am: return control to caller, for housekeeping

diff --git a/bisect.c b/bisect.c
index 1aad49b1a642a308f449001960958ccab79347d1..374d9e24bd0a18b0453f3e948db93251a859ba18 100644 (file)
--- a/bisect.c
+++ b/bisect.c
@@ -525,9 +525,9 @@ struct commit_list *filter_skipped(struct commit_list *list,
  * is increased by one between each call, but that should not matter
  * for this application.
  */
-static int get_prn(int count) {
+static unsigned get_prn(unsigned count) {
        count = count * 1103515245 + 12345;
-       return ((unsigned)(count/65536) % PRN_MODULO);
+       return (count/65536) % PRN_MODULO;
 }
 
 /*
@@ -956,3 +956,41 @@ int bisect_next_all(const char *prefix, int no_checkout)
        return bisect_checkout(bisect_rev_hex, no_checkout);
 }
 
+static inline int log2i(int n)
+{
+       int log2 = 0;
+
+       for (; n > 1; n >>= 1)
+               log2++;
+
+       return log2;
+}
+
+static inline int exp2i(int n)
+{
+       return 1 << n;
+}
+
+/*
+ * Estimate the number of bisect steps left (after the current step)
+ *
+ * For any x between 0 included and 2^n excluded, the probability for
+ * n - 1 steps left looks like:
+ *
+ * P(2^n + x) == (2^n - x) / (2^n + x)
+ *
+ * and P(2^n + x) < 0.5 means 2^n < 3x
+ */
+int estimate_bisect_steps(int all)
+{
+       int n, x, e;
+
+       if (all < 3)
+               return 0;
+
+       n = log2i(all);
+       e = exp2i(n);
+       x = all - e;
+
+       return (e < 3 * x) ? n : n - 1;
+}