You can write: auto z = int{}; because int{} is a valid initializer.

Once one realizes this, the next attempt would be:

auto z = int[5]{};

Note that your int y[5] does not have any initializer. If it had then you would have jumped straight here.

Unfortunately this does not work either for obscure syntax reasons. Instead you must find a legal way to name the array type in an initializer. For example, a typedef name can be used in an initializer. A handy reusable template type alias eliminates the burdensome requirement of a new typedef for every array type:

template<typename T, int N> using raw_array = T[N];

auto z = raw_array<int,5>{};

Aside: You can use template type aliases to fix the weird 'inside-out' syntax of C++, allowing you to name any compound type in an orderly, left-to-right fashion, by using this proposal.

Unfortunately due to the design bug in C and C++ which causes array-to-pointer conversions at the drop of a hat, the deduced type of the variable z is int* rather int[5]. The resulting variable becomes a dangling pointer when the temporary array is destroyed.

C++14 introduces decltype(auto) which uses different type deduction rules, correctly deducing an array type:

decltype(auto) z = raw_array<int,5>{};

But now we run into another design bug with arrays; they do not behave as proper objects. You can't assign, copy construct, do pass by value, etc., with arrays. The above code is like saying:

int g[5] = {};
int h[5] = g;

By all rights this should work, but unfortunately built-in arrays behave bizarrely in C and C++. In our case, the specific problem is that arrays are not allowed to have just any kind of initializer; they are strictly limited to using initializer lists. An array temporary, initialized by an initializer list, is not itself an initializer list.

Answer 1:

At this point Johannes Schaub makes the excellent suggestion that we can use temporary lifetime extension.

auto &&z = raw_array<int,5>{};

decltype(auto) isn't needed because the addition of && changes the deduced type, so Johannes Schaub's suggestion works in C++11. This also avoids the limitation on array initializers because we're initializing a reference instead of an array.

If you want the array to deduce its length from an initializer, you can use an incomplete array type:

template<typename T> using unsized_raw_array = T[];

auto &&z = unsized_raw_array<int>{1, 2, 3};

Although the above does what you want you may prefer to avoid raw arrays entirely, due to the fact that raw arrays do not behave like proper C++ objects, and the obscurity of their behavior and the techniques used above.

Answer 2:

The std::array template in C++11 does act like a proper object, including assignment, being passable by value, etc., and just generally behaving sanely and consistently where built-in arrays do not.

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

However, with this you miss out on being able to have the array type infer its own length from an initializer. Instead You can write a make_array template function that does the inference. Here's a really simple version I haven't tested and which doesn't do things you might want, such as verify that all the arguments are the same type, or let you explicitly specify the type.

template<typename... T>
std::array<typename std::common_type<T...>::type, sizeof...(T)>
make_array(T &&...t) {
    return {std::forward<T>(t)...};

auto z = make_array(1,2,3,4,5);
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