convert to little endian

[mike]

Trying to convert the following to little endian. I have set up my
long bits but dont know how to shift the bits correctly. Heres what I
have at the moment.

long lgn = 0;
long l = Double.doubleToLongBits(dbl);
for (int shift = 0; shift < 64; shift+=8)
{
lgn |= l & 0xff << shiftBy;
}
Any help would be great.

 

[Thomas Fritsch]

Trying to convert the following to little endian. I have set up my
long bits but dont know how to shift the bits correctly. Heres what I
have at the moment.

Convert to little-endian from what? From little-endian or from
big-endian? Currently your algorithm converts from little-endian to
little-endian; your end-result is simply lgn == l.

long lgn = 0;
long l = Double.doubleToLongBits(dbl);
for (int shift = 0; shift < 64; shift+=8)
{
lgn |= l & 0xff << shiftBy;

You probably mean 'shift' here, not 'shiftby' which isn't declared.
Also: Please use parentheses here, to clarify what you intend.
lgn |= (l & 0xff) << shiftBy;
or lgn |= l & (0xff << shiftBy);
(Because the rules of precedence are easy to remember for compilers, but
hard for humans.)

}
Any help would be great.

Please refine your problem description! What do you want to achieve?

 

[hilz]

"Thomas:Fritsch$ops:de".replace(':','.').replace('$','@')

nice!

 

[Thomas G. Marshall]

hilz coughed up:

nice!

What he said. That's kinda nifty.