On 07.03.2024 14:44, Oleksii wrote:
> On Thu, 2024-03-07 at 14:24 +0100, Jan Beulich wrote:
>> On 07.03.2024 14:01, Oleksii wrote:
>>> On Wed, 2024-03-06 at 15:13 +0100, Jan Beulich wrote:
>>>>> +/* Generic IO read/write. These perform native-endian
>>>>> accesses.
>>>>> */
>>>>> +static inline void __raw_writeb(uint8_t val, volatile void
>>>>> __iomem
>>>>> *addr)
>>>>> +{
>>>>> + asm volatile ( "sb %0, 0(%1)" : : "r" (val), "r" (addr) );
>>>>> +}
>>>>
>>>> I realize this is like Linux has it, but how is the compiler to
>>>> know
>>>> that
>>>> *addr is being access here?
>>> Assembler syntax told compiler that. 0(%1) - means that the memory
>>> location pointed to by the address in register %1.
>>
>> No, the compiler doesn't decompose the string to figure how operands
>> are used. That's what the constraints are for. The only two things
>> the
>> compiler does with the string is replace % operators and count line
>> separators.
> It looks like I am missing something.
>
> addr -> a some register ( because of "r" contraint ).
> val -> is also register ( because of "r" contraint ).
>
> So the compiler will update instert an instruction:
> sb reg1, 0(reg2)
>
> what means *(uint_8 *)reg2 = (uint8_t)reg1.
>
> What am I missing?
The fact that the compiler will not know that *(uint_8 *)reg2 actually
changes across this asm(). It may therefore continue to hold a cached
value in a register, without knowing that its contents went stale.
Jan
