## digitalmars.D.learn - Re: little/big endian conversions

thank you all for your help!!

Bill Baxter Wrote:

You don't necessarily need to put it into a byte array.  Just cast:

void swap(ref byte a, ref byte b) {
byte tmp; tmp=a; a=b; b=tmp;
}
float f;
byte[4] fbytes = (cast(byte*)&f)[0..4];
swap(fbytes[0],fbytes[3]);
swap(fbytes[1],fbytes[2]);
float fswapped = *(cast(float*)fbytes.ptr);

Or instead of casts you can use a union.

union FC { float f;      ubyte[4] c; }
FC fs;
fs.f = f;
swap(fs.c[0],fs.c[3]);
swap(fs.c[1],fs.c[2]);
float fswapped = fs.f;

--bb

lurker wrote:
so i need to put the float/double into an byte array and just swap?

Regan Heath Wrote:

Bill Baxter wrote:
Regan Heath wrote:
lurker wrote:
does anybody know how to convert float and doubles to little/big endian?

http://en.wikipedia.org/wiki/IEEE_754 You'll see the internal representation of a float, given that and a little guess work I've come up with: import std.stdio; int extractSign(float f) { return (*(cast(int*)&f) & 0x80000000) ? -1 : 1; } ubyte extractExp(float f) { return (*(cast(int*)&f) << 1) & 0xFF000000; } int extractFraction(float f) { return *(cast(int*)&f) & 0x007FFFFF; } void main() { float f = -1.25f; auto sign = extractSign(f); auto exp = extractExp(f); auto fraction = extractFraction(f); writefln(f); writefln(sign); writefln(exp); writefln(fraction); } which will extract the various parts of a float. Now, I have no idea how they might change on a big/little endian system but I suspect each part would have it's byte order swapped. In which case, byte order swapping the extracted parts then re-assembling might give you a byte order swapped float. Like I said, I'm guessing. What you want is 2 systems with different ordering and then you want to dump the content of the float like this: writefln("%032b", *(cast(int*)&f)); then compare. Regan

It doesn't matter that it's in IEEE 745 format. You just swap the bytes like it was any old kind of data.

Regan

Apr 09 2008