Developer Information

This document provides information for developers who maintain or extend BTrees.


BTrees are defined using a “template”, roughly akin to a C++ template. To create a new family of BTrees, create a source file that defines macros used to handle differences in key and value types:

Configuration Macros


A string to hold an RCS/CVS Id key to be included in compiled binaries.


A string (like “IO” or “OO”) that provides the prefix used for the module. This gets used to generate type names and the internal module name string.

Macros for Keys


The C type declaration for keys (e.g., int or PyObject*).


Define if KEY_TYPE is a PyObject*`, else ``undef.


Tests whether the PyObject* K can be converted to the (C) key type (KEY_TYPE). The macro should return a boolean (zero for false, non-zero for true). When it returns false, its caller should probably set a TypeError exception.


Like KEY_CHECK, but only checked during __setitem__.


Like Python’s cmp(). Compares K(ey) to T(arget), where K and T are C values of type KEY_TYPE. V is assigned an int value depending on the outcome:

 < 0 if K < T
== 0 if K == T
 > 0 if K > T

This macro acts like an if, where the following statement is executed only if a Python exception has been raised because the values could not be compared.


K is a value of KEY_TYPE. If KEY_TYPE is a flavor of PyObject*, write this to do Py_DECREF(K). Else (e.g., KEY_TYPE is int) make it a nop.


K is a value of KEY_TYPE. If KEY_TYPE is a flavor of PyObject*, write this to do Py_INCREF(K). Else (e.g., KEY_TYPE is int) make it a nop.


Like K=E. Copy a key from E to K, both of KEY_TYPE. Note that this doesn’t decref K or incref E when KEY_TYPE is a PyObject*; the caller is responsible for keeping refcounts straight.


Roughly like O=K. O is a PyObject*, and the macro must build a Python object form of K, assign it to O, and ensure that O owns the reference to its new value. It may do this by creating a new Python object based on K (e.g., PyInt_FromLong(K) when KEY_TYPE is int), or simply by doing Py_INCREF(K) if KEY_TYPE is a PyObject*.


Copy an argument to the target without creating a new reference to ARG. ARG is a PyObject*, and TARGET is of type KEY_TYPE. If this can’t be done (for example, KEY_CHECK(ARG) returns false), set a Python error and set status to 0. If there is no error, leave status alone.

Macros for Values


The C type declaration for values (e.g., int or PyObject*).


Define if VALUE_TYPE is a PyObject*, else undef.


Like Python’s cmp(). Compares X to Y, where X & Y are C values of type VALUE_TYPE. The macro returns an int, with value:

 < 0 if X < Y
== 0 if X == Y
 > 0 if X > Y

Bug: There is no provision for determining whether the comparison attempt failed (set a Python exception).


Like DECREF_KEY, except applied to values of VALUE_TYPE.


Like INCREF_KEY, except applied to values of VALUE_TYPE.


Like COPY_KEY, except applied to values of VALUE_TYPE.


Like COPY_KEY_TO_OBJECT, except applied to values of VALUE_TYPE.


Like COPY_KEY_FROM_ARG, except applied to values of VALUE_TYPE.


Normalize the value, V, using the parameter MIN. This is almost certainly a YAGNI. It is a no-op for most types. For integers, V is replaced by V/MIN only if MIN > 0.

Macros for Set Operations


A value of VALUE_TYPE specifying the value to associate with set elements when sets are merged with mappings via weighed union or weighted intersection.

MERGE(O1, w1, O2, w2)

Performs a weighted merge of two values, O1 and O2, using weights w1 and w2. The result must be of VALUE_TYPE. Note that weighted unions and weighted intersections are not enabled if this macro is left undefined.


Computes a weighted value for O. The result must be of VALUE_TYPE. This is used for “filling out” weighted unions, i.e. to compute a weighted value for keys that appear in only one of the input mappings. If left undefined, MERGE_WEIGHT defaults to:

#define MERGE_WEIGHT(O, w) (O)


The value doesn’t matter. If defined, SetOpTemplate.c compiles code for a multiunion() function (compute a union of many input sets at high speed). This currently makes sense only for structures with integer keys.


There are two tunable values exposed on BTree and TreeSet classes. Their default values are found in and shared across C and Python.


An int giving the maximum bucket size (number of key/value pairs). When a bucket gets larger than this due to an insertion into a BTREE, it splits. Inserting into a bucket directly doesn’t split, and functions that produce a bucket output (e.g., union()) also have no bound on how large a bucket may get. This used to come from the C macro DEFAULT_MAX_BUCKET_SIZE.


An int giving the maximum size (number of children) of an internal btree node. This used to come from the C macro DEFAULT_MAX_BTREE_SIZE

BTree Clues

More or less random bits of helpful info.

