Hi Ayan,
On 21/10/2022 16:31, Ayan Kumar Halder wrote:
Defined readq_relaxed()/writeq_relaxed() to read and write 64 bit regs.
This in turn calls readl_relaxed()/writel_relaxed() twice for the lower
and upper 32 bits.
This needs an explanation why we can't use "strd/ldrd". And the same
would have to be duplicated in the code below.
Signed-off-by: Ayan Kumar Halder <ayank...@amd.com>
---
xen/arch/arm/include/asm/arm32/io.h | 4 ++++
1 file changed, 4 insertions(+)
diff --git a/xen/arch/arm/include/asm/arm32/io.h
b/xen/arch/arm/include/asm/arm32/io.h
index 73a879e9fb..6a5f563fbc 100644
--- a/xen/arch/arm/include/asm/arm32/io.h
+++ b/xen/arch/arm/include/asm/arm32/io.h
@@ -80,10 +80,14 @@ static inline u32 __raw_readl(const volatile void __iomem
*addr)
__raw_readw(c)); __r; })
#define readl_relaxed(c) ({ u32 __r = le32_to_cpu((__force __le32) \
__raw_readl(c)); __r; })
+#define readq_relaxed(c) ({ u64 __r = (le64_to_cpu(readl_relaxed(c+4)) << 32)
| \
+ readl_relaxed(c); __r; })
All the read*_relaxed are provide atomic read. This is not guaranteed by
your new helper. The name should be different (maybe
readq_relaxed_non_atomic()) to make clear of the difference.
I also don't quite understand the implementation. The value returned by
readl_relaxed() is already in the CPU endianess. So why do you call
le64_to_cpu() on top?
#define writeb_relaxed(v,c) __raw_writeb(v,c)
#define writew_relaxed(v,c) __raw_writew((__force u16) cpu_to_le16(v),c)
#define writel_relaxed(v,c) __raw_writel((__force u32) cpu_to_le32(v),c)
+#define writeq_relaxed(v,c) writel_relaxed(((uint64_t)v&0xffffffff), c); \
+ writel_relaxed((((uint64_t)v)>>32), (c+4));
This needs to be surrounded with do { } while (0), otherwise the
following would not properly work:
if ( foo )
writeq_relaxed(v, c);
Similarly, if 'v' is a call, then it will end up to be called twice.
#define readb(c) ({ u8 __v = readb_relaxed(c); __iormb(); __v; })
#define readw(c) ({ u16 __v = readw_relaxed(c); __iormb();
__v; })
Cheers,
--
Julien Grall
