Rewrite the comment explaining why swiotlb copies the original buffer to
the TLB buffer before initiating DMA *from* the device, i.e. before the
device DMAs into the TLB buffer.  The existing comment's argument that
preserving the original data can prevent a kernel memory leak is bogus.

If the driver that triggered the mapping _knows_ that the device will
overwrite the entire mapping, or the driver will consume only the written
parts, then copying from the original memory is completely pointless.

If neither of the above holds true, then copying from the original adds
value only if preserving the data is necessary for functional
correctness, or the driver explicitly initialized the original memory.
If the driver didn't initialize the memory, then copying the original
buffer to the TLB buffer simply changes what kernel data is leaked to
user space.

Writing the entire TLB buffer _does_ prevent leaking stale TLB buffer
data from a previous bounce, but that can be achieved by simply zeroing
the TLB buffer when grabbing a slot.

The real reason swiotlb ended up initializing the TLB buffer with the
original buffer is that it's necessary to make swiotlb operate as
transparently as possible, i.e. to behave as closely as possible to
hardware, and to avoid corrupting the original buffer, e.g. if the driver
knows the device will do partial writes and is relying on the unwritten
data to be preserved.
Reviewed-by: Robin Murphy <robin.murphy@arm.com>
Link: https://lore.kernel.org/all/ZN5elYQ5szQndN8n@google.comSigned-off-by: Sean Christopherson <seanjc@google.com>
Signed-off-by: Christoph Hellwig <hch@lst.de>