Commit 325e4114 authored by Benjamin Herrenschmidt's avatar Benjamin Herrenschmidt Committed by Michael Ellerman

powerpc/powernv: Properly fix LPC debugfs endianness

Endian is hard, especially when I designed a stupid FW interface, and
I should know better... oh well, this is attempt #2 at fixing this
properly. This time it seems to work with all access sizes and I
can run my flashing tool (which exercises all sort of access sizes
and types to access the SPI controller in the BMC) just fine.
Signed-off-by: default avatarBenjamin Herrenschmidt <benh@kernel.crashing.org>
CC: stable@vger.kernel.org
Signed-off-by: default avatarMichael Ellerman <mpe@ellerman.id.au>
parent 808be314
...@@ -216,14 +216,54 @@ static ssize_t lpc_debug_read(struct file *filp, char __user *ubuf, ...@@ -216,14 +216,54 @@ static ssize_t lpc_debug_read(struct file *filp, char __user *ubuf,
&data, len); &data, len);
if (rc) if (rc)
return -ENXIO; return -ENXIO;
/*
* Now there is some trickery with the data returned by OPAL
* as it's the desired data right justified in a 32-bit BE
* word.
*
* This is a very bad interface and I'm to blame for it :-(
*
* So we can't just apply a 32-bit swap to what comes from OPAL,
* because user space expects the *bytes* to be in their proper
* respective positions (ie, LPC position).
*
* So what we really want to do here is to shift data right
* appropriately on a LE kernel.
*
* IE. If the LPC transaction has bytes B0, B1, B2 and B3 in that
* order, we have in memory written to by OPAL at the "data"
* pointer:
*
* Bytes: OPAL "data" LE "data"
* 32-bit: B0 B1 B2 B3 B0B1B2B3 B3B2B1B0
* 16-bit: B0 B1 0000B0B1 B1B00000
* 8-bit: B0 000000B0 B0000000
*
* So a BE kernel will have the leftmost of the above in the MSB
* and rightmost in the LSB and can just then "cast" the u32 "data"
* down to the appropriate quantity and write it.
*
* However, an LE kernel can't. It doesn't need to swap because a
* load from data followed by a store to user are going to preserve
* the byte ordering which is the wire byte order which is what the
* user wants, but in order to "crop" to the right size, we need to
* shift right first.
*/
switch(len) { switch(len) {
case 4: case 4:
rc = __put_user((u32)data, (u32 __user *)ubuf); rc = __put_user((u32)data, (u32 __user *)ubuf);
break; break;
case 2: case 2:
#ifdef __LITTLE_ENDIAN__
data >>= 16;
#endif
rc = __put_user((u16)data, (u16 __user *)ubuf); rc = __put_user((u16)data, (u16 __user *)ubuf);
break; break;
default: default:
#ifdef __LITTLE_ENDIAN__
data >>= 24;
#endif
rc = __put_user((u8)data, (u8 __user *)ubuf); rc = __put_user((u8)data, (u8 __user *)ubuf);
break; break;
} }
...@@ -263,12 +303,31 @@ static ssize_t lpc_debug_write(struct file *filp, const char __user *ubuf, ...@@ -263,12 +303,31 @@ static ssize_t lpc_debug_write(struct file *filp, const char __user *ubuf,
else if (todo > 1 && (pos & 1) == 0) else if (todo > 1 && (pos & 1) == 0)
len = 2; len = 2;
} }
/*
* Similarly to the read case, we have some trickery here but
* it's different to handle. We need to pass the value to OPAL in
* a register whose layout depends on the access size. We want
* to reproduce the memory layout of the user, however we aren't
* doing a load from user and a store to another memory location
* which would achieve that. Here we pass the value to OPAL via
* a register which is expected to contain the "BE" interpretation
* of the byte sequence. IE: for a 32-bit access, byte 0 should be
* in the MSB. So here we *do* need to byteswap on LE.
*
* User bytes: LE "data" OPAL "data"
* 32-bit: B0 B1 B2 B3 B3B2B1B0 B0B1B2B3
* 16-bit: B0 B1 0000B1B0 0000B0B1
* 8-bit: B0 000000B0 000000B0
*/
switch(len) { switch(len) {
case 4: case 4:
rc = __get_user(data, (u32 __user *)ubuf); rc = __get_user(data, (u32 __user *)ubuf);
data = cpu_to_be32(data);
break; break;
case 2: case 2:
rc = __get_user(data, (u16 __user *)ubuf); rc = __get_user(data, (u16 __user *)ubuf);
data = cpu_to_be16(data);
break; break;
default: default:
rc = __get_user(data, (u8 __user *)ubuf); rc = __get_user(data, (u8 __user *)ubuf);
......
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