从二进制文件中读取 double（字节顺序？）

Question

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

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 ...

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

I have used

#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);
}

However, that prints value: -261668255698743527401808385063734961309220864.000000

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.

Answer 1

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;
}

Then the output is

12.073796
-261668255698743530000000000000000000000000000.000000

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.

Answer 2

Endianness is convention . Reader and writer should agree on what endianness to use and stick to it.

You should read your number as int64, convert endianness and then cast to double.