On Fri, Jul 31, 2020 at 11:36:04AM -0300, Jason Gunthorpe wrote:
> On Fri, Jul 31, 2020 at 04:21:48PM +0200, Greg Kroah-Hartman wrote:
>
> > > The spec was updated in C11 to require zero'ing padding when doing
> > > partial initialization of aggregates (eg = {})
> > >
> > > """if it is an aggregate, every member is initialized (recursively)
> > > according to these rules, and any padding is initialized to zero
> > > bits;"""
> >
> > But then why does the compilers not do this?
>
> Do you have an example?
At the moment, no, but we have had them in the past due to security
issues we have had to fix for this.
> > > Considering we have thousands of aggregate initializers it
> > > seems likely to me Linux also requires a compiler with this C11
> > > behavior to operate correctly.
> >
> > Note that this is not an "operate correctly" thing, it is a "zero out
> > stale data in structure paddings so that data will not leak to
> > userspace" thing.
>
> Yes, not being insecure is "operate correctly", IMHO :)
>
> > > Does this patch actually fix anything? My compiler generates identical
> > > assembly code in either case.
> >
> > What compiler version?
>
> I tried clang 10 and gcc 9.3 for x86-64.
>
> #include <string.h>
>
> void test(void *out)
> {
> struct rds_rdma_notify {
> unsigned long user_token;
> unsigned int status;
> } foo = {};
> memcpy(out, &foo, sizeof(foo));
> }
>
> $ gcc -mno-sse2 -O2 -Wall -std=c99 t.c -S
>
> test:
> endbr64
> movq $0, (%rdi)
> movq $0, 8(%rdi)
> ret
>
> Just did this same test with gcc 4.4 and it also gave the same output..
>
> Made it more complex with this:
>
> struct rds_rdma_notify {
> unsigned long user_token;
> unsigned char status;
> unsigned long user_token1;
> unsigned char status1;
> unsigned long user_token2;
> unsigned char status2;
> unsigned long user_token3;
> unsigned char status3;
> unsigned long user_token4;
> unsigned char status4;
> } foo;
>
> And still got the same assembly vs memset on gcc 4.4.
>
> I tried for a bit and didn't find a way to get even old gcc 4.4 to not
> initialize the holes.
Odd, so it is just the "= {0};" that does not zero out the holes?
thanks,
greg k-h
