C++ – How to declare array with auto

autocc++11

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

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

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

Best Answer

TL;DR

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);

Setting a bit

Use the bitwise OR operator (|) to set a bit.

number |= 1UL << n;

That will set the nth bit of number. n should be zero, if you want to set the 1st bit and so on upto n-1, if you want to set the nth bit.

Use 1ULL if number is wider than unsigned long; promotion of 1UL << n doesn't happen until after evaluating 1UL << n where it's undefined behaviour to shift by more than the width of a long. The same applies to all the rest of the examples.

Clearing a bit

Use the bitwise AND operator (&) to clear a bit.

number &= ~(1UL << n);

That will clear the nth bit of number. You must invert the bit string with the bitwise NOT operator (~), then AND it.

Toggling a bit

The XOR operator (^) can be used to toggle a bit.

number ^= 1UL << n;

That will toggle the nth bit of number.

Checking a bit

You didn't ask for this, but I might as well add it.

To check a bit, shift the number n to the right, then bitwise AND it:

bit = (number >> n) & 1U;

That will put the value of the nth bit of number into the variable bit.

Changing the nth bit to x

Setting the nth bit to either 1 or 0 can be achieved with the following on a 2's complement C++ implementation:

number ^= (-x ^ number) & (1UL << n);

Bit n will be set if x is 1, and cleared if x is 0. If x has some other value, you get garbage. x = !!x will booleanize it to 0 or 1.

To make this independent of 2's complement negation behaviour (where -1 has all bits set, unlike on a 1's complement or sign/magnitude C++ implementation), use unsigned negation.

number ^= (-(unsigned long)x ^ number) & (1UL << n);

unsigned long newbit = !!x;    // Also booleanize to force 0 or 1
number ^= (-newbit ^ number) & (1UL << n);

It's generally a good idea to use unsigned types for portable bit manipulation.

number = (number & ~(1UL << n)) | (x << n);

(number & ~(1UL << n)) will clear the nth bit and (x << n) will set the nth bit to x.

It's also generally a good idea to not to copy/paste code in general and so many people use preprocessor macros (like the community wiki answer further down) or some sort of encapsulation.

C++ – How to iterate over the words of a string

I use this to split string by a delimiter. The first puts the results in a pre-constructed vector, the second returns a new vector.

#include <string>
#include <sstream>
#include <vector>
#include <iterator>

template <typename Out>
void split(const std::string &s, char delim, Out result) {
    std::istringstream iss(s);
    std::string item;
    while (std::getline(iss, item, delim)) {
        *result++ = item;
    }
}

std::vector<std::string> split(const std::string &s, char delim) {
    std::vector<std::string> elems;
    split(s, delim, std::back_inserter(elems));
    return elems;
}

Note that this solution does not skip empty tokens, so the following will find 4 items, one of which is empty:

std::vector<std::string> x = split("one:two::three", ':');