PIC programming question - Electronic Circuits Projects Diagrams Free

ljcox · 17th January 2009, 07:36 AM

I downloaded a PIC programme from the internet for a PIC16F88.

PORTA 1, 2 & 3 are used as analogue inputs.

But there is an instruction in the programme that I don’t understand.

movlw (1<<ana1)|(1<<ana2)|(1<<ana3)

I’m familiar with the normal movlw instruction, eg. movlw 0x1A, but not with the one above.

In the definitions, the parameters are defined as:-

ana3 equ 3 ; analog input AN3

ana2 equ 2 ; analog input AN2

ana1 equ 1 ; analog input AN1

Any assistance will be appreciated.

be80be · 17th January 2009, 07:45 AM

Your setting the bit to 1

be80be · 17th January 2009, 07:51 AM

Bitwise inclusive OR

Pommie · 17th January 2009, 07:54 AM

The "<<" directive shifts the number left by the number to the right.

So,
1<<3 = 8
1<<6 = 64

and so the movlw instruction above if the same as movlw b'00001110'.

Mike.

be80be · 17th January 2009, 08:23 AM

So it is moving a 1 right so it would put

Code:

 b'00001110'

but why the pipe | is it for stacking

Code:

movlw (1<<ana1)|(1<<ana2)|(1<<ana3)

which is the same as this

Code:

movlw b'00001110

And this set a 0

Code:

 
movlw b'11111111'-(1<<ana1)-(1<<ana2)-(1<<ana3)

which is

Code:

movlw b'11110001

futz · 17th January 2009, 08:33 AM

Quote:

Originally Posted by be80be

So it is moving a 1 right so it would put

Code:

 b'00001110'

but why the pipe | is it for stacking

The pipe means OR. I've seen people do this in C too. What it usually means is that they don't know how to convert to binary. This gets the bits in there in a roundabout way.

Pommie · 17th January 2009, 08:35 AM

The | is inclusive or and so,

b'00000010' or b'00000100' or b'00001000' = b'00001110'

Mike.

ljcox · 17th January 2009, 07:56 PM

Thanks for the prompt responses. Much appreciated.

I understand. eg. 1<<n means shift left n places.

| means inclusive OR.

It seems a complicated way to do something simple, but I guess he had his reasons.

**Nigel Goodwin** · 17th January 2009, 08:23 PM

Quote:

Originally Posted by ljcox

Thanks for the prompt responses. Much appreciated.

I understand. eg. 1<<n means shift left n places.

| means inclusive OR.

It seems a complicated way to do something simple, but I guess he had his reasons.

Isn't it just standard C? - and I don't even do C.

be80be · 17th January 2009, 10:23 PM

Quote:

Arithmetic operators may be used with directives and their variables.
These operators cannot be used with program variables. They are for use with directives only.

From Mplab Assembler

Pommie · 18th January 2009, 02:47 AM

The reason it is done this way is so that if sometime in the future the design was changed and instead of A2, A4 was used then it is a simple matter to change,

ana2 equ 4 ; analog input AN2

rather than trying to find all occurrences in the code.

I use to use this method to setup SFRs. EG,

Code:

	movlw	(1<<GIE|1<<PEIE|0<<TMR0IE|0<<INTE|0<<RBIE|0<<TMR0IF|0<<INTF|0<<RBIF)
	movwf	INTCON;		enable Peripheral interrupts

but it confused people and so I stopped using it.

Mike.

ljcox · 18th January 2009, 08:53 PM

Quote:

Originally Posted by Nigel Goodwin

Isn't it just standard C? - and I don't even do C.

It may be, but I know nothing at all about C.

ljcox · 18th January 2009, 08:57 PM

Quote:

Originally Posted by Pommie

The reason it is done this way is so that if sometime in the future the design was changed and instead of A2, A4 was used then it is a simple matter to change,

ana2 equ 4 ; analog input AN2

rather than trying to find all occurrences in the code.

I use to use this method to setup SFRs. EG,

Code:

	movlw	(1<<GIE|1<<PEIE|0<<TMR0IE|0<<INTE|0<<RBIE|0<<TMR0IF|0<<INTF|0<<RBIF)
	movwf	INTCON;		enable Peripheral interrupts

but it confused people and so I stopped using it.

Mike.

Thanks Mike.

be80be · 18th January 2009, 11:23 PM

I like It. It works for me and makes a lot of sense to me.
_______________________________________
But I have never been confused

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17th January 2009, 07:36 AM	#1
ljcox	PIC programming question I downloaded a PIC programme from the internet for a PIC16F88. PORTA 1, 2 & 3 are used as analogue inputs. But there is an instruction in the programme that I don’t understand. movlw (1<<ana1)\|(1<<ana2)\|(1<<ana3) I’m familiar with the normal movlw instruction, eg. movlw 0x1A, but not with the one above. In the definitions, the parameters are defined as:- ana3 equ 3 ; analog input AN3 ana2 equ 2 ; analog input AN2 ana1 equ 1 ; analog input AN1 Any assistance will be appreciated. __________________ Len

17th January 2009, 07:45 AM	#2
be80be	Your setting the bit to 1 Last edited by be80be; 17th January 2009 at 07:46 AM.

17th January 2009, 07:54 AM	#4
Pommie	The "<<" directive shifts the number left by the number to the right. So, 1<<3 = 8 1<<6 = 64 and so the movlw instruction above if the same as movlw b'00001110'. Mike.

17th January 2009, 08:23 AM	#5
be80be	So it is moving a 1 right so it would put Code: b'00001110' but why the pipe \| is it for stacking Code: movlw (1<<ana1)\|(1<<ana2)\|(1<<ana3) which is the same as this Code: movlw b'00001110 And this set a 0 Code: movlw b'11111111'-(1<<ana1)-(1<<ana2)-(1<<ana3) which is Code: movlw b'11110001

17th January 2009, 08:35 AM	#7
Pommie	The \| is inclusive or and so, b'00000010' or b'00000100' or b'00001000' = b'00001110' Mike.

17th January 2009, 07:56 PM	#8
ljcox	Thanks for the prompt responses. Much appreciated. I understand. eg. 1<<n means shift left n places. \| means inclusive OR. It seems a complicated way to do something simple, but I guess he had his reasons. __________________ Len

18th January 2009, 02:47 AM	#11
Pommie	The reason it is done this way is so that if sometime in the future the design was changed and instead of A2, A4 was used then it is a simple matter to change, ana2 equ 4 ; analog input AN2 rather than trying to find all occurrences in the code. I use to use this method to setup SFRs. EG, Code: movlw (1<<GIE\|1<<PEIE\|0<<TMR0IE\|0<<INTE\|0<<RBIE\|0<<TMR0IF\|0<<INTF\|0<<RBIF) movwf INTCON; enable Peripheral interrupts but it confused people and so I stopped using it. Mike.

18th January 2009, 11:23 PM	#14
be80be	I like It. It works for me and makes a lot of sense to me. _______________________________________ But I have never been confused