How to convert from ubyte[] to and from float?

[Charles Hixson via Digitalmars-d-learn]

What is the best way to convert from a part of a ubyte[] to a float?

I've tried converting the ubyte[] into a uint, but neither casting the
uint to a float nor to!float work.

I suppose I could use a "trick record" union, but that seems inelegant.
If I use pointers, the alignment may (unpredictably) not be proper
(whatever that means these days).

[Ali Çehreli via Digitalmars-d-learn]

Post by Charles Hixson via Digitalmars-d-learn
What is the best way to convert from a part of a ubyte[] to a float?
I've tried converting the ubyte[] into a uint, but neither casting the
uint to a float nor to!float work.
I suppose I could use a "trick record" union, but that seems inelegant.
If I use pointers, the alignment may (unpredictably) not be proper
(whatever that means these days).

This is what I understood:

import std.exception;

ubyte[float.sizeof] toBytes(float f)
{
ubyte* beg = cast(ubyte*)&f;
return beg[0..f.sizeof];
}

float toFloat(const(ubyte)[] bytes)
{
enforce(bytes.length >= float.sizeof);
return *cast(float*)bytes.ptr;
}

void main()
{
float f = 1.5;

auto bytes = toBytes(f);
float f2 = toFloat(bytes);

assert(f2 == f);
}

There are no alignment issues because f and ubyte[float.sizeof] are not
related. If you meant that ubyte[] should be a reference to an existing
float, then toBytes must take by 'ref float' and then it can return a
ubyte[]. However, it would be the responsibility of the caller to ensure
that the float would live long enough.

If that happened, then there would be no alignment issues because we
would have started with a float anyway and the ubyte[] would be
referring to that float in memory.

Ali

[uri via Digitalmars-d-learn]

On Sunday, 19 October 2014 at 03:14:26 UTC, Charles Hixson via

Post by Charles Hixson via Digitalmars-d-learn
What is the best way to convert from a part of a ubyte[] to a
float?
I've tried converting the ubyte[] into a uint, but neither
casting the uint to a float nor to!float work.
I suppose I could use a "trick record" union, but that seems
inelegant. If I use pointers, the alignment may
(unpredictably) not be proper (whatever that means these days).

Is this what you're after?
http://dlang.org/phobos/std_bitmanip.html#.peek
http://dlang.org/phobos/std_bitmanip.html#.read
http://dlang.org/phobos/std_bitmanip.html#.write

These accept indices, or you can just slice the ubyte[] to the
part you need.

---

import std.stdio;
import std.bitmanip;
import std.system;
void main()
{
// UBYTE[] to FLOAT
//
// 6.5535000E+004 00-FF-7F-47
ubyte[] ubval = [0, 0xff, 0x7f, 0x47];
auto fval = ubval.peek!(float, Endian.littleEndian);
writefln("%s as float: %s", ubval, fval);
writefln("%s as float: %s", ubval, ubval.read!(float,
Endian.littleEndian));

// 16383.8 00-FF-7F-46
ubval = [0, 0xff, 0x7f, 0x46];
fval = ubval.peek!(float, Endian.littleEndian);
writefln("%s as float: %s", ubval, fval);
writefln("%s as float: %s", ubval, ubval.read!(float,
Endian.littleEndian));

// FLOAT to UBYTE[]
ubval = [0, 0, 0, 0];
std.bitmanip.write!(float, Endian.littleEndian)(ubval,
65535.0f, 0);
writefln("%s as ubyte[]: %s", fval, ubval);
ubval = [0, 0, 0, 0];
std.bitmanip.write!(float, Endian.littleEndian)(ubval, fval,
0);
writefln("%s as ubyte[]: %s", fval, ubval);


}

---