  • In papers and textbooks, this flavor of BTree is usually called a B+-Tree, where “+” is a superscript.

  • All keys and all values live in the bucket leaf nodes. Keys in interior (BTree) nodes merely serve to guide a search efficiently toward the correct leaf.

  • When a key is deleted, it’s physically removed from the bucket it’s in, but this doesn’t propagate back up the tree: since keys in interior nodes only serve to guide searches, it’s OK– and saves time –to leave “stale” keys in interior nodes.

  • No attempt is made to rebalance the tree after a deletion, unless a bucket thereby becomes entirely empty. “Classic BTrees” do rebalance, keeping all buckets at least half full (provided there are enough keys in the entire tree to fill half a bucket). The tradeoffs are murky. Pathological cases in the presence of deletion do exist. Pathologies include trees tending toward only one key per bucket, and buckets at differing depths (all buckets are at the same depth in a classic BTree).

  • max_leaf_size and max_internal_size are chosen mostly to “even out” pickle sizes in storage. That’s why, e.g., an IIBTree has larger values than an OOBTree: pickles store ints more efficiently than they can store arbitrary Python objects.

  • In a non-empty BTree, every bucket node contains at least one key, and every BTree node contains at least one child and a non-NULL firstbucket pointer. However, a BTree node may not contain any keys.

  • An empty BTree consists solely of a BTree node with len==0 and firstbucket==NULL.

  • Although a BTree can become unbalanced under a mix of inserts and deletes (meaning both that there’s nothing stronger that can be said about buckets than that they’re not empty, and that buckets can appear at different depths), a BTree node always has children of the same kind: they’re all buckets, or they’re all BTree nodes.


For notational ease, consider a fixed BTree node x, and let

K(i) mean x->data.key[i]
C(i) mean all the keys reachable from x->data.child[i]

For each i in 0 to x->len-1 inclusive,

K(i) <= C(i) < K(i+1)

is a BTree node invariant, where we pretend that K(0) holds a key smaller than any possible key, and K(x->len) holds a key larger than any possible key. (Note that K(x->len) doesn’t actually exist, and K(0) is never used although space for it exists in non-empty BTree nodes.)

When searching for a key k, then, the child pointer we want to follow is the one at index i such that K(i) <= k < K(i+1). There can be at most one such i, since the K(i) are strictly increasing. And there is at least one such i provided the tree isn’t empty (so that 0 < len). For the moment, assume the tree isn’t empty (we’ll get back to that later).

The macro’s chief loop invariant is

K(lo) < k < K(hi)

This holds trivially at the start, since lo is set to 0, and hi to x->len, and we pretend K(0) is minus infinity and K(len) is plus infinity. Inside the loop, if K(i) < k we set lo to i, and if K(i) > k we set hi to i. These obviously preserve the invariant. If K(i) == k, the loop breaks and sets the result to i, and since K(i) == k in that case i is obviously the correct result.

Other cases depend on how i = floor((lo + hi)/2) works, exactly. Suppose lo + d = hi for some d >= 0. Then i = floor((lo + lo + d)/2) = floor(lo + d/2) = lo + floor(d/2). So:

  1. [d == 0] (lo == i == hi) if and only if (lo == hi).

  2. [d == 1] (lo == i  < hi) if and only if (lo+1 == hi).

  3. [d  > 1] (lo  < i  < hi) if and only if (lo+1  < hi).

If the node is empty (x->len == 0), then lo==i==hi==0 at the start, and the loop exits immediately (the first i > lo test fails), without entering the body.

Else lo < hi at the start, and the invariant K(lo) < k < K(hi) holds.

If lo+1 < hi, we’re in case (c): i is strictly between lo and hi, so the loop body is entered, and regardless of whether the body sets the new lo or the new hi to i, the new lo is strictly less than the new hi, and the difference between the new lo and new hi is strictly less than the difference between the old lo and old hi. So long as the new lo + 1 remains < the new hi, we stay in this case. We can’t stay in this case forever, though: because hi-lo decreases on each trip but remains > 0, lo+1 == hi must eventually become true. (In fact, it becomes true quickly, in about log2(x->len) trips; the point is more that lo doesn’t equal hi when the loop ends, it has to end with lo+1==hi and i==lo).

Then we’re in case (b): i==lo==hi-1 then, and the loop exits. The invariant still holds, with lo==i and hi==lo+1==i+1:

K(i) < k < K(i+1)

so i is again the correct answer.

Optimization points

  • Division by 2 is done via shift rather via “/2”. These are signed ints, and almost all C compilers treat signed int division as truncating, and shifting is not the same as truncation for signed int division. The compiler has no way to know these values aren’t negative, so has to generate longer-winded code for “/2”. But we know these values aren’t negative, and exploit it.

