Question

I want to write a byte to register with specific memory address (0x1228A432)

But, this register has a following structure:

 Bits  |   Access   |   Name   | Reset  | Description |
[31:8] | Read only  | -------- | ------ |  Reserved   |
[7:0]  | Read-write | REG[7:0] | 0xXX   | ----------- |

Please tell me, how to write a byte to this register without “touching” the Reserved bits?

EDIT1: My target is Cortex A9.
I could successfully read/write to onboard DDR2 memory using 256-bit values (such as 0xFF)

EDIT2: I used to work with DDR2 memory in the following way :

// First stage
static unsigned char *p = 0;
char * argv1="0x60000000";
unsigned long address=strtoul(argv1, 0, 0);
p = (unsigned char *) argv1;

// Second stage
char * argv4="FF";
int value=strtol(argv4,0,16);

// Third stage
int offset = 9;
p[offset]=value;

EDIT3: I found out the following information:

All registers are 32 bits wide and do not support byte writes.
Write operations must be word-wide and bits marked as reserved must be preserved
using read-modify-write.

Answer 1

One way to preserve bits [31:8], assuming 32-bit wide access, is to read the value, zero-out bits [7:0], bitwise-or it with the value needed and then write it back to the register.

Something like (stealing from RedX a bit 😉 ):

uint8_t your_8_bit_value = 0x42;
uint32_t volatile * const mem_map_register = (uint32_t volatile *) 0x1228a432;
*mem_map_register = (*mem_map_register & 0xFFFFFF00) | your_8_bit_value;

Yet I think there should be more info available about your hardware. I’ve seen several datasheets saying e.g. that you have to write all 1 to reserved bits (meaning that reserved bits are reserved for future use, and 1 is a safe default), etc. So it is not always obvious, that leaving reserved bits untouched is the right thing to do.

You should find more details about your hardware – are byte-wide writes supported, are writes to reserved bits ignored perhaps, or should be all 0/1, etc.