Why does c++ initialise a std::vector with zeros, but not a std::array?

Question

Isn't it a waste of time to initialize a vector with zeros, when you don't want it?

I try this code:

#include <iostream>
#include <vector>
#include <array>

#define SIZE 10

int main()
{
#ifdef VECTOR

  std::vector<unsigned> arr(SIZE);

#else

  std::array<unsigned, SIZE> arr;

#endif // VECTOR

  for (unsigned n : arr)
    printf("%i ", n);
  printf("\n");

  return 0;
}

and I get the output:

with vector

$ g++ -std=c++11 -D VECTOR test.cpp -o test && ./test 
0 0 0 0 0 0 0 0 0 0 

with an array

g++ -std=c++11  test.cpp -o test && ./test 
-129655920 32766 4196167 0 2 0 4196349 0 1136 0 

And I also try with clang++

So why zeros? And by the way, could I declare a vector without initializing it?

Answer 1

The more common way to declare a vector is without specifying the size:

std::vector<unsigned> arr;

This doesn't allocate any space for the vector contents, and doesn't have any initialization overhead. Elements are usually added dynamically with methods like .push_back() . If you want to allocate memory you can use reserve() :

arr.reserve(SIZE);

This doesn't initialize the added elements, they're not included in the size() of the vector, and trying to read them is undefined behavior. Compare this with

arr.resize(SIZE);

which grows the vector and initializes all the new elements.

std::array , on the other hand, always allocates the memory. It implements most of the same behaviors as C-style arrays, except for the automatic decay to a pointer. This includes not initializing the elements by default.

Answer 2

The default allocator is doing the zero-initialization. You can use a different allocator that does not do that. I wrote an allocator that uses default construction rather than initialization when feasible. More precisely, it is an allocator-wrapper called ctor_allocator . Then I define a vector template.

dj:vector<unsigned> vec(10); does exactly what you want. It's an std::vector<unsigned> (10) that is not initialized to zeros.

--- libdj/vector.h ----
#include <libdj/allocator.h>
#include <vector>

namespace dj {
template<class T>
    using vector = std::vector<T, dj::ctor_allocator<T>>;
}

--- libdj/allocator.h  ----
#include <memory>

namespace dj {

template <typename T, typename A = std::allocator<T>>
    class ctor_allocator : public A 
    {
        using a_t = std::allocator_traits<A>;
    public:
        using A::A; // Inherit constructors from A

        template <typename U> struct rebind 
        {
            using other =
                ctor_allocator
                <  U, typename a_t::template rebind_alloc<U>  >;
        };

        template <typename U>
        void construct(U* ptr)
            noexcept(std::is_nothrow_default_constructible<U>::value) 
        {
            ::new(static_cast<void*>(ptr)) U;
        }

        template <typename U, typename...Args>
        void construct(U* ptr, Args&&... args) 
        {
            a_t::construct(static_cast<A&>(*this),
                ptr, std::forward<Args>(args)...);
        }
    };
}

Answer 3

Suppose we have some class:

class MyClass {
    int value;

public:
    MyClass() {
        value = 42;
    }
    // other code
};

std::vector<MyClass> arr(10); will default construct 10 copies of MyClass , all with value = 42 .

But suppose it didn't default construct the 10 copies. Now if I wrote arr[0].some_function() , there's a problem: MyClass 's constructor has not yet run, so the invariants of the class aren't set up. I might have assumed in the implementation of some_function() that value == 42 , but since the constructor hasn't run, value has some indeterminate value. This would be a bug.

That's why in C++, there's a concept of object lifetimes . The object doesn't exist before the constructor is called, and it ceases to exist after the destructor is called. std::vector<MyClass> arr(10); calls the default constructor on every element so that all the objects exist.

It's important to note that std::array is somewhat special, since it is initialized following the rules of aggregate initialization . This means that std::array<MyClass, 10> arr; also default constructs 10 copies of MyClass all with value = 42 . However, for non-class types such as unsigned , the values will be indeterminate.

---

There is a way to avoid calling all the default constructors: std::vector::reserve . If I were to write:

std::vector<MyClass> arr;
arr.reserve(10);

The vector would allocate its backing array to hold 10 MyClass s, and it won't call the default constructors. But now I can't write arr[0] or arr[5] ; those would be out-of-bounds access into arr ( arr.size() is still 0, even though the backing array has more elements). To initialize the values, I'd have to call push_back or emplace_back :

arr.push_back(MyClass{});

This is often the right approach. For example, if I wanted to fill arr with random values from std::rand , I can use std::generate_n along with std::back_inserter :

std::vector<unsigned> arr;
arr.reserve(10);
std::generate_n(std::back_inserter(arr), 10, std::rand);

It's also worth noting that if I already have the values I want for arr in a container, I can just pass the begin()/end() in with the constructor:

std::vector<unsigned> arr{values.begin(), values.end()};