There are some discussions about the same question but I would like to ask some more ,
1) How portable is the below code for a double byte swapping
int ReadDouble(FILE *fptr,double *n)
{
unsigned char *cptr,tmp;
if (fread(n,8,1,fptr) != 1)
return(FALSE);
cptr = (unsigned char *)n;
tmp = cptr[0];
cptr[0] = cptr[7];
cptr[7] = tmp;
tmp = cptr[1];
cptr[1] = cptr[6];
cptr[6] = tmp;
tmp = cptr[2];
cptr[2] = cptr[5];
cptr[5] =tmp;
tmp = cptr[3];
cptr[3] = cptr[4];
cptr[4] = tmp;
return(TRUE);
}
2) Should I keep the 3 important parts of a floating point number, sign bit, mantissa, exponent as integers and then try to manipulate them somehow.
I know the basics of floating point representations, not that deeply as a mechanical engineer, however I need to read some big-endian file where my machine is little endian. I can maybe worry about the portability issues later on. But I would like to learn about them perhaps you can direct me to some more direct things on this because there is too much information on this, I was confused which one to read.
So after some comments this should more or less do that in a portable way right? Sorry for the C file pointers…
double_t ReadDouble(ifstream& source) {
// read
char buf[sizeof(double_t)];
source.read(buf, sizeof(double_t));
// reverse and return
reverse( buf, buf+sizeof(double_t) );
return *(reinterpret_cast<double_t*>(buf));
}
Best,
Umut
It’s not as easy as that. Just because an architecture is big-endian for integers doesn’t mean it’s big-endian for floating point numbers. I’ve heard of platforms that store integers big-endian and floats little-endian.
So first you should discover what the actual memory representation of double on your source platform is.
As for the swap itself, it’s inefficient and way too much code. An additional 8-byte buffer won’t kill you, so why not do this:
Way less code, even if you decide to manually unroll the loop.