16bit unsigned bit fields

Nitin Manohar

I'm porting a vendor's source code to TI DSP 6410 platform.

I get "nonstandard type for a bit field" warning from compiler when following structure declaration is compiled:

typedef unsigned short int cs_uint16 ;

typedef unsigned int cs_uint32 ;

typedef union {
struct {
    cs_uint16 HSIFe                : 1 ; /* bits 0:0 */
    cs_uint16 XFIe                 : 1 ; /* bits 1:1 */
    cs_uint16 N40Ge                : 1 ; /* bits 2:2 */
    cs_uint16 OHPPe                : 1 ; /* bits 3:3 */
    cs_uint16 N10G0e               : 1 ; /* bits 4:4 */
    cs_uint16 N10G1e               : 1 ; /* bits 5:5 */
    cs_uint16 N10G2e               : 1 ; /* bits 6:6 */
    cs_uint16 N10G3e               : 1 ; /* bits 7:7 */
    cs_uint16 PP10G0e              : 1 ; /* bits 8:8 */
    cs_uint16 PP10G1e              : 1 ; /* bits 9:9 */
    cs_uint16 PP10G2e              : 1 ; /* bits 10:10 */
    cs_uint16 PP10G3e              : 1 ; /* bits 11:11 */
    cs_uint16 PP40Ge               : 1 ; /* bits 12:12 */
    cs_uint16 CFPe                 : 1 ; /* bits 13:13 */
    cs_uint16 CUPLLe               : 1 ; /* bits 14:14 */
    cs_uint16 rsrvd1               : 1 ;
} bf ;

cs_uint16 wrd ;
} TEN_MPIF_MODULE_B_INTENABLE_t;

In place of cs_unit16, if I use cs_unit32, problem goes away.

Does TI compiler not support 16bit packed bit-fields?

I can't change this structure to 32 bit wide because this structure is used to R/W hardware memory which is 16 bit wide.

Please help.

thanks,

Nitin

over 14 years ago

0 AartiG over 14 years ago

TI__Guru**** 173310 points

According to the C89 standard, which our compiler adheres to, "unsigned short" is not a standard type for bit fields. Some compilers may support it as a non standard extension, however that will make the code not guaranteed to be portable between compilers. Some discussion about it is here: http://processors.wiki.ti.com/index.php/C6000_EABI_Migration#Bit-Field_Layout

The -pr option(relaxed ANSI check) will help avoid the warning as it tells the compiler to not perform strict ANSI C checking.

0 Paul Knueven over 14 years ago in reply to AartiG

TI__Expert 4280 points

Aarti never says it explicitly, but note that the TI compiler does support bit-fields with type unsigned short; it just warns that it is nonstandard (with respect to C89).

Bit-fields themselves are not all that portable because the rules for layout in memory allow some variation. Most compilers will support this kind of bit-field (and it is standard in C++). So portability is not likely to be further compromised by non-standard integral types in most situations.

Note that you can suppress just that specific warning with the -pds=232 option (long form: --diag_suppress=232) or by using the equivalent pragma in the source.

0 Nikolay Kokorin over 13 years ago in reply to Paul Knueven

Intellectual 875 points

Hi Paul,

So, If I defined bit-field structure

typedef struct {

char fieldA : 4;

char fieldB : 4;

char fieldC : 4;

char fieldD : 4;

}MyField;

How will be order of fields in the C67xx memory (little endian) ?

Such as below or not ?

| Memory Bits | Offset

| 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |

| fieldA | fieldB | 0 byte

| fieldC | fieldD | 1 byte

Thanks

0 _kirem rahmani_ over 13 years ago in reply to Nikolay Kokorin

Prodigy 75 points

Hi Nikolay ,

Sorry no-one has replied to your question.

Supposing you have a union defined with your structure the memory allocation would be as shown below.

typedef union uTest_u

{
 unsigned char b[2];
 struct {
 char     fieldA   :   4;
 char     fieldB   :   4;
 char     fieldC   :   4;
 char     fieldD   :   4;
 } bits;

} utest_u

uTest_u uTest;

|         Memory  Bits          |  Offset | 
| 7 | 6 | 5 | 4 | 3 | 2 | 1 | 0 |
|     fieldB    |     fieldA      |  0 byte | uTest.b[0]
|     fieldD    |     fieldC     |  1 byte | uTest.b[1]

Regards,

Kirem

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16bit unsigned bit fields