C++中是否有二进制内存流

Are there binary memory streams in C++

提问by FireAphis

I usually use stringstreamto write into in-memory string. Is there a way to write to a char buffer in binary mode? Consider the following code:

我通常stringstream用来写入内存中的字符串。有没有办法以二进制模式写入字符缓冲区？考虑以下代码：

stringstream s;
s << 1 << 2 << 3;
const char* ch = s.str().c_str();

The memory at chwill look like this: 0x313233 - the ASCII codes of the characters 1, 2 and 3. I'm looking for a way to write the binary values themselves. That is, I want 0x010203 in the memory. The problem is that I want to be able to write a function

内存ch将如下所示：0x313233 - 字符 1、2 和 3 的 ASCII 代码。我正在寻找一种方法来编写二进制值本身。也就是说，我想要内存中的 0x010203。问题是我希望能够写一个函数

void f(ostream& os)
{
    os << 1 << 2 << 3;
}

And decide outside what kind of stream to use. Something like this:

并在外部决定使用哪种流。像这样的东西：

mycharstream c;
c << 1 << 2 << 3; // c.data == 0x313233;
mybinstream b;
b << 1 << 2 << 3; // b.data == 0x010203;

Any ideas?

有任何想法吗？

回答by KeithB

To read and write binary data to streams, including stringstreams, use the read() and write() member functions. So

要将二进制数据读取和写入流，包括字符串流，请使用 read() 和 write() 成员函数。所以

unsigned char a(1), b(2), c(3), d(4);
std::stringstream s;
s.write(reinterpret_cast<const char*>(&a), sizeof(unsigned char));
s.write(reinterpret_cast<const char*>(&b), sizeof(unsigned char));
s.write(reinterpret_cast<const char*>(&c), sizeof(unsigned char));
s.write(reinterpret_cast<const char*>(&d), sizeof(unsigned char));

s.read(reinterpret_cast<char*>(&v), sizeof(unsigned int)); 
std::cout << std::hex << v << "\n";

This gives 0x4030201on my system.

这给0x4030201了我的系统。

Edit: To make this work transparently with the insertion and extraction operators (<< and >>), your best bet it to create a derived streambuf that does the right thing, and pass that to whatever streams you want to use.

编辑：为了使用插入和提取运算符（<< 和 >>）使这项工作透明化，最好创建一个派生的流缓冲，它可以做正确的事情，并将其传递给您想要使用的任何流。

回答by Samuel Powell

You can do this sort of thing with templates. E.g:

你可以用模板做这种事情。例如：

//struct to hold the value:
template<typename T> struct bits_t { T t; }; //no constructor necessary
//functions to infer type, construct bits_t with a member initialization list
//use a reference to avoid copying. The non-const version lets us extract too
template<typename T> bits_t<T&> bits(T &t) { return bits_t<T&>{t}; }
template<typename T> bits_t<const T&> bits(const T& t) { return bits_t<const T&>{t}; }
//insertion operator to call ::write() on whatever type of stream
template<typename S, typename T>
S& operator<<(S &s, bits_t<T> b) {
    return s.write((char*)&b.t, sizeof(T));
}
//extraction operator to call ::read(), require a non-const reference here
template<typename S, typename T>
S& operator>>(S& s, bits_t<T&> b) {
    return s.read((char*)&b.t, sizeof(T));
}

It could use some cleanup, but it's functional. E.g:

它可以使用一些清理，但它的功能。例如：

//writing
std::ofstream f = /*open a file*/;
int a = 5, b = -1, c = 123456;
f << bits(a) << bits(b) << bits(c);

//reading
std::ifstream f2 = /*open a file*/;
int a, b, c;
f >> bits(a) >> bits(b) >> bits(c);

回答by Jean Daniel Pauget

overloading some unusual operators works rather well. Here below I choosed to overload <=because it has the same left-to-right associativity as <<and has somehow a close look-and-feel ...

重载一些不寻常的运算符效果很好。在下面，我选择重载<=因为它具有与<<相同的从左到右关联性，并且在某种程度上具有接近的外观......

#include <iostream>
#include <stdint.h>
#include <arpa/inet.h>

using namespace std;

ostream & operator<= (ostream& cout, string const& s) {
    return cout.write (s.c_str(), s.size());
}
ostream & operator<= (ostream& cout, const char *s) {
    return cout << s;
}
ostream & operator<= (ostream&, int16_t const& i) {
    return cout.write ((const char *)&i, 2);
}
ostream & operator<= (ostream&, int32_t const& i) {
    return cout.write ((const char *)&i, 4);
}
ostream & operator<= (ostream&, uint16_t const& i) {
    return cout.write ((const char *)&i, 2);
}
ostream & operator<= (ostream&, uint32_t const& i) {
    return cout.write ((const char *)&i, 4);
}

int main() {
    string s("some binary data follow : ");

    cout <= s <= " (machine ordered) : " <= (uint32_t)0x31323334 <= "\n"
         <= s <= " (network ordered) : " <= htonl(0x31323334) ;
    cout << endl;

    return 0;
}

There are several drawbacks :

有几个缺点：

• the new meaning of <=may confuse readers or lead to unexpected results :

  cout <= 31 <= 32;
  

  won't give the same result as

  cout <= (31 <= 32);
  

• the endianess isn't clearly mentionned at reading the code, as illustrated in the above example.

• it cannot mix simply with <<because it doesn't belong to the same group of precedence. I usually use parenthesis to clarify such as :

  ( cout <= htonl(a) <= htonl(b) ) << endl;
  

• <=的新含义可能会混淆读者或导致意想不到的结果：

  cout <= 31 <= 32;
  

  不会给出相同的结果

  cout <= (31 <= 32);
  

• 如上例所示，在阅读代码时没有明确提及字节序。

• 它不能简单地与<<混合，因为它不属于同一组优先级。我通常使用括号来澄清，例如：

  ( cout <= htonl(a) <= htonl(b) ) << endl;
  

回答by kamikaze

For this use case I implemented myself a "raw shift operator":

对于这个用例，我自己实现了一个“原始移位运算符”：

template <typename T, class... StreamArgs>
inline std::basic_ostream<StreamArgs...> &
operator <= (std::basic_ostream<StreamArgs...> & out, T const & data) {
        out.write(reinterpret_cast<char const *>(&data), sizeof(T));
        return out;
}

Put it somewhere convenient and use it like this:

把它放在方便的地方并像这样使用它：

std::cout <= 1337 <= 1337ULL <= 1337. <= 1337.f;

Advantages:

好处：

• chainable
• automatic sizeof()
• takes arrays and struct/class instances, too

• 可链接的
• 自动的 sizeof()
• 也接受数组和结构/类实例

Disadvantages:

缺点：

• unsafe for non-POD objects: leaks pointers and padding
• output is platform specific: padding, endianess, integer types

• 对非 POD 对象不安全：泄漏指针和填充
• 输出是特定于平台的：填充、字节序、整数类型

回答by Luká? Lalinsky

Well, just use characters, not integers.

好吧，只使用字符，而不是整数。

s << char(1) << char(2) << char(3);