"Phil Endecott" <spam_from_gcc_help@xxxxxxxxxxxx> writes: > Could you help me out with an example? Say I have a uint8_t* that I > want to write to: > > uint8_t* ptr = ......; > *ptr = 123; > > How do I change that so that it does a read/modify/write? Are you > saying that I can do something like > > volatile uint32_t* ptr = .....; > uint8_t* ptr2 = ptr; > *ptr2 = 123; > > with appropriate extra casts and 'volatile's, and it will do what I want? Oh, I see, you want the compiler to figure stuff out for you. Unfortunately, that won't work. You need to write the 32-bit read/modify/write instructions yourself. volatile uint32_t* ptr = .....; uint32_t v = *ptr; v = (v & 0xffffff00) | 123; *ptr = v; Note that if you don't use the volatile qualifier, gcc is wholly capable of optimizing that into an 8-bit memory write. There is nothing in gcc which will force it to use 32-bit memory accesses if you don't explicitly write 32-bit memory accesses. Ian
