I cant detect anything with the following code (8051 and the keypad i
connected to the port 0)


#define K0_mask 0xED
#define K1_mask 0x7E
#define K2_mask 0x7D
#define K3_mask 0x7B
#define K4_mask 0xBE
#define K5_mask 0xBD
#define K6_mask 0xBB
#define K7_mask 0xDE
#define K8_mask 0xDD
#define K9_mask 0xDB
.....
unsigned char readKeyboard()
{
//Enter if a key has been pressed
if (portState!=0xFF)
{
//need a 21ms debounce period
timerWait25();

//check the key pressed
if (portState == K0_mask)
{
//Display the digit
writeToLCD(0,6);
//wait for the touch to be released
while (portState ==K0_mask);
return(0);
}
....


DOESNT WORK....why ??? thanks for any help

zebulon wrote:
I assume portState is defined somewhere as the input port. If it's not,
you never read the port. You only display anything at all if you get the
right input, so how about displaying (somehow) whatever you do get, e.g.
using a serial port, and work on from there?

paul Burke

On Mon, 15 May 2006 05:08:13 -0500, in comp.arch.embedded "zebulon"
<lyoncfpl@hotmail.com> wrote:

port 0 needs pullup resistors, do you have them ?


martin

sorry i forgot to put this...

//portState points to the port 1
sfr portState=0x90;

yep there are all electronic stuff that should make it running (dont as
me for electronics things , i'm just a trainee that should develop a phon
software...)

When i press a 0 , i should read 11101101 , i.e 0xED...no ?

actually it's connected to port 1 ....
I always confuse , but anyways , doesnt change anything to m
problem...arghh embedded....

zebulon wrote:
- Ark

zebulon wrote:
Get a scope, voltmeter or logic probe and look at the port pin.

I assume you wrote 0xFF to the port to initialize it.
And, that somewhere you defined portState as P1.

zebulon wrote:
Stuff I forgot: as well as the pullups someone else mentioned, you have
to set port 0 to 0xff (all high) when you iniyialise the system, as it's
a sort of open- collector output.

But seriously, you should be poking round with a scope probe more than
asking us, and printing out what the port actually reads. Check for
changes, and send out using the UART.

Paul Burke

i changed the whole code , the new one , doesnt work too....i'm fed up wit
this...

for(i=0;i<4;i++)
{
//Put the i'th column to 0
P1 = 0xff & ~(1<<i);
delay(100);

//check the pin
if(P1^7==0)
{
timerWait25(); //debounce period
if (P1^7==0) //check again
{
//Switching to know what key was pressed...
if (i==0)
digit=1;
if (i==1)
digit=2;
if (i==2)
digit=3;
if (i==3)
digit=F1;
//Print the output , DIGIT only !
if (i!=3)
writeToLCD(digit,posToDisplay);
//Wait for the release
while (P1^7==0);
return(digit);
}
}
if(P1^6==0)
{
//timerWait25();
delay(100);
if(P1^6==0)
{
if (i==0)
digit=4;
if (i==1)
digit=5;
if (i==2)
digit=6;
if (i==3)
digit=F2;
if (i!=3)
writeToLCD(digit,posToDisplay);
while (P1^6==0);
return(digit);
}
}
if(P1^5==0)
{
// timerWait25();
delay(100);
if(P1^5==0)
{
if (i==0)
digit=7;
if (i==1)
digit=8;
if (i==2)
digit=9;
if (i==3)
digit=P;
//print the output , DIGIT only !
if (i!=3)
writeToLCD(digit,posToDisplay);
while (P1^5==0);
return(digit);
}
}

if(P1^4==0)
{
// timerWait25();
delay(100);
if(P1^4==0)
{
if (i==0)
digit=STAR;
if (i==1)
digit=0;
if (i==2)
digit=SHARP;
if (i==3)
digit=REDIAL;
//print the output , DIGIT only !
if (i==1)
writeToLCD(digit,posToDisplay);
while (P1^4==0);
return(digit);
}
}
}
return(0xFF);//code for no key pressed

"zebulon" <lyoncfpl@hotmail.com> wrote in message
news:CMadnR6TyMYmDffZRVn-tw@giganews.com...
What do you expect this to do? (P1^7)==0 is the same as P1 == 7. For the
first three cycles you are pulling down one of these bits so its bound to
fail. Four the fourth cycle you'd need to press four keys to zero the top
four bits...

