Post by thread:[OT]: CCS C COMPILER

My question is: Will affect the remainder bits? The remainders bits are configurated as INPUT and OUTPUT... Miguel ----- Original Message ----- From: Andrew Kunz <spam_OUTakunzTakeThisOuTTDIPOWER.COM> To: <.....PICLISTKILLspam@spam@MITVMA.MIT.EDU> Sent: Tuesday, June 13, 2000 4:42 PM Subject: Re: [OT]: CCS C COMPILER > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > Andy

It depends on how much you are draining/sourcing through the pins. If you want to be sure, you need to use shadow registers instead of directly reading the hardware. UINT8 shadowB; shadowB = (shadowB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble PORTB = shadowB; // Send updated shadow to hardware Andy WF <wfKILLspamBLUSOFT.ORG.BR> on 06/13/2000 03:42:33 PM Please respond to pic microcontroller discussion list <.....PICLISTKILLspam.....MITVMA.MIT.EDU> To: EraseMEPICLISTspam_OUTTakeThisOuTMITVMA.MIT.EDU cc: (bcc: Andrew Kunz/TDI_NOTES) Subject: Re: [OT]: CCS C COMPILER My question is: Will affect the remainder bits? The remainders bits are configurated as INPUT and OUTPUT... Miguel ----- Original Message ----- From: Andrew Kunz <akunzspam_OUTTDIPOWER.COM> To: <@spam@PICLISTKILLspamMITVMA.MIT.EDU> Sent: Tuesday, June 13, 2000 4:42 PM Subject: Re: [OT]: CCS C COMPILER > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > Andy

It will affect the output bits, but not the inputs. All the bits are read in the first instance, then the high or low value of the bits you are keeping are re-written to portB, along with the low or high value of the bits your are writing to portb. This can cause a problem in this way: Say you've set bit 7 as output, and are driving it high. If an external load is driving the pin low (and doing a better job than the PIC) then when you read in PORTB it will report that that pin is low, even though you previously told it to output high. When you write the modified value back to the port it will set that pin low. See the instructions BCF and BSF from the data sheet. In a nutshell, ports do not exist as registers. If you only modify a portion of a port you will read the actual value on the pin, not the intended value. -Adam WF wrote: {Quote hidden}> > My question is: > > Will affect the remainder bits? The remainders bits are configurated as > INPUT and OUTPUT... > > Miguel > ----- Original Message ----- > From: Andrew Kunz <KILLspamakunzKILLspamTDIPOWER.COM> > To: <RemoveMEPICLISTTakeThisOuTMITVMA.MIT.EDU> > Sent: Tuesday, June 13, 2000 4:42 PM > Subject: Re: [OT]: CCS C COMPILER > > > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > > > Andy

You are correct...today i do BIT manipulation...but, i want to reduce my code... ----- Original Message ----- From: M. Adam Davis <spamBeGoneadavisspamBeGoneUBASICS.COM> To: <TakeThisOuTPICLISTEraseMEspam_OUTMITVMA.MIT.EDU> Sent: Tuesday, June 13, 2000 4:59 PM Subject: Re: [OT]: CCS C COMPILER > It will affect the output bits, but not the inputs. All the bits are read in > the first instance, then the high or low value of the bits you are keeping are > re-written to portB, along with the low or high value of the bits your are > writing to portb. > > This can cause a problem in this way: > Say you've set bit 7 as output, and are driving it high. > If an external load is driving the pin low (and doing a better job than the PIC) > then when you read in PORTB it will report that that pin is low, even though you > previously told it to output high. When you write the modified value back to > the port it will set that pin low. > > See the instructions BCF and BSF from the data sheet. In a nutshell, ports do > not exist as registers. If you only modify a portion of a port you will read {Quote hidden}> the actual value on the pin, not the intended value. > > -Adam > > WF wrote: > > > > My question is: > > > > Will affect the remainder bits? The remainders bits are configurated as > > INPUT and OUTPUT... > > > > Miguel > > ----- Original Message ----- > > From: Andrew Kunz <RemoveMEakunzTakeThisOuTTDIPOWER.COM> > > To: <PICLISTEraseME.....MITVMA.MIT.EDU> > > Sent: Tuesday, June 13, 2000 4:42 PM > > Subject: Re: [OT]: CCS C COMPILER > > > > > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > > > > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > > > > > Andy

