ARM Technical Support Knowledge Articles

ORDER OF BITFIELDS

Applies to: C51 C Compiler

Answer


Information in this article applies to:


QUESTION

What direction does the compiler define bitfields? Is it LSB to MSB or the reverse.

ANSWER

The compiler defines bitfields from the least significant bit (LSB) to the most significant bit (MSB).

In the following example listing, you can see that the bits are arranged from the highest order byte to the lowest order byte and within each byte, from the LSB to the MSB.

The Module Information shows that the data size is 2 bytes (16 bits). This is because the structure was declared with int (or unsigned int). This causes 1-byte to be wasted if you use 8 bits or less. Declaring the structure with "unsigned char" will result in a data size of 1 byte (8 bits).

stmt level    source

   1          typedef struct
   2          {
   3            unsigned int bit_0 : 1;        // Byte 1, Bit 0
   4            unsigned int bit_1 : 1;        // Byte 1, Bit 1
   5            unsigned int bit_2 : 1;        // Byte 1, Bit 2
   6            unsigned int bit_3 : 1;        // Byte 1, Bit 3
   7            unsigned int bit_4 : 1;        // Byte 1, Bit 4
   8            unsigned int bit_5 : 1;        // Byte 1, Bit 5
   9            unsigned int bit_6 : 1;        // Byte 1, Bit 6
  10            unsigned int bit_7 : 1;        // Byte 1, Bit 7
  11            unsigned int bit_8 : 1;        // Byte 0, Bit 0
  12            unsigned int bit_9 : 1;        // Byte 0, Bit 1
  13            unsigned int bit_10: 1;        // Byte 0, Bit 2
  14            unsigned int bit_11: 1;        // Byte 0, Bit 3
  15            unsigned int bit_12: 1;        // Byte 0, Bit 4
  16            unsigned int bit_13: 1;        // Byte 0, Bit 5
  17            unsigned int bit_14: 1;        // Byte 0, Bit 6
  18            unsigned int bit_15: 1;        // Byte 0, Bit 7
  19          }bit_st;
  20
  21          bit_st x;
  22
  23
  24          void main(void)
  25          {
  26   1        x.bit_0 = 1;
  27   1        x.bit_1 = 1;
  28   1        x.bit_8 = 1;
  29   1        x.bit_15 = 1;
  30   1
  31   1        while(1);
  32   1      }

             ; FUNCTION main (BEGIN)
                                           ; SOURCE LINE # 24
                                           ; SOURCE LINE # 25
                                           ; SOURCE LINE # 26
0000 850000      R     MOV     x,x
0003 E500        R     MOV     A,x+01H
0005 4401              ORL     A,#01H
0007 F500        R     MOV     x+01H,A
                                           ; SOURCE LINE # 27
0009 850000      R     MOV     x,x
000C 4402              ORL     A,#02H
000E F500        R     MOV     x+01H,A
                                           ; SOURCE LINE # 28
0010 FF                MOV     R7,A
0011 E500        R     MOV     A,x
0013 4401              ORL     A,#01H
0015 F500        R     MOV     x,A
0017 8F00        R     MOV     x+01H,R7
                                           ; SOURCE LINE # 29
0019 4480              ORL     A,#080H
001B F500        R     MOV     x,A
001D 8F00        R     MOV     x+01H,R7
001F         ?C0001:
                                           ; SOURCE LINE # 31
001F 80FE              SJMP    ?C0001
             ; FUNCTION main (END)

SEE ALSO

Article last edited on: 2006-03-12 09:16:22

Rate this article

[Bad]
|
|
[Good]
Disagree? Move your mouse over the bar and click

Did you find this article helpful? Yes No

How can we improve this article?

Link to this article
Copyright © 2011 ARM Limited. All rights reserved. External (Open), Non-Confidential