bit manipulation

[Marc]

I have 3 registry bits:
ADCONbits.CHS2=1
ADCONbits.CHS1=0
ADCONbits.CHS0=0

How can I define something which contains them so that I can write
CHS=100
and those bits take the values I want

Thank You

[kevcon]

You didn't say what processor you were using, so I wrote this is for the PIC18F65J10.

you should be able to convert it though.

[Marc]

Thank you but if I have:
unsigned char ADCON2;
#locate ADCON2=0x0FC0
struct {
unsigned ADCS0:1;
unsigned ADCS1:1;
unsigned ADCS2:1;
unsigned ACQT0:1;
unsigned ACQT1:1;
unsigned ACQT2:1;
unsigned :1;
unsigned ADFM:1;
} ADCON2bits;
#locate ADCON2bits=0x0FC0

without touching this structure, how can I use ADCS=110 ?

[future]

You can set bit by bit.

ADCON2.ADCS0 = 0;
ADCON2.ADCS1 = 1;
ADCON2.ADCS2 = 1;

or

ADCON2 &= 0b11111000;
ADCON2 |= 0b00000110;

[Marc]

Perhaps I can explain.
I need something that modify only three bits and have a name ADCS, I don't know but I think a structure or union that contains those bits

[PCM programmer]

Apparently you don't want to use the "dot" syntax to access the bitfield.
Instead, you want to access it with the symbolic name of "ADCS".

You can do this with a #define statement. Use Kevcon's structure
and define a symbolic name for that bitfield. Then access it by that
name in your main program. Example:

[Marc]

So with command

#define ADCS (ADCON0.CHS)

the compiler knows that ADCON0.CHS means ADCON0.CHS0 plus ADCON0.CHS1, ADCON0.CHS2, ADCON0.CHS3 (ADCON0.CHSx) ?
Or it works only in this situation? Because I'm speaking about grouping 3 bits in a structure which contains 8 bits each of them with a name