PIC programming question

[ljcox]

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]

Your setting the bit to 1

 

[be80be]

Bitwise inclusive OR

 

[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.

 

[be80be]

So it is moving a 1 right so it would put

 b'00001110'

but why the pipe | is it for stacking

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

which is the same as this

movlw b'00001110

And this set a 0

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

which is

movlw b'11110001

 

[futz]

So it is moving a 1 right so it would put

 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]

The | is inclusive or and so,

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

Mike.

 

[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.

 

[Nigel Goodwin]

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]

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]

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,

	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]

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]

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,

	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]

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