Fix divide by zero if macro used with wrong key

If PSA_CIPHER_ENCRYPT_OUTPUT_SIZE was called on a non symmetric key,
then a divide by zero could happen, as PSA_CIPHER_BLOCK_LENGTH will
return 0 for such a key, and PSA_ROUND_UP_TO_MULTIPLE will divide by the
block length.

Signed-off-by: Paul Elliott <paul.elliott@arm.com>
diff --git a/include/psa/crypto_sizes.h b/include/psa/crypto_sizes.h
index 2811c23..836f9ee 100644
--- a/include/psa/crypto_sizes.h
+++ b/include/psa/crypto_sizes.h
@@ -992,9 +992,11 @@
  */
 #define PSA_CIPHER_ENCRYPT_OUTPUT_SIZE(key_type, alg, input_length)             \
     (alg == PSA_ALG_CBC_PKCS7 ?                                                 \
+     (((key_type) & PSA_KEY_TYPE_CATEGORY_MASK)                                 \
+                                           == PSA_KEY_TYPE_CATEGORY_SYMMETRIC ? \
      PSA_ROUND_UP_TO_MULTIPLE(PSA_BLOCK_CIPHER_BLOCK_LENGTH(key_type),          \
                               (input_length) + 1) +                             \
-     PSA_CIPHER_IV_LENGTH((key_type), (alg)) :                                  \
+     PSA_CIPHER_IV_LENGTH((key_type), (alg)) : 0) :                             \
      (PSA_ALG_IS_CIPHER(alg) ?                                                  \
       (input_length) + PSA_CIPHER_IV_LENGTH((key_type), (alg)) :                \
      0))