Subject: Re: [linux-audio-dev] CDDA interface - How to convert to float? Any library?
From: Mark Rages (markrages_AT_mlug.missouri.edu)
Date: Sun Aug 11 2002 - 02:32:36 EEST
On Sat, Aug 10, 2002 at 04:05:40PM -0700, Martin Wolters wrote:
> On Saturday 10 August 2002 09:49, Mark Rages wrote:
> > Anyhow, try something like this:
> >
> > u16 *u16_buffer = (u16 *)buf;
> > float sample[samplesInBuffer]
> >
> > for (i=0; i<samplesInBuffer; i++)
> > sample[i]=u16_buffer[i]; // implicit conversion to float
>
> Well, that doesn't seem to work for me. If I use "s16" instead, it sounds a
> little more like music (I actually can recognize a song). But it's far from
> being undistored. I am wondering, if a byte-swap might be required and how to
> do that.
>
> -M
>
> PS.: I assume "u16" is equivalent to "__u16" as defined in <linux/types.h>,
> correct? I now use the following lines:
> #include <linux/types.h>
> #define s16 __s16
> #define u16 __u16
>
Ahh, I forgot CDDA is signed.
Yes, I meant the u16 in the general sense that you picked up on.
Well, you can try a byte-swap like this:
#include <string.h> // for swab()
#include <asm/byteorder.h> // for __LITTLE_ENDIAN macro
float sample[samplesInBuffer];
#ifdef __LITTLE_ENDIAN
s16 s16_buffer[samplesInBuffer];
swab(buf, s16_buffer, samplesInBuffer*sizeof(s16));
#else
s16 *s16buffer = (s16 *)buf;
#endif
for (i=0; i<samplesInBuffer; i++)
sample[i]=s16_buffer[i]; // implicit conversion to float
Regards,
Mark
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