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Accesing bit addressable variables

Elliot Chisholm

I was wondering what the "best" way to set a bit in a variable declared in the bit addressable area.

Originally I declared each bit as an sbit and therefore can set and clear them as follows

uint8_t bdata status ;
sbit enable = staus ^ 0 ;

enable = 1 ;
enable = 0 ;
I think it looks clearer to do as follows
#define ENABLE 0x01

uint8_t bdata status ;

status |= ENABLE ;
status &= ~ENABLE ;

but I am worried that this could less maintainable because there will be many #defines with the same values but related to different variables and confusion could ensue.

I was wondering what other people do and if there is a standard method for doing this?

Parents
  • 0 in reply to Andy Neil

    The compiler does not generate the same code!

    ----- FUNCTION main (BEGIN) -----
 FILE: 'main.c'
    9: void main( void )
   10: {
   11:     enable = 1;
   12:     enable = 0;
00000F D200              SETB    enable
   13:
000011 C200              CLR     enable
   14:     status |= ENABLE;
000013 432001            ORL     status,#01H
   15:     status &= ~ENABLE;
000016 5320FE            ANL     status,#0FEH
   16: }
000019 22                RET
----- FUNCTION main (END) -------
    I am pretty sure that the compiler always treads variables stored in bdata identically with those in data. The only difference is that it is possible to locate a bit in a byte of bdata by using the sbit keyword.

Reply
  • 0 in reply to Andy Neil

    The compiler does not generate the same code!

    ----- FUNCTION main (BEGIN) -----
 FILE: 'main.c'
    9: void main( void )
   10: {
   11:     enable = 1;
   12:     enable = 0;
00000F D200              SETB    enable
   13:
000011 C200              CLR     enable
   14:     status |= ENABLE;
000013 432001            ORL     status,#01H
   15:     status &= ~ENABLE;
000016 5320FE            ANL     status,#0FEH
   16: }
000019 22                RET
----- FUNCTION main (END) -------
    I am pretty sure that the compiler always treads variables stored in bdata identically with those in data. The only difference is that it is possible to locate a bit in a byte of bdata by using the sbit keyword.

Children
  • 0 in reply to Graham Cole

    It looks like mine does. Maybe it has something to do with the way i have defined the variable.

    $NOMACRO
$SEGMENTED CASE MOD167
;
; 'main.SRC' GENERATED FROM INPUT FILE 'main.c'
; COMPILER INVOKED BY:
;	C:\Keil\C166\BIN\C166.EXE main.c BROWSE MOD167 DEBUG
;
$MODINF (43)
		 NAME MAIN

          NCODE  CGROUP  ?PR?MAIN
          SDATA  DGROUP  ?BD0?MAIN,SYSTEM

                 ASSUME  DPP3 : SDATA

      ?BD0?MAIN  SECTION  DATA BITADDRESSABLE 'BDATA0'
         status  DSB  1
         PUBLIC  status
      ?BD0?MAIN  ENDS

          EXTRN  ?C_STARTUP : NEAR
         enable  BIT   status.0
         PUBLIC  enable


                 REGDEF   R0 - R15


       ?PR?MAIN  SECTION  CODE WORD 'NCODE'
; line 1: #include "types.h"
; line 2:
; line 3: #define ENABLE 0x01
; line 4:
; line 5: uint8_t bdata status ;
; line 6:
; line 7: sbit enable = status ^ 0 ;
; line 8:
; line 9: #pragma SRC
; line 10:
; line 11: void main ( void )

	main  PROC  NEAR
	GLOBAL  main
; FUNCTION main (BEGIN  RMASK = @0x8000)
; line 12: {
; line 13:
; line 14:
; line 15:
; line 16: 	enable = 1 ;
	BSET	enable
; line 17: 	enable = 0 ;
	BCLR	enable
; line 18:
; line 19: 	status |= ENABLE ;
	BSET	(status).0
; line 20: 	status &= ~ENABLE ;
	BCLR	(status).0
; line 21:
; line 22: 	while(1)
; line 23: 	{
; line 24:
; line 25: 	}
?C0001:
	JMP	cc_UC,?C0001
; FUNCTION main (END    RMASK = @0x8000)
	main  ENDP
       ?PR?MAIN  ENDS


; line 26:
; line 27: }

	END

  • 0 in reply to Elliot Chisholm

    "Maybe it has something to do with the way i have defined the variable."

    Or because you are using a C166 compiler as opposed to the (apparent) use of C51 by other posters. In fairness to everyone, your posts did not specify a product and I would hazard a guess that when none is specified, C51 is assumed.

  • 0 in reply to Dan Henry

    "Or because you are using a C166 compiler as opposed to the (apparent) use of C51 by other posters."

    Which does, of course, beg the question: If C166 can do it, why can't (or doesn't) C51?

  • 0 in reply to Andy Neil

    A: Because the C167 has a different output structure.