[Mesa-dev] [PATCH] st/mesa: init hash keys with memset(), not designated initializers

[Eric Anholt]

Brian Paul <brianp at vmware.com> writes:

> On 03/11/2019 04:47 PM, Eric Anholt wrote:
>> Brian Paul <brianp at vmware.com> writes:
>> 
>>> Since the compiler may not zero-out padding in the object.
>>> Add a couple comments about this to prevent misunderstandings in
>>> the future.
>>>
>>> Fixes: 67d96816ff5 ("st/mesa: move, clean-up shader variant key decls/inits")
>>> ---
>>>   src/mesa/state_tracker/st_atom_shader.c |  9 +++++++--
>>>   src/mesa/state_tracker/st_program.c     | 13 ++++++++++---
>>>   2 files changed, 17 insertions(+), 5 deletions(-)
>>>
>>> diff --git a/src/mesa/state_tracker/st_atom_shader.c b/src/mesa/state_tracker/st_atom_shader.c
>>> index ac7a1a5..a4475e2 100644
>>> --- a/src/mesa/state_tracker/st_atom_shader.c
>>> +++ b/src/mesa/state_tracker/st_atom_shader.c
>>> @@ -112,7 +112,10 @@ st_update_fp( struct st_context *st )
>>>          !stfp->variants->key.bitmap) {
>>>         shader = stfp->variants->driver_shader;
>>>      } else {
>>> -      struct st_fp_variant_key key = {0};
>>> +      struct st_fp_variant_key key;
>>> +
>>> +      /* use memset, not an initializer to be sure all memory is zeroed */
>>> +      memset(&key, 0, sizeof(key));
>> 
>> Wait, what?  We rely on this form of initialization all over, what's
>> changed?
>
> The question is do all compilers, when presented with
>
> struct st_fp_variant_key key = {0};

You seem to be saying that not all compilers do, but throughout the
compiler we're relying on all compilers doing so.  Can we get some more
information about what compiler you're fixing here?
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