# HG changeset patch
# User Louis Opter <kalessin@kalessin.fr>
# Date 1373770031 25200
# Node ID 9748b64edb26606431ab00bda1192920c698b2b5
# Parent  1f4c37678f692992e0be00cbca86737f0a3bf042
Add the bit count exercise

diff -r 1f4c37678f69 -r 9748b64edb26 arrays/bit_count/bit_count.c
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/arrays/bit_count/bit_count.c	Sat Jul 13 19:47:11 2013 -0700
@@ -0,0 +1,37 @@
+#include <stdio.h>
+#include <stdlib.h>
+
+/*
+ * There is different solutions but this one is simple and easy to remember
+ * (it's very similar to !(a & (a - 1)) used to check if a is a power of two).
+ * If you want to look smart, say that this solution is perfect for embedded
+ * systems.
+ *
+ * Another simple one, more optimized, is to have a lookup table that gives you
+ * the number of 1 bit in each integer for integers from 0 to 255 (or 0 to
+ * 2**16 - 1).
+ */
+
+unsigned int
+bit_count(const unsigned char *a, size_t n)
+{
+    unsigned int c = 0;
+
+    while (n--)
+        for (unsigned char v = a[n]; v; v &= v - 1)
+            c++;
+
+    return c;
+}
+
+int
+main(void)
+{
+    int a[] = { 1, 2, 3, 4 };
+    int b = 42;
+
+    printf("%d\n", bit_count((unsigned char *)a, sizeof(a)));
+    printf("%d\n", bit_count((unsigned char *)&b, sizeof(b)));
+
+    return 0;
+}
diff -r 1f4c37678f69 -r 9748b64edb26 arrays/bit_count/hints.rst
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/arrays/bit_count/hints.rst	Sat Jul 13 19:47:11 2013 -0700
@@ -0,0 +1,4 @@
+Hints
+=====
+
+This is actually probably easier in C.
diff -r 1f4c37678f69 -r 9748b64edb26 arrays/bit_count/question.rst
--- /dev/null	Thu Jan 01 00:00:00 1970 +0000
+++ b/arrays/bit_count/question.rst	Sat Jul 13 19:47:11 2013 -0700
@@ -0,0 +1,5 @@
+Bit Count
+=========
+
+Write a function that returns the number of bits set to 1 in the given memory
+area of size n.