You are correct...today i do BIT manipulation...but, i want to reduce my code... ----- Original Message ----- From: M. Adam Davis <EraseMEadavisUBASICS.COM> To: <RemoveMEPICLISTEraseMEEraseMEMITVMA.MIT.EDU> Sent: Tuesday, June 13, 2000 4:59 PM Subject: Re: [OT]: CCS C COMPILER > It will affect the output bits, but not the inputs. All the bits are read in > the first instance, then the high or low value of the bits you are keeping are > re-written to portB, along with the low or high value of the bits your are > writing to portb. > > This can cause a problem in this way: > Say you've set bit 7 as output, and are driving it high. > If an external load is driving the pin low (and doing a better job than the PIC) > then when you read in PORTB it will report that that pin is low, even though you > previously told it to output high. When you write the modified value back to > the port it will set that pin low. > > See the instructions BCF and BSF from the data sheet. In a nutshell, ports do > not exist as registers. If you only modify a portion of a port you will read {Quote hidden}> the actual value on the pin, not the intended value. > > -Adam > > WF wrote: > > > > My question is: > > > > Will affect the remainder bits? The remainders bits are configurated as > > INPUT and OUTPUT... > > > > Miguel > > ----- Original Message ----- > > From: Andrew Kunz <RemoveMEakunzspam_OUTKILLspamTDIPOWER.COM> > > To: <RemoveMEPICLISTTakeThisOuTspamMITVMA.MIT.EDU> > > Sent: Tuesday, June 13, 2000 4:42 PM > > Subject: Re: [OT]: CCS C COMPILER > > > > > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > > > > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > > > > > Andy

It will affect the output bits, but not the inputs. All the bits are read in the first instance, then the high or low value of the bits you are keeping are re-written to portB, along with the low or high value of the bits your are writing to portb. This can cause a problem in this way: Say you've set bit 7 as output, and are driving it high. If an external load is driving the pin low (and doing a better job than the PIC) then when you read in PORTB it will report that that pin is low, even though you previously told it to output high. When you write the modified value back to the port it will set that pin low. See the instructions BCF and BSF from the data sheet. In a nutshell, ports do not exist as registers. If you only modify a portion of a port you will read the actual value on the pin, not the intended value. -Adam WF wrote: {Quote hidden}> > My question is: > > Will affect the remainder bits? The remainders bits are configurated as > INPUT and OUTPUT... > > Miguel > ----- Original Message ----- > From: Andrew Kunz <EraseMEakunzspamspamBeGoneTDIPOWER.COM> > To: <RemoveMEPICLISTKILLspamMITVMA.MIT.EDU> > Sent: Tuesday, June 13, 2000 4:42 PM > Subject: Re: [OT]: CCS C COMPILER > > > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > > > Andy

It depends on how much you are draining/sourcing through the pins. If you want to be sure, you need to use shadow registers instead of directly reading the hardware. UINT8 shadowB; shadowB = (shadowB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble PORTB = shadowB; // Send updated shadow to hardware Andy WF <wfSTOPspamspam_OUTBLUSOFT.ORG.BR> on 06/13/2000 03:42:33 PM Please respond to pic microcontroller discussion list <spamBeGonePICLISTSTOPspamEraseMEMITVMA.MIT.EDU> To: KILLspamPICLISTspamBeGoneMITVMA.MIT.EDU cc: (bcc: Andrew Kunz/TDI_NOTES) Subject: Re: [OT]: CCS C COMPILER My question is: Will affect the remainder bits? The remainders bits are configurated as INPUT and OUTPUT... Miguel ----- Original Message ----- From: Andrew Kunz <EraseMEakunzEraseMETDIPOWER.COM> To: <@spam@PICLIST@spam@spam_OUTMITVMA.MIT.EDU> Sent: Tuesday, June 13, 2000 4:42 PM Subject: Re: [OT]: CCS C COMPILER > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > Andy

My question is: Will affect the remainder bits? The remainders bits are configurated as INPUT and OUTPUT... Miguel ----- Original Message ----- From: Andrew Kunz <spamBeGoneakunzKILLspamTDIPOWER.COM> To: <.....PICLISTspam_OUTMITVMA.MIT.EDU> Sent: Tuesday, June 13, 2000 4:42 PM Subject: Re: [OT]: CCS C COMPILER > PORTB = (PORTB & 0x0F) | (NEWVALUE & 0xF0); // Updates high nybble > > PORTB = (PORTB & 0xF0) | (NEWVALUE & 0x0F); // Updates low nybble > > Andy