optimize allocation for GNU systems. That is allocate only blocks of a size malloc handles.

src/memory.c

@@ -623,65 +623,13 @@ deleteElement(struct memSegment ** tree, struct memSegment * element)
     return del_node;
 }
 
-
-//static
-//void
-//traverse(struct memSegment * tree, void (*cb)(struct memSegment *, int))
-//{
-//    struct memSegment * previous = tree;
-//    struct memSegment * node     = tree;
-//    int              depth    = 1;
-//
-//    /*
-//     * I think this has something like O(n+log(n)) on a ballanced
-//     * tree because I have to traverse back the rightmost leaf to
-//     * the root to get a break condition.
-//     */
-//    while (node) {
-//        /*
-//         * If we come from the right so nothing and go to our
-//         * next parent.
-//         */
-//        if (previous == node->right) {
-//            previous = node;
-//            node     = node->parent;
-//            depth--;
-//            continue;
-//        }
-//
-//        if ((NULL == node->left || previous == node->left)) {
-//            /*
-//             * If there are no more elements to the left or we
-//             * came from the left, process data.
-//             */
-//            cb(node, depth);
-//            previous = node;
-//
-//            if (NULL != node->right) {
-//                node = node->right;
-//                depth++;
-//            } else {
-//                node = node->parent;
-//                depth--;
-//            }
-//        } else {
-//            /*
-//             * if there are more elements to the left go there.
-//             */
-//            previous = node;
-//            node     = node->left;
-//            depth++;
-//        }
-//    }
-//}
-
 static
 void
 post(struct memSegment * tree, void (*cb)(struct memSegment *, int))
 {
     struct memSegment * previous = tree;
     struct memSegment * node     = tree;
-    int              depth    = 1;
+    int                 depth    = 1;
 
     /*
      * I think this has something like O(n+log(n)) on a ballanced
@@ -732,42 +680,6 @@ post(struct memSegment * tree, void (*cb)(struct memSegment *, int))
     }
 }
 
-//void
-//printElement(struct memSegment * node, int depth)
-//{
-//    int  i;
-//
-//    printf("%s %010zu:%p(%02d)",
-//            (node->color==rbRed)?"R":"B",
-//            node->size,
-//            node->ptr,
-//            depth);
-//    for (i=0; i<depth; i++) printf("-");
-//    puts("");
-//
-//    node = node->next;
-//    while (NULL != node) {
-//        printf("  %s %010zu:%p(%02d)",
-//                (node->color==rbRed)?"R":"B",
-//                node->size,
-//                node->ptr,
-//                depth);
-//        for (i=0; i<depth; i++) printf("-");
-//        puts("");
-//        node = node->next;
-//    }
-//}
-
-//void
-//cleanup(struct memSegment * node, int depth)
-//{
-//    while (NULL != node) {
-//        struct memSegment * next = node->next;
-//        free(node);
-//        node = next;
-//    }
-//}
-
 static
 struct memSegment * segments = NULL;
 
@@ -793,20 +705,55 @@ TR_reference(void * mem)
 }
 
 /*
- * This will always allocate a multiple of PAGESIZE
+ * This tries to reflect the memory management behaviour of the
+ * GNU version of malloc. For other versions this might need
+ * to be changed to be optimal.
+ *
+ * However, GNU malloc keeps separate pools for each power of
+ * 2 memory size up to page size. So one page consists all of
+ * memory blocks of the same sizei (a power of 2).
+ *
+ * Also as far as I understand the smallest allocatable block is
+ * 8 bytes. At least the adresses are alwayse a multiple of 8.
+ *
+ * So lets say page size is 4096. There is nothing allocated
+ * right now. We allocate a block of 8 bytes. This will request
+ * a memory page from the OS. Then define it as a page containing
+ * 8 byte blocks and return the address of the first one of these.
+ * Any subsequent call to malloc for 8 bytes will return one of the
+ * blocks within this page as long as there are some left.
+ *
+ * So what we do here is up to page size round the request size up
+ * to the next power of 2 >= 8.
+ * Sizes greater then pagesize will be round up to the next
+ * multiple of pagesize. As far as I understand these are not
+ * pooled anyway.
+ *
+ * For now this assumes we are on a little endian machine.
  */
 void *
 TR_malloc(size_t size)
 {
-	struct memSegment * seg = NULL;
-	//long                psize = sysconf(_SC_PAGESIZE);
-	long                psize = 64;
+	struct memSegment * seg   = NULL;
+	long                psize = sysconf(_SC_PAGESIZE);
+	size_t              check;
 
 	size += sizeof(struct memSegment);
 
-	/* allocate only blocks of a multiple of pagesize, similar to cbuf */
-	size  = (0>=size)?1:(0!=size%psize)?(size/psize)+1:size/psize;
-	size *= psize;
+	if (size > psize) {
+		if (0 != (size % psize)) {
+			// size if not a multiple of pagesize so bring it to one.
+			size = ((size / psize) + 1) * psize;
+		}
+	} else {
+		check = size >> 1;
+		check = (size | check) - check;
+
+		if (check != size) {
+			// size is not a power of 2 so bring it to one.
+			size = ((size << 1) | size) - size;
+		}
+	}
 
 #ifdef MEM_OPT
 	seg = findElement(segments, size);