10

I have a struct representing a binary message. I want to write a function to get the next such record from a buffer (whether a file or a socket, doesn't matter):

template <typename Record>
Record getNext();

Now, I could write this like:

template <typename Record>
Record getNext() {
    Record r;
    populateNext(reinterpret_cast<char*>(&r),  // maybe ::read()
                 sizeof(r));                   // or equivalent
    return r;
}

which is nice and gives me the benefits of RVO. However, it will invoke the default constructor of Record, which may be composed of types with non-trival default constructors which do work that I would like to avoid - these are not necessarily POD types, but they are standard layout.

Is there a way to write getNext() such that we avoid any constructors (default or copy/move) on Record? Ideally, when the user calls:

auto record = getNext<Record>();

The buffer is read directly into the memory of record. Is this possible?

4

2 回答 2

5

no_init是一个类型的常数no_init_t。

如果你从 a 构造一个 pod no_init_t，你会得到一个未初始化的 pod，并且（假设省略）没有什么可做的。

如果从 a 构造非 pod no_init_t，则必须覆盖构造函数，并使其不初始化数据。通常class_name(no_init_t):field1(no_init), field2(no_init){}会这样做，有时class_name(no_init_t){}会这样做（假设所有内容都是 pod）。

然而，从每个成员开始构建no_init可以作为成员确实是 pod 的健全性检查。在您编写构造函数之前，构造自的非 pod 类no_init将无法编译no_init_t。

这（必须对no_init每个成员构造函数）确实会产生一些烦人的 DRY 失败，但我们没有得到反思，所以你会重复自己并喜欢它。

namespace {
  struct no_init_t {
    template<class T, class=std::enable_if_t<std::is_pod<T>{}>>
    operator T()const{
      T tmp;
      return tmp;
    }
    static no_init_t instance() { return {}; }
    no_init_t(no_init_t const&) = default;
  private:
    no_init_t() = default;
  };
  static const no_init = no_init_t::instance();
}


struct Foo {
  char buff[1000];
  size_t hash;
  Foo():Foo(""){}
  template<size_t N, class=std::enable_if_t< (N<=sizeof(buff)) >>
  Foo( char const(&in)[N] ) {
    // some "expensive" copy and hash
  }
  Foo(no_init_t) {} // no initialization!
};
struct Record {
  int x;
  Foo foo;
  Record()=default;
  Record(no_init_t):
    x(no_init), foo(no_init)
  {}
};

现在我们可以构造Recordwithno_init并且它不会被初始化。

每个 POD 类都没有初始化。每个非 POD 类都必须提供一个no_init_t构造函数（并且可能尽可能实现非初始化）。

然后你memcpy就过去了。

这需要修改您的类型及其包含的类型，以支持非初始化。

于 2015-06-09T17:29:57.160 回答
1

像这样的东西？

编辑：

  1. 解决关于对齐的评论。现在使用匿名联合来确保正确对齐。

  2. TestRecord 现在包含另一种标准布局类型egg

  3. 添加了证明，即使egg有一个默认构造函数，该类在被填充之前也没有被构造populateNextRecord()

我认为这是尽可能快的，不是吗？

#include <iostream>
#include <array>
#include <algorithm>

struct egg {
    egg(int i) : _val(i) {}
    egg() {}
    int _val = 6;    
    friend std::ostream& operator<<(std::ostream& os, const egg& e) {
        return os << e._val; 
    }
};

struct TestRecord {
    egg x;
    double y;
};

void populateNext(uint8_t* first, size_t length)
{
    // do work here
    TestRecord data_source { 10, 100.2 };
    auto source = reinterpret_cast<uint8_t*>(&data_source);
    std::copy(source, source + length, first);
}

template<class Record>
struct RecordProxy
{
    RecordProxy() {}

  uint8_t* data() {
      return _data;
  }

  static constexpr size_t size() {
      return sizeof(Record);
  }

  Record& as_record() {
      return _record;
  }

    union {
        Record _record;
        uint8_t _data[sizeof(Record)];
    };
};


template <typename Record>
RecordProxy<Record> getNext() {
    RecordProxy<Record> r;
    populateNext(r.data(),  // maybe ::read()
                 r.size());                   // or equivalent
    return r;
}

using namespace std;
int main()
{
    RecordProxy<TestRecord> prove_not_initialised;
    auto& r1 = prove_not_initialised.as_record();
    cout << "x = " << r1.x << ", y = " << r1.y << endl;

    auto buffer = getNext<TestRecord>();
    auto& actual_record = buffer.as_record();
    cout << "x = " << actual_record.x << ", y = " << actual_record.y << endl;
   return 0;
}
于 2015-06-09T18:06:30.397 回答