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如何使用C或C ++读取FORTRAN二进制文件?

[英]How to read FORTRAN binary file with c or c++?

我有FORTRAN 77二进制文件(在Sun Sparc机器上创建,大端)。 我想在我的小端机器上阅读它。 我遇到了这个

http://paulbourke.net/dataformats/reading/

Paul已经为C或C ++编写了这些宏,但是我不明白它们的实际作用。

#define SWAP_2(x) ( (((x) & 0xff) << 8) | ((unsigned short)(x) >> 8) )
#define SWAP_4(x) ( ((x) << 24) | (((x) << 8) & 0x00ff0000) | \
         (((x) >> 8) & 0x0000ff00) | ((x) >> 24) )
#define FIX_SHORT(x) (*(unsigned short *)&(x) = SWAP_2(*(unsigned short *)&(x)))
#define FIX_LONG(x) (*(unsigned *)&(x) = SWAP_4(*(unsigned *)&(x)))
#define FIX_FLOAT(x) FIX_LONG(x)

我知道文件的每个记录都包含

x,y,z,t,d,i

我是整数* 2,所有其他变量都是实数* 4。 前512个字节的十六进制转储

0000000 0000 1800 0000 0000 0000 0000 0000 0000
0000010 0000 0000 0000 0000 ffff ffff 0000 1800
0000020 0000 1800 003f 0000 0000 0000 233c 0ad7
0000030 0000 0000 233c 0ad7 0000 0100 0000 1800
0000040 0000 1800 803f 0000 0000 0000 233c 0ad7
0000050 0000 0000 233c 0ad7 0000 0100 0000 1800
0000060 0000 1800 c03f 0000 0000 0000 233c 0ad7
0000070 0000 0000 233c 0ad7 0000 0100 0000 1800
0000080 0000 1800 0040 0000 0000 0000 233c 0ad7
0000090 0000 0000 233c 0ad7 0000 0100 0000 1800
00000a0 0000 1800 2040 0000 0000 0000 233c 0ad7
00000b0 0000 0000 233c 0ad7 0000 0100 0000 1800
00000c0 0000 1800 4040 0000 0000 0000 233c 0ad7
00000d0 0000 0000 233c 0ad7 0000 0100 0000 1800
00000e0 0000 1800 6040 0000 0000 0000 233c 0ad7
00000f0 0000 0000 233c 0ad7 0000 0100 0000 1800
0000100 0000 1800 8040 0000 0000 0000 233c 0ad7
0000110 0000 0000 233c 0ad7 0000 0100 0000 1800
0000120 0000 1800 9040 0000 0000 0000 233c 0ad7
0000130 0000 0000 233c 0ad7 0000 0100 0000 1800
0000140 0000 1800 a040 0000 0000 0000 233c 0ad7
0000150 0000 0000 233c 0ad7 0000 0100 0000 1800
0000160 0000 1800 b040 0000 0000 0000 233c 0ad7
0000170 0000 0000 233c 0ad7 0000 0100 0000 1800
0000180 0000 1800 c040 0000 0000 0000 233c 0ad7
0000190 0000 0000 233c 0ad7 0000 0100 0000 1800
00001a0 0000 1800 d040 0000 0000 0000 233c 0ad7
00001b0 0000 0000 233c 0ad7 0000 0100 0000 1800
00001c0 0000 1800 e040 0000 0000 0000 233c 0ad7
00001d0 0000 0000 233c 0ad7 0000 0100 0000 1800
00001e0 0000 1800 f040 0000 0000 0000 233c 0ad7
00001f0 0000 0000 233c 0ad7 0000 0100 0000 1800
0000200

我的代码读取文件

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

int main() 
{
    FILE *file;
    char *buffer;
    char *rec;
    long fileLen;

    file = fopen("rec.in", "rb");


    fseek(file, 0, SEEK_END);
    fileLen=ftell(file);
    fseek(file, 0, SEEK_SET);

    buffer=(char *)malloc(fileLen+1);

    fread(buffer, fileLen, 1, file);
    fclose(file);
    free(buffer);

char *curr = buffer;
char *end = buffer + fileLen;

constexpr int LINE_SIZE = sizeof(float)*5 + sizeof(uint16_t); //based upon your "x,y,z,t,d,i" description

while(curr < end) {
    uint32_t temp = be32toh(*reinterpret_cast<uint32_t*>(*curr));
    float x = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+sizeof(float))));
    float y = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+2*sizeof(float))));
    float z = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+3*sizeof(float))));
    float t = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+4*sizeof(float))));
    float d = *reinterpret_cast<float*>(&temp);

    uint16_t i = be16toh(*reinterpret_cast<uint16_t*>(*(curr+5*sizeof(float))));

    curr += LINE_SIZE;

}

}

我有两个错误r.cc:在函数'int main()'中:

r.cc:29:1: error: ‘constexpr’ was not declared in this scope
 constexpr int LINE_SIZE = sizeof(float)*5 + sizeof(uint16_t); //based upon your "x,y,z,t,d,i" description
 ^
r.cc:49:13: error: ‘LINE_SIZE’ was not declared in this scope
     curr += LINE_SIZE;

如果您正在Linux机器上读取文件,则在endian.h标头( 此处的文档)中提供了一些用于此目的的库函数。 要将16位整数转换为主机顺序(在您的情况下为little-endian):

uint16_t hostInteger = be16toh(bigEndianIntegerFromFile);

对于浮点数,您可以执行类似的操作,但要合并重新解释:

float hostFloat = reinterpret_cast<float>(be32toh(reinterpret_cast<uint32_t>(bigEndianFloatFromFile)));

或者,如果您首先将其作为unsigned int读取,则不需要内部的reinterpret_cast

float hostFloat = reinterpret_cast<float>(be32toh(bigEndianUint32FromFile));

更新:给定代码,您可以通过在fclosefree调用之间插入此文件来读取文件:

char *curr = buffer;
char *end = buffer + fileLen;

constexpr int LINE_SIZE = sizeof(float)*5 + sizeof(uint16_t); //based upon your "x,y,z,t,d,i" description

while(curr < end) {
    uint32_t temp = be32toh(*reinterpret_cast<uint32_t*>(*curr));
    float x = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+sizeof(float))));
    float y = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+2*sizeof(float))));
    float z = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+3*sizeof(float))));
    float t = *reinterpret_cast<float*>(&temp);

    temp = be32toh(*reinterpret_cast<uint32_t*>(*(curr+4*sizeof(float))));
    float d = *reinterpret_cast<float*>(&temp);

    uint16_t i = be16toh(*reinterpret_cast<uint16_t*>(*(curr+5*sizeof(float))));

    curr += LINE_SIZE;

    ...
    //do something with these values
    ...
}

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