从二进制文件读取double（字节顺序？） [英] Reading double from binary file (byte order?)

问题描述

我有一个二进制文件，我想从中读取一个double。

I have a binary file, and I want to read a double from it.

在十六进制表示中，我在文件中有这8个字节（然后是一些之后更多）：

In hex representation, I have these 8 bytes in a file (and then some more after that):

40 28 25 c8 9b 77 27 c9 40 28 98 8a 8b 80 2b d5 40 ...

40 28 25 c8 9b 77 27 c9 40 28 98 8a 8b 80 2b d5 40 ...

这应该对应于大约10的双倍值（基于该条目的含义）。

This should correspond to a double value of around 10 (based on what that entry means).

我使用过

#include<stdio.h>
#include<assert.h>

int main(int argc, char ** argv) {
   FILE * f = fopen(argv[1], "rb");
   assert(f != NULL);
   double a;
   fread(&a, sizeof(a), 1, f);
   printf("value: %f\n", a);
}

然而，打印
值：-261668255698743527401808385063734961309220864.000000

However, that prints value: -261668255698743527401808385063734961309220864.000000

很明显，字节不能正确转换为double。到底是怎么回事？
使用ftell，我可以确认正在读取8个字节。

So clearly, the bytes are not converted into a double correctly. What is going on? Using ftell, I could confirm that 8 bytes are being read.

推荐答案

就像整数类型一样，浮点类型受平台字节序限制。当我在小端机器上运行这个程序时：

Just like integer types, floating point types are subject to platform endianness. When I run this program on a little-endian machine:

#include <stdio.h>
#include <stdint.h>

uint64_t byteswap64(uint64_t input) 
{
    uint64_t output = (uint64_t) input;
    output = (output & 0x00000000FFFFFFFF) << 32 | (output & 0xFFFFFFFF00000000) >> 32;
    output = (output & 0x0000FFFF0000FFFF) << 16 | (output & 0xFFFF0000FFFF0000) >> 16;
    output = (output & 0x00FF00FF00FF00FF) << 8  | (output & 0xFF00FF00FF00FF00) >> 8;
    return output;
}

int main() 
{
    uint64_t bytes = 0x402825c89b7727c9;
    double a = *(double*)&bytes;
    printf("%f\n", a);

    bytes = byteswap64(bytes);
    a = *(double*)&bytes;
    printf("%f\n", a);

    return 0;
}

然后输出

12.073796
-261668255698743530000000000000000000000000000.000000

这表明您的数据存储在文件中小端格式，但您的平台是大端。因此，您需要在读取值后执行字节交换。上面的代码显示了如何做到这一点。

This shows that your data is stored in the file in little endian format, but your platform is big endian. So, you need to perform a byte swap after reading the value. The code above shows how to do that.

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