The PSA Protected Storage credential backend (subsys/net/lib/tls_credentials/tls_credentials_trusted.c) declared its credential-store mutex as a plain zero-filled static struct k_mutex credential_lock; and never called k_mutex_init() on it. A statically zero-filled k_mutex has an uninitialized wait queue (its dlist head/tail are NULL instead of the self-referential sentinels that k_mutex_init/K_MUTEX_DEFINE install). The uncontended lock path does not touch the wait queue, so the defect is latent and serialized use behaves correctly.
When two execution contexts contend on the lock, k_mutex_lock() pends the blocking thread on the wait queue via z_pend_curr(), which calls sys_dlist_append() on the zeroed list and dereferences a NULL tail pointer (tail->next = node), faulting the kernel. The lock is held during TLS handshake credential loading and by all credential add/get/delete operations, so a deployment performing concurrent TLS handshakes (for example a server handling multiple simultaneous connections from a remote peer) or a credential-management operation concurrent with a handshake can trigger the dereference.
The impact is a denial of service: a deterministic kernel panic / device reset on the first contention. There is no memory corruption beyond the NULL dereference and no confidentiality or integrity impact; mutual exclusion on the fast path remains correct. Exposure is limited to builds with CONFIG_TLS_CREDENTIALS_BACKEND_PROTECTED_STORAGE enabled (PSA Protected Storage / TF-M platforms); the default volatile RAM backend initializes its lock correctly and is unaffected.
The fix initializes the mutex statically with K_MUTEX_DEFINE(credential_lock), providing a valid wait queue so the contended path no longer touches a NULL list.
Weakness
The product does not initialize or incorrectly initializes a resource, which might leave the resource in an unexpected state when it is accessed or used.
Potential Mitigations
- Use a language that does not allow this weakness to occur or provides constructs that make this weakness easier to avoid.
- For example, in Java, if the programmer does not explicitly initialize a variable, then the code could produce a compile-time error (if the variable is local) or automatically initialize the variable to the default value for the variable’s type. In Perl, if explicit initialization is not performed, then a default value of undef is assigned, which is interpreted as 0, false, or an equivalent value depending on the context in which the variable is accessed.