C++ vector that *doesn't* initialize its members?

Question

I'm making a C++ wrapper for a piece of C code that returns a large array, and so I've tried to return the data in a vector<unsigned char>.

Now the problem is, the data is on the order of megabytes, and vector unnecessarily initializes its storage, which essentially turns out to cut down my speed by half.

How do I prevent this?

Or, if it's not possible -- is there some other STL container that would avoid such needless work? Or must I end up making my own container?

(Pre-C++11)

Note:

I'm passing the vector as my output buffer. I'm not copying the data from elsewhere.
It's something like:

vector<unsigned char> buf(size);   // Why initialize??
GetMyDataFromC(&buf[0], buf.size());

Answer 1

For default and value initialization of structs with user-provided default constructors which don't explicitly initialize anything, no initialization is performed on unsigned char members:

struct uninitialized_char {
    unsigned char m;
    uninitialized_char() {}
};

// just to be safe
static_assert(1 == sizeof(uninitialized_char), "");

std::vector<uninitialized_char> v(4 * (1<<20));

GetMyDataFromC(reinterpret_cast<unsigned char*>(&v[0]), v.size());

I think this is even legal under the strict aliasing rules.

When I compared the construction time for v vs. a vector<unsigned char> I got ~8 µs vs ~12 ms. More than 1000x faster. Compiler was clang 3.2 with libc++ and flags: -std=c++11 -Os -fcatch-undefined-behavior -ftrapv -pedantic -Weverything -Wno-c++98-compat -Wno-c++98-compat-pedantic -Wno-missing-prototypes

C++11 has a helper for uninitialized storage, std::aligned_storage. Though it requires a compile time size.

---

Here's an added example, to compare total usage (times in nanoseconds):

VERSION=1 (vector<unsigned char>):

clang++ -std=c++14 -stdlib=libc++ main.cpp -DVERSION=1 -ftrapv -Weverything -Wno-c++98-compat -Wno-sign-conversion -Wno-sign-compare -Os && ./a.out

initialization+first use: 16,425,554
array initialization: 12,228,039
first use: 4,197,515
second use: 4,404,043

VERSION=2 (vector<uninitialized_char>):

clang++ -std=c++14 -stdlib=libc++ main.cpp -DVERSION=2 -ftrapv -Weverything -Wno-c++98-compat -Wno-sign-conversion -Wno-sign-compare -Os && ./a.out

initialization+first use: 7,523,216
array initialization: 12,782
first use: 7,510,434
second use: 4,155,241

#include <iostream>
#include <chrono>
#include <vector>

struct uninitialized_char {
  unsigned char c;
  uninitialized_char() {}
};

void foo(unsigned char *c, int size) {
  for (int i = 0; i < size; ++i) {
    c[i] = '\0';
  }
}

int main() {
  auto start = std::chrono::steady_clock::now();

#if VERSION==1
  using element_type = unsigned char;
#elif VERSION==2
  using element_type = uninitialized_char;
#endif

  std::vector<element_type> v(4 * (1<<20));

  auto end = std::chrono::steady_clock::now();

  foo(reinterpret_cast<unsigned char*>(v.data()), v.size());

  auto end2 = std::chrono::steady_clock::now();

  foo(reinterpret_cast<unsigned char*>(v.data()), v.size());

  auto end3 = std::chrono::steady_clock::now();

  std::cout.imbue(std::locale(""));
  std::cout << "initialization+first use: " << std::chrono::nanoseconds(end2-start).count() << '\n';
  std::cout << "array initialization: " << std::chrono::nanoseconds(end-start).count() << '\n';
  std::cout << "first use: " << std::chrono::nanoseconds(end2-end).count() << '\n';
  std::cout << "second use: " << std::chrono::nanoseconds(end3-end2).count() << '\n';
}

I'm using clang svn-3.6.0 r218006

Answer 2

Sorry, there's no way to avoid it.

C++11 adds a constructor that takes only a size, but even that will value-initialize the data.

Your best bet is to just allocate an array on the heap, stick it in a unique_ptr (where available), and use it from there.

If you're willing to, as you say, "hacking into STL," you could always grab a copy of EASTL to work from. It's a variation of certain STL containers that allows for more restricted memory conditions. A proper implementation of what you're trying to do would be to give its constructor a special value that means "default initialize the members," which for POD types means to do nothing to initialize the memory. This requires using some template metaprogramming to detect if it is a POD type, of course.

Answer 3

The optimal solution is to simply change the allocator to do nothing for a zero-arguments construct. This means that the underlying type is the same, which dodges any kind of nasty reinterpret_casting and potential aliasing violations and can non-intrusively uninitialize any type.

template<typename T> struct no_initialize : std::allocator<T> {
    void construct(T* p) {}
    template<typename... Args> void construct(T* p, Args&&... args) {
        new (p) T(std::forward<Args>(args)...);
    }
};

Answer 4

1 It seems that using std::vector is neither necessary not sensible in your situation. You only want some object to manage some raw memory for you. This can be easily achieved by

std::unique_ptr<void, void(*)(void*)> p(std::malloc(n), std::free);

2 If you really want to use std::vector<> you could use the trick described here.

Answer 5

How about using vector.reserve() to only allocate the storage but not initialize it?