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'8 + 8bits variable'
1999\11\09@100423
by
erto Fonseca Iannini
What's the c syntax to perform an 16-bit sum like:
CCPR1 = CCPR1 + TIMER1;
Note that TIMER1 wasn't defined by the 1687x.inc. Instead it was made as
TIMER1H and TIME1L (each one 8 bits type). The same is true to CCPR1.
Summarizing: How can I define or use a new 16 bit variable called
TIMER1 that is formed by TIMERH and TIMERL (MSB and LSB)?
(this code is for 16F877)
See you guys,
Beto.
1999\11\09@110238
by
Edson Brusque
Your syntax is correct.
To make TIMER1 you could use:
unsigned long (uns16, etc) TIMER1;
TIMER1 = TIME1L + TIME1H * 256;
Regards,
Brusque
{Original Message removed}
1999\11\09@110913
by
Michael Rigby-Jones
|
part 0 5002 bytes
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">You could use a union as follows:</FONT>
</P>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">union intchar</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">{</FONT>
<BR> <FONT COLOR="#0000FF" SIZE=2 FACE="Arial">struct{</FONT>
<BR> <FONT COLOR="#0000FF" SIZE=2 FACE="Arial">unsigned char lo;</FONT>
<BR> <FONT COLOR="#0000FF" SIZE=2 FACE="Arial">unsigned char hi;</FONT>
<BR> <FONT COLOR="#0000FF" SIZE=2 FACE="Arial">}byte;</FONT>
<BR> <FONT COLOR="#0000FF" SIZE=2 FACE="Arial">unsigned int total;</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">}</FONT>
</P>
<BR>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">union intchar timer1;</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">union intchar myvar;</FONT>
</P>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">timer1.lo = TIMER1L;</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">timer.hi = TIMER1H;</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">myvar.lo = CCPR1L;</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">myvar.hi = CCPR1H;</FONT>
</P>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">myvar.total += timer1.total;</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">CCPR1L = myvar.lo</FONT>
<BR><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">CCPR1H = myvar.hi</FONT>
</P>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">Probably not very efficient though...</FONT>
</P>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">Regards</FONT>
</P>
<P><FONT COLOR="#0000FF" SIZE=2 FACE="Arial">Mike Rigby-Jones</FONT>
</P>
<BR>
<UL>
<P><FONT SIZE=1 FACE="Arial">{Original Message removed}
1999\11\09@120115
by
Dag Bakken
|
The (more or less) correct syntax would be something like:
Timer1 = OFF;
Shadow16bitT1 = TMR1L + TMR1H*256;
Timer1 = ON;
CCPR1 += Shadow16bitT1;
First of all you must switch off timer 1 so you have no byte-rollover
within the timer while you read. This period of time should be made
as short as possible, so just copy the timer to a shadow register.
Add it all up in the end.
If you do this instead:
Timer1 = OFF;
CCPR1 += TMR1L + TMR1H*256;
Timer1 = ON;
...the timer will be switched off longer because then you have an
extra 16bit addition to perform.
If you choose not to switch timer 1 off, you _will_ get occational
errors with this method.
If you can't switch timer 1 off, you have to test the values you read
from the timer to make sure you had no rollover. Something like this:
do {
L = TMR1L;
H = TMR1H;
} while (TMR1L < L); // if this is true, lo-byte rolled over,
// and the concatenated value will be wrong
// if it rolled over before TMR1H were copied.
With all this in mind, it's not just to make timer 1 a 16bit variable.
The code looks better with it, but works better without it.
-DS
Tuesday, November 09, 1999, 5:00:23 PM, you wrote:
EB> Your syntax is correct.
EB> To make TIMER1 you could use:
EB> unsigned long (uns16, etc) TIMER1;
EB> TIMER1 = TIME1L + TIME1H * 256;
EB> Regards,
EB> Brusque
EB> {Original Message removed}
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