Messages in this thread | | | Date | Tue, 13 Feb 2007 12:36:51 -0500 | From | Chuck Ebbert <> | Subject | Re: [PATCH] fix atl1 braino |
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Al Viro wrote: > Spot the bug... > > Signed-off-by: Al Viro <viro@zeniv.linux.org.uk> > --- > > diff --git a/drivers/net/atl1/atl1_hw.c b/drivers/net/atl1/atl1_hw.c > index 08b2d78..e28707a 100644 > --- a/drivers/net/atl1/atl1_hw.c > +++ b/drivers/net/atl1/atl1_hw.c > @@ -357,7 +357,7 @@ void atl1_hash_set(struct atl1_hw *hw, u32 hash_value) > */ > hash_reg = (hash_value >> 31) & 0x1; > hash_bit = (hash_value >> 26) & 0x1F; > - mta = ioread32((hw + REG_RX_HASH_TABLE) + (hash_reg << 2)); > + mta = ioread32((hw->hw_addr + REG_RX_HASH_TABLE) + (hash_reg << 2)); > mta |= (1 << hash_bit); > iowrite32(mta, (hw->hw_addr + REG_RX_HASH_TABLE) + (hash_reg << 2)); > }
The comment above is funny:
349 /* 350 * The HASH Table is a register array of 2 32-bit registers. 351 * It is treated like an array of 64 bits. We want to set 352 * bit BitArray[hash_value]. So we figure out what register 353 * the bit is in, read it, OR in the new bit, then write 354 * back the new value. The register is determined by the 355 * upper 7 bits of the hash value and the bit within that 356 * register are determined by the lower 5 bits of the value.
How can you use seven bits to choose between two registers? And they're not using the lower 5 bits, but rather bits 30..26.
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