Use snprintf in an expected way

If we want to get the size of buffer needed for snprintf, we
can call snprintf with NULL, we don't need any temporary array.
Addressing:
/home/jcerny/openscap/src/XCCDF/benchmark.c: In function ‘xccdf_benchmark_gen_id’:
/home/jcerny/openscap/src/XCCDF/benchmark.c:689:23: warning: ‘%03d’ directive output truncated writing 3 bytes into a region of size 1 [-Wformat-truncation=]
  689 |  const char *fmt = "%s%03d";
      |                       ^~~~
/home/jcerny/openscap/src/XCCDF/benchmark.c:691:15: note: ‘snprintf’ output 4 or more bytes into a destination of size 1
  691 |  int length = snprintf(foo, 1, fmt, prefix, 0);
      |               ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Author: jan-cerny

src/XCCDF/benchmark.c

@@ -687,8 +687,7 @@ char *xccdf_benchmark_gen_id(struct xccdf_benchmark *benchmark, xccdf_type_t typ
 	assert(prefix != NULL);
 
 	const char *fmt = "%s%03d";
-	char foo[2];
-	int length = snprintf(foo, 1, fmt, prefix, 0);
+	int length = snprintf(NULL, 0, fmt, prefix, 0);
 	if (length < 0)
 		return NULL;
 	length++;