Resizing a C++ std::vector<char> without initializing data [duplicate] - c++

Question

This question already has answers here:
Using vector<char> as a buffer without initializing it on resize()
(6 answers)
Closed 6 years ago.
With vectors, one can assume that elements are stored contiguously in memory, allowing the range [&vec[0], &vec[vec.capacity()) to be used as a normal array. E.g.,
vector<char> buf;
buf.reserve(N);
int M = read(fd, &buf[0], N);
But now the vector doesn't know that it contains M bytes of data, added externally by read(). I know that vector::resize() sets the size, but it also clears the data, so it can't be used to update the size after the read() call.
Is there a trivial way to read data directly into vectors and update the size after? Yes, I know of the obvious workarounds like using a small array as a temporary read buffer, and using vector::insert() to append that to the end of the vector:
char tmp[N];
int M = read(fd, tmp, N);
buf.insert(buf.end(), tmp, tmp + M)
This works (and it's what I'm doing today), but it just bothers me that there is an extra copy operation there that would not be required if I could put the data directly into the vector.
So, is there a simple way to modify the vector size when data has been added externally?

Answer 1

vector<char> buf;
buf.reserve(N);
int M = read(fd, &buf[0], N);
This code fragment invokes undefined behavior. You can't write beyond than size() elements, even if you have reserved the space.
The correct code is like:
vector<char> buf;
buf.resize(N);
int M = read(fd, &buf[0], N);
buf.resize(M);
PS. Your statement "With vectors, one can assume that elements are stored contiguously in memory, allowing the range [&vec[0], &vec[vec.capacity()) to be used as a normal array" isn't true. The allowable range is [&vec[0], &vec[vec.size()).

Answer 2

It looks like you can do what you want in C++11 (though I haven't tried this myself). You'll have to define a custom allocator for the vector, then use emplace_back().
First, define
struct do_not_initialize_tag {};
Then define your allocator with this member function:
class my_allocator {
void construct(char* c, do_not_initialize_tag) const {
// do nothing
}
// details omitted
// ...
}
Now you can add elements to your array without initializing them:
std::vector<char, my_allocator> buf;
buf.reserve(N);
for (int i = 0; i != N; ++i)
buf.emplace_back(do_not_initialize_tag());
int M = read(fd, buf.data(), N);
buf.resize(M);
The efficiency of this depends on the compiler's optimizer. For instance, the loop may increment the size member variable N times.

Answer 3

Another, newer, question, a duplicate of this one, has an answer, which looks like exactly what is asked here. Here's its copy (of v3) for quick reference:
It is a known issue that initialization can not be turned off even
explicitly for std::vector.
People normally implement their own pod_vector<> that does not do
any initialization of the elements.
Another way is to create a type which is layout-compatible with char,
whose constructor does nothing:
struct NoInitChar
{
char value;
NoInitChar() {
// do nothing
static_assert(sizeof *this == sizeof value, "invalid size");
static_assert(__alignof *this == __alignof value, "invalid alignment");
}
};
int main() {
std::vector<NoInitChar> v;
v.resize(10); // calls NoInitChar() which does not initialize
// Look ma, no reinterpret_cast<>!
char* beg = &v.front().value;
char* end = beg + v.size();
}

Answer 4

Writing into and after the size()th element is an undefined behavior.
Next example copies whole file into a vector in a c++ way (no need to know the file's size and no need to preallocate the memory in the vector):
#include <algorithm>
#include <fstream>
#include <iterator>
#include <vector>
int main()
{
typedef std::istream_iterator<char> istream_iterator;
std::ifstream file("example.txt");
std::vector<char> input;
file >> std::noskipws;
std::copy( istream_iterator(file),
istream_iterator(),
std::back_inserter(input));
}

Answer 5

Your program fragment has entered the realm of undefined behavior.
when buf.empty() is true, buf[0] has undefined behavior, and therefore &buf[0] is also undefined.
This fragment probably does what you want.
vector<char> buf;
buf.resize(N); // preallocate space
int M = read(fd, &buf[0], N);
buf.resize(M); // disallow access to the remainder