Get BNs Z coordinates from the BN_CTX

While there likely won't be enough BNs already available in the ctx, and
thus it won't greatly reduce the amount of allocated BNs, it simplifies
the exit path quite a bit.

review feedback from jsing

1 files changed, 3 insertions, 8 deletions

diff --git a/src/lib/libcrypto/ec/ecp_methods.c b/src/lib/libcrypto/ec/ecp_methods.c
index a4713bdcdc..b394206aea 100644
--- a/src/lib/libcrypto/ec/ecp_methods.c
+++ b/src/lib/libcrypto/ec/ecp_methods.c
@@ -1,4 +1,4 @@
-/* $OpenBSD: ecp_methods.c,v 1.10 2024/11/30 16:18:01 tb Exp $ */
+/* $OpenBSD: ecp_methods.c,v 1.11 2024/11/30 16:34:34 tb Exp $ */
 /* Includes code written by Lenka Fibikova <fibikova@exp-math.uni-essen.de>
  * for the OpenSSL project.
  * Includes code written by Bodo Moeller for the OpenSSL project.
@@ -1141,7 +1141,7 @@ ec_points_make_affine(const EC_GROUP *group, size_t num, EC_POINT *points[],
         if ((prod_Z = calloc(num, sizeof *prod_Z)) == NULL)
                 goto err;
         for (i = 0; i < num; i++) {
-                if ((prod_Z[i] = BN_new()) == NULL)
+                if ((prod_Z[i] = BN_CTX_get(ctx)) == NULL)
                         goto err;
         }
 
@@ -1253,12 +1253,7 @@ ec_points_make_affine(const EC_GROUP *group, size_t num, EC_POINT *points[],
 
  err:
         BN_CTX_end(ctx);
-
-        if (prod_Z != NULL) {
-                for (i = 0; i < num; i++)
-                        BN_free(prod_Z[i]);
-                free(prod_Z);
-        }
+        free(prod_Z);
 
         return ret;
 }