ODR IDR question

[xtian]

Posted on January 21, 2013 at 00:35

Hi guys,

I have this question, /* GPIOA Periph clock enable */   RCC_APB2PeriphClockCmd(RCC_APB2Periph_GPIOA, ENABLE);   /* Configure PA0 .. PA7 in output pushpull mode */   GPIO_InitStructure.GPIO_Pin = GPIO_Pin_0 | GPIO_Pin_1 | GPIO_Pin_2 | GPIO_Pin_3 | GPIO_Pin_4 | GPIO_Pin_5 | GPIO_Pin_6 | GPIO_Pin_7;   GPIO_InitStructure.GPIO_Speed = GPIO_Speed_50MHz;   GPIO_InitStructure.GPIO_Mode = GPIO_Mode_Out_PP;   GPIO_Init(GPIOA, &GPIO_InitStructure);   /* Configure PA8 .. PA15 in input, pulled up mode */   GPIO_InitStructure.GPIO_Pin = GPIO_Pin_8 | GPIO_Pin_9 | GPIO_Pin_10 | GPIO_Pin_11 | GPIO_Pin_12 | GPIO_Pin_13 | GPIO_Pin_14 | GPIO_Pin_15;   GPIO_InitStructure.GPIO_Mode = GPIO_Mode_IPU;   GPIO_Init(GPIOA, &GPIO_InitStructure);   GPIOA->ODR = GPIOA->IDR >> 8; // Mirror high ordered inputs to low ordered outputs could you explain to me the bold one especially ''8'' and IDR for example I want to mirror only the first four bits from high ordered to low ordered or last four bits from high ordered to low ordered. Thank you very much!

[Tesla DeLorean]

Posted on January 21, 2013 at 00:54

It's a shift right by 8 bits, So bit 8 goes to bit 0, and bit 15 to bit 7. In all A[8..15] to A[0..7]. It is a basic C operator.

So reading IDR as  XXXXXXXX-------- And writing ODR as 00000000XXXXXXXX So A[12..15] reflected to A[0..3] GPIOA->ODR = (GPIOA->ODR & 0x00F0) | ((GPIOA->IDR >> 8) & 0x000F);

[xtian]

Posted on January 21, 2013 at 01:44

Thanks clive... sorry for my ignorance but im reading books in c now but in parallel

So A[12..15] reflected to A[0..3] GPIOA->ODR = (GPIOA->ODR & 0x00F0) | ((GPIOA->IDR >> 8) & 0x000F); -> means? the value of GPIOA is pointed to the value of ODR, where ODR is equal to (GPIOA pointed to the values of 0000XXXX <12-15>) of (GPIOA pointed to the values of XXXX0000 <0-3>) correct?

[Tesla DeLorean]

Posted on January 21, 2013 at 03:18

Sorry that should have read

''So A[8..11] reflected to A[0..3]'' Whereas A[12..15] reflected to A[0..3] would be GPIOA->ODR = (GPIOA->ODR & 0x00F0) | ((GPIOA->IDR >> 12) & 0x000F);

[xtian]

Posted on January 21, 2013 at 03:24 Thanks clive >> 12 to apply the mirror on the 12 bit.. OK.. thanks much

[xtian]

Posted on January 21, 2013 at 03:26

just to finalize my questions.. hmm what does | do..

[xtian]

Posted on January 21, 2013 at 03:35

ahh clive is this what implements the pipeline in the ARM M series architecture?

[Tesla DeLorean]

Posted on January 21, 2013 at 04:14

ahh clive is this what implements the pipeline in the ARM M series architecture?

No, logical OR. The & is a logical AND You really need to get a decent text on the C language, like K&R