在C数据序列化？ [英] Data serialization in C?

问题描述

我有这样的结构，我想写一个文件：

I have this structure which I want to write to a file:

typedef struct
{
    char* egg;
    unsigned long sausage;
    long bacon;
    double spam;
} order;

这个文件必须是二进制，并且必须由具有任何机器可读
C99编译器。

This file must be binary and must be readable by any machine that has a C99 compiler.

我看着各种方法这个问题，如ASN.1，XDR，XML，
ProtocolBuffers和许​​多其他人，但没有一个适合我的要求：

I looked at various approaches to this matter such as ASN.1, XDR, XML, ProtocolBuffers and many others, but none of them fit my requirements:

• 小


• 简单


• 写在 C

• small
• simple
• written in C

然后我决定让我自己的数据的协议。我可以处理
下面的整数类型的再presentations：

I decided then to make my own data protocol. I could handle the following representations of integer types:

• 无符号

在
• 签署的的补的

在
• 签署的补的


• 在签署的符号和幅度的

• unsigned
• signed in one's complement
• signed in two's complement
• signed in sign and magnitude

在一个有效，简单和干净的方式（IM pressive，不是吗？）。然而，
真正类型现在是一个痛苦。

in a valid, simple and clean way (impressive, no?). However, the real types are being a pain now.

我应该如何读浮动和双击从字节流？标准
说，按位运算符（至少＆安培; ， | ，＆LT;＆LT; 和＆GT;＆GT; ）是
整数只类型，其中给我留下没有希望。唯一的办法我可以
认为是：

How should I read float and double from a byte stream? The standard says that bitwise operators (at least &, |, << and >>) are for integer types only, which left me without hope. The only way I could think was:

int sign;
int exponent;
unsigned long mantissa;

order my_order;

sign = read_sign();
exponent = read_exponent();
mantissa = read_mantissa();

my_order.spam = sign * mantissa * pow(10, exponent);

但似乎并没有真正有效的。我也找不到
重新presentation说明双和浮动。应该怎样
这前一继续？

but that doesn't seem really efficient. I also could not find a description of the representation of double and float. How should one proceed before this?

推荐答案

如果你想尽可能便携用浮漂，您可以使用frexp和ldexp：

If you want to be as portable as possible with floats you can use frexp and ldexp:

void WriteFloat (float number)
{
  int exponent;
  unsigned long mantissa;

  mantissa = (unsigned int) (INT_MAX * frexp(number, &exponent);

  WriteInt (exponent);
  WriteUnsigned (mantissa);
}

float ReadFloat ()
{
  int exponent = ReadInt();
  unsigned long mantissa = ReadUnsigned();

  float value = (float)mantissa / INT_MAX;

  return ldexp (value, exponent);
}

这背后的想法是，ldexp，frexp和INT_MAX是标准C.另外一个unsigned long的precision通常至少高达尾数（没有保证的宽度，但它是一个有效假设，我不知道一个单一的架构，不同的是这里）。

The Idea behind this is, that ldexp, frexp and INT_MAX are standard C. Also the precision of an unsigned long is usually at least as high as the width of the mantissa (no guarantee, but it is a valid assumption and I don't know a single architecture that is different here).

因此​​，转换工作没有precision损失。用INT_MAX司/乘法可能会失去一个位转换期间precision的，但是这是一个妥协的人可以住在一起。

Therefore the conversion works without precision loss. The division/multiplication with INT_MAX may loose a bit of precision during conversion, but that's a compromise one can live with.

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