Lpc2000

This was a great idea !

I am now at around 5,8956 MBytes / second, which is close to 5,898 MBytes /
sec. ( = Fosc * 4 / 10).
So the two operations ( ldr ip, [r0, #0] and strb ip, [r2], #1) seems to
take in sum 10 cycles.
Is this correct ?

Re-adding my shift instruction: gives me around 5,360400 MBytes / s, so 11
cycles. So shift itself takes 1 cycle ?

BTW: Is there a cycle table somewhere for LPC21xx on the net or in an
appnote / manual ? Is it the same as for original ARMs ?

Greetings,

          Martin

----- Original Message ----- 
From: "Ben Dooks" <ben@...>
To: <lpc2100@yahoogroups.com>
Sent: Sunday, February 01, 2004 11:55 AM
Subject: Re: [lpc2100] Optimization of Capture Routine


> On Sat, Jan 31, 2004 at 09:01:05PM +0100, capiman@... wrote:
> > Hello,
> >
> > i want to read in 1 byte multiple times from the port pins as fast as
possible:
> > Currently i have the following C code:
> >
> > unsigned char Data[60000];
> >
> > void CaptureBuffer()
> > {
> >     unsigned char *ptr = &Data[0];
> >     unsigned char *ptrend = &Data[60000];
> >
> >     while(ptr < ptrend)
> >     {
> >         (*ptr) = (IOPIN >> LA_D0_BIT) & 0xff;
> >         ptr++;
> >     }
> > }
> >
> > When i compile it with gcc and option -O3, i can capture with around 3,9
MBytes/sec.
> > Avoiding the shift (by using P0.0 - P0.7) gives me 4,2 MBytes/sec.
> > Leaving out the (*ptr) = (IOPIN...) instruction gives me 11,8
MBytes/sec, but no more functionality :-)
> >
> > Can i improve the speed with inline assembler ? Produced assembler code
already looks very compact...
> >
> > .L142:
> >  ldr ip, [r0, #0]
> >  strb ip, [r2], #1
> >  cmp r2, r1
> >  bcc .L142
> >
> > Are there any other tricks ?
>
> unrolling the loop a bit may help, as it reduces the number of branch
> instructions needed.
>
> -- 
> Ben
>
> Q:      What's a light-year?
> A:      One-third less calories than a regular year.
>
>
>
>
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