[snip similar]

Peter

I cant understand what you're trying to tell me...
if(P1^7==0) : doesnt it check the value of the 7th pin ??? why should i
be like writting 7 to port 1 ????

i was told to put one bit (column) to 0 , everything else to 1 , the
check the rows to find a 0 which means thaht a key pressed whose value
will be determined both by col and rows...
so i dont understant your answer at al ???

"zebulon" <lyoncfpl@hotmail.com> wrote in message
news:H-qdnX--QsSpZ_fZRVn-jg@giganews.com...
I think you need to spend some time learning the language instead of
worrying about hardware. Your code exors the value read from port 1 with the
value 7 and checks the result is 0. The only value of port 1 that when
exored with 7 gives 0 is 7.

Yes. Output scanning zeros on the bottom four bits and check the top four -
assuming a 4x4 matrix keypad.

My answer was that you code was gibberish, You said it didn't work, so I may
be right :-)

for (i=0; i<4; i++)
{
P1 = 0xFE<<i;
Wait();
switch (P1 & 0xF0)
{
case 0x70:
...
case 0xB0:
...

Peter

I know that i'm not experimented , but i must understand...

where do you see XOR ??

P1 = 0xff & (~(1<<i));

This just put one of the 4 lower bits to 0

if(P1^7==0)
{
Then check the 7 th pin , no?

where is the error in this ?

zebulon wrote:
If you quoted things properly it would be much easier. At any
rate, you have repeated the error in "if (P1 ^ 7 == 0) {". Why do
you want the xor of P1 (whatever that may be) and the constant 7?
BTW, there are no prizes given for obfuscating code by eliding
spaces.

--
"If you want to post a followup via groups.google.com, don't use
the broken "Reply" link at the bottom of the article. Click on
"show options" at the top of the article, then click on the
"Reply" at the bottom of the article headers." - Keith Thompson
More details at: <http://cfaj.freeshell.org/google/>
Also see <http://www.safalra.com/special/googlegroupsreply/>

On Wed, 17 May 2006 20:55:59 -0500, "zebulon" <lyoncfpl@hotmail.com>
wrote:

'^' is the exclusive-OR operator in C
'7' in binary is 0...000111

"zebulon" <lyoncfpl@hotmail.com> wrote:
'^' is the exclusive-OR operator in C
'7' in binary is 0...000111

P1^7 just toggles P1's 3 lowest bits.

From the Keil C51 help files... :

sbit name = sfr-name ^ bit-position;

The previously declared SFR (sfr-name) is the base address for the sbit
It must be evenly divisible by 8. The bit-position (which must be a numbe
from 0-7) follows the carat symbol ('^') and specifies the bit position t
access. For example:



........

btw , this version works :


for(i=0;i<4;i++)
{
//Put the i'th column to 0
P1 = 0xff & (~(1<<i));
timerWait25();

//check the pin
if(P1==0x7E || P1==0x7D || P1==0x7B || P1==0x77)
{
timerWait25(); //debounce period
if (P1==0x7E || P1==0x7D || P1==0x7B || P1==0x77) //check again
{
//Switching to know what key was pressed...
if (i==0)
digit=1;
if (i==1)
digit=2;
if (i==2)
digit=3;
if (i==3)
digit=F1;
//Print the output , DIGIT only !
if (i!=3)
writeToLCD(digit,posToDisplay);
//Wait for the release
while (P1==0x7E || P1==0x7D || P1==0x7B || P1==0x77);
return(digit);
}
}




Thanks everybody

"zebulon" <lyoncfpl@hotmail.com> wrote in message
news:ZaudndOm96DBcfbZRVn-ug@giganews.com...
Correct, but this construct only works when you define an sbit, not in a
regular C expression.

sbit mypin = P1^7 /* this is a non-standard C */

defines mypin to be bit 7 of port 1 while

if (P1^7 == 0) /* standard C */

EXORs the contents of P1 with 7

Meindert

i should have precised that i was working under Keil C51...sorry

"zebulon" <lyoncfpl@hotmail.com> wrote in message
news:ze-dndMCV8sNpPHZ4p2dnA@giganews.com...
I doesn't matter. Even with the Keil compiler the 'if ( P1^7==0 )' code does
not have the interpretation you give it. sbit is a Keil extension and ^ does
not have this meaning elsewhere in C, even Keil C.

Peter