bitband technique

Hello Mohsin,

You can refer to the following E2E thread

http://e2e.ti.com/support/microcontrollers/tiva_arm/f/908/p/314176/1093601.aspx#1093601

Regards

Amit Ashara

Hi Amit,

Thanks for your reply. I can understand the following statement

HWREGBITW(&g_ui32Flags, 0) ^= 1;

but i fail to understand how following line of code

for(ui32Idx = 0; ui32Idx < 32; ui32Idx++)
{

HWREGBITW(&g_ui32Value, 31 - ui32Idx) = (0xdecafbad >> (31 - ui32Idx)) & 1;

 UARTprintf("\r%08x", g_ui32Value);

}

gives hex values

00000000 80000000 c0000000 c0000000 d0000000 d8000000 dc000000 de000000 de000000 de800000 dec00000 dec00000 dec00000 dec80000 dec80000 deca0000 deca0000 deca8000 decac000 decae000 decaf000 decaf800 decaf800 decafa00 decafb00 decafb80 decafb80 decafba0 decafba0 decafba8 decafbac decafbac decafbad

From my calculation first should be 0x80000000  instead of 00000000 as after right shifting 0xdecafbad by 31 we should get 0x00000001 & 1 = 1 on bit31 of ofg_ui32Value. Hence become 0x80000000  

Similarly after right shifting 0xdecafbad by 30 we should get 0x00000003 & 1 = 1 for bit30 ofg_ui32Value. hence become 0xC0000000 and so on

Kindly could you please explain me a more about this.

Regards,

Mohsin

Hello Mohsin,

Please note that you are writing to a bit-banded region. So the shift operator is selecting the bit value to be written to the SRAM bit banded address in the HWREGBITW which is itslef computed by using the SRAM address for g_ui32Value and to the bit banded location specified by 31-ui32Idx

Regards

Amit