bpf: Relax scalar id equivalence for state pruning

bpf: Relax scalar id equivalence for state pruning

Scalar register IDs are used by the verifier to track relationships
between registers and enable bounds propagation across those
relationships. Once an ID becomes singular (i.e. only a single
register/stack slot carries it), it can no longer contribute to bounds
propagation and effectively becomes stale. The previous commit makes the
verifier clear such ids before caching the state.

When comparing the current and cached states for pruning, these stale
IDs can cause technically equivalent states to be considered different
and thus prevent pruning.

For example, in the selftest added in the next commit, two registers -
r6 and r7 are not linked to any other registers and get cached with
id=0, in the current state, they are both linked to each other with
id=A.  Before this commit, check_scalar_ids would give temporary ids to
r6 and r7 (say tid1 and tid2) and then check_ids() would map tid1->A,
and when it would see tid2->A, it would not consider these state
equivalent.

Relax scalar ID equivalence by treating rold->id == 0 as "independent":
if the old state did not rely on any ID relationships for a register,
then any ID/linking present in the current state only adds constraints
and is always safe to accept for pruning. Implement this by returning
true immediately in check_scalar_ids() when old_id == 0.

Maintain correctness for the opposite direction (old_id != 0 && cur_id
== 0) by still allocating a temporary ID for cur_id == 0. This avoids
incorrectly allowing multiple independent current registers (id==0) to
satisfy a single linked old ID during mapping.

Signed-off-by: Puranjay Mohan <puranjay@kernel.org>
Link: https://lore.kernel.org/r/20260203165102.2302462-5-puranjay@kernel.org
Signed-off-by: Alexei Starovoitov <ast@kernel.org>