将字符串数据附加到std :: vector <std::byte> &gt;

Question

I'm implementing a HTTP server and the API I follow define the raw response data as a std::vector<std::byte>> .

I store http responses headers as std::string in my code and at some point I have to write them to to raw response data before sending it back.

The thing is, I cannot find a clean way to write/append data from a std::string to my std::vector<std::byte>> (by clean way I mean not looping on the string and appending each char).

What is the best way to do that ?

Side question: What is the best way to read a string from a std::vector<std::byte>> ?

Answer 1

The char do cannot be converted to std::byte c++17 . Its defined as scoped enum:

enum class byte : unsigned char {} ;

cppreference.com std::byte

A numeric value n can be converted to a byte value using std::byte{n} , due to C++17 relaxed enum class initialization rules.

What you can do is use a helper function or lambda:

std::string headers;
std::vector<std::byte> response;

response.reserve(response.size() + headers.size());  // Optional
std::transform(headers.begin(), headers.end(), std::back_inserter(response),
    [](unsigned char c) { return std::byte{c}; }              // or `encode(c)`
);  

You can also resize the response and skip the back_inserter :

const auto response_size = response.size();
response.resize(response_size + headers.size());
std::transform(headers.begin(), headers.end(), std::next(response.begin(), response_size),
    [](unsigned char c) { return std::byte{c}; }
);  

Actually the whole will be optimized by the compiler to something similar to std::copy .

Or just replace std::byte with a char and use std::vector::insert() or std::copy() .

Answer 2

Just use the ranged- insert overload ( #4 ):

void extend(std::vector<std::byte>& v, std::string const& s) {
    auto bytes = reinterpret_cast<std::byte const*>(s.data());
    v.insert(v.end(), bytes, bytes + s.size());
}

You can read char as byte , it's a permitted alias.

Answer 3

First start with a gsl::span or similar.

template<class T>
struct span {
  T* b =0, *e = 0;
  T* begin() const { return b; }
  T* end() const { return e; }
  std::size_t size() const { return end()-begin(); }
  bool empty() const { return end()==begin(); }
  span( T* s, T* f ):b(s),e(f) {}
  span( T* s, std::size_t len ):span(s, s+len) {}

  template<class Uptr>
  using is_compatible = std::is_convertible< Uptr*, T* >;

  template<class R,
    std::enable_if_t<!std::is_same<std::decay_t<R>, span>{}, bool> = true,
    std::enable_if_t<is_compatible<decltype(std::declval<R&>().data())>{}, bool> = true
  >
  span( R&& r ):
    span(r.data(), r.size())
  {}
  template<class U, std::size_t N,
    std::enable_if_t<is_compatible<U*>{}, bool> = true
  >
  span( U(&arr)[N] ):span(arr, N) {}
};

now we have an abstraction for "possibly mutable view into contiguous T s".

std::vector<std::byte> concat_string( std::vector<std::byte> lhs, span<char const> rhs ) {
  lhs.reserve(lhs.size()+rhs.size());
  lhs.insert( lhs.end(), (std::byte const*)rhs.begin(), (std::byte const*)rhs.end() );
  return rhs;
}

this assumes you don't want to embed the '\\0' .

Answer 4

you'll want something like the following. I realize this still uses copy but there is only the one memory allocation which is the expensive part.

std::vector<std::byte> data;
std::string input;
...
data.reserve(data.size() + input.size());
std::copy(input.begin(), input.end(), std::back_inserter(data));