@@ -216,14 +216,54 @@ static ssize_t lpc_debug_read(struct file *filp, char __user *ubuf,
&data, len);
if (rc)
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) {
case 4:
rc = __put_user((u32)data, (u32 __user *)ubuf);
break;
case 2:
+#ifdef __LITTLE_ENDIAN__
+ data >>= 16;
+#endif
rc = __put_user((u16)data, (u16 __user *)ubuf);
break;
default:
+#ifdef __LITTLE_ENDIAN__
+ data >>= 24;
+#endif
rc = __put_user((u8)data, (u8 __user *)ubuf);
break;
}
@@ -263,12 +303,31 @@ static ssize_t lpc_debug_write(struct file *filp, const char __user *ubuf,
else if (todo > 1 && (pos & 1) == 0)
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) {
case 4:
rc = __get_user(data, (u32 __user *)ubuf);
+ data = cpu_to_be32(data);
break;
case 2:
rc = __get_user(data, (u16 __user *)ubuf);
+ data = cpu_to_be16(data);
break;
default:
rc = __get_user(data, (u8 __user *)ubuf);
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: Benjamin Herrenschmidt <benh@kernel.crashing.org> CC: <stable@vger.kernel.org>