  • The order of _cmp comparisons matters. We’re in an interior BTree node, and are looking at only a tiny fraction of all the keys that exist. So finding the key exactly in this node is unlikely, and checking _cmp == 0 is a waste of time to the same extent. It doesn’t matter whether we check for _cmp < 0 or _cmp > 0 first, so long as we do both before worrying about equality.

  • At the start of a routine, it’s better to run this macro even if x->len is 0 (check for that afterwards). We just called a function and so probably drained the pipeline. If the first thing we do then is read up self->len and check it against 0, we just sit there waiting for the data to get read up, and then another immediate test-and-branch, and for a very unlikely case (BTree nodes are rarely empty). It’s better to get into the loop right away so the normal case makes progress ASAP.


This has a different job than BTREE_SEARCH: the key 0 slot is legitimate in a bucket, and we want to find the index at which the key belongs. If the key is larger than the bucket’s largest key, a new slot at index len is where it belongs, else it belongs at the smallest i with keys[i] >= the key we’re looking for. We also need to know whether or not the key is present (BTREE_SEARCH didn’t care; it only wanted to find the next node to search).

The mechanics of the search are quite similar, though. The primary loop invariant changes to (say we’re searching for key k):

K(lo-1) < k < K(hi)

where K(i) means keys[i], and we pretend K(-1) is minus infinity and K(len) is plus infinity.

If the bucket is empty, lo=hi=i=0 at the start, the loop body is never entered, and the macro sets INDEX to 0 and ABSENT to true. That’s why _cmp is initialized to 1 (_cmp becomes ABSENT).

Else the bucket is not empty, lo<hi at the start, and the loop body is entered. The invariant is obviously satisfied then, as lo=0 and hi=len.

If K[i]<k, lo is set to i+1, preserving that K(lo-1) = K[i] < k.

If K[i]>k, hi is set to i, preserving that K[hi] = K[i] > k.

If the loop exits after either of those, _cmp != 0, so ABSENT becomes true.

If K[i]=k, the loop breaks, so that INDEX becomes i, and ABSENT becomes false (_cmp=0 in this case).

The same case analysis for BTREE_SEARCH on lo and hi holds here:

  1. (lo == i == hi) if and only if (lo   == hi).

  2. (lo == i  < hi) if and only if (lo+1 == hi).

  3. (lo  < i  < hi) if and only if (lo+1  < hi).

So long as lo+1 < hi, we’re in case (c), and either break with equality (in which case the right results are obviously computed) or narrow the range. If equality doesn’t obtain, the range eventually narrows to cases (a) or (b).

To go from (c) to (a), we must have lo+2==hi at the start, and K[i]=K[lo+1]<k. Then the new lo gets set to i+1 = lo+2 = hi, and the loop exits with lo=hi=i and _cmp<0. This is correct, because we know that k != K(i) (loop invariant! we actually know something stronger, that k < K(hi); since i=hi, this implies k != K(i)).

Else (c) eventually falls into case (b), lo+1==hi and i==lo. The invariant tells us K(lo-1) < k < K(hi) = K(lo+1), so if the key is present it must be at K(lo). i==lo in this case, so we test K(lo) against k. As always, if equality obtains we do the right thing, else case #b becomes case (a).

When (b) becomes (a), the last comparison was non-equal, so _cmp is non-zero, and the loop exits because lo==hi==i in case (a). The invariant then tells us K(lo-1) < k < K(lo), so the key is in fact not present, it’s correct to exit with _cmp non-zero, and i==lo is again the index at which k belongs.

Optimization points

  • As for BTREE_SEARCH, shifting of signed ints is cheaper than division.

  • Unlike as for BTREE_SEARCH, there’s nothing special about searching an empty bucket, and the macro computes thoroughly sensible results in that case.

  • The order of _cmp comparisons differs from BTREE_SEARCH. When searching a bucket, it’s much more likely (than when searching a BTree node) that the key is present, so testing __cmp==0 isn’t a systematic waste of cycles. At the extreme, if all searches are successful (key present), on average this saves one comparison per search, against leaving the determination of _cmp==0 implicit (as BTREE_SEARCH does). But even on successful searches, __cmp != 0 is a more popular outcome than __cmp == 0 across iterations (unless the bucket has only a few keys), so it’s important to check one of the inequality cases first. It turns out it’s better on average to check K(i) < key (than to check K(i) > key), because when it pays it narrows the range more (we get a little boost from setting lo=i+1 in this case; the other case sets hi=i, which isn’t as much of a narrowing).