Solve Exercise 2-10: lower()

Also added more information to 2-09's comments.

2 files changed, 33 insertions, 1 deletions

diff --git a/ch2/2-09_bitcount.c b/ch2/2-09_bitcount.c
index d824ca6..6ad6086 100644
--- a/ch2/2-09_bitcount.c
+++ b/ch2/2-09_bitcount.c
@@ -7,7 +7,7 @@
  * faster version of bitcount.
  *
  * Answer: Subtracting 1 from a number reverses the rightmost 1 bit and
- * replaces all lower-order bits to 0. So for example:
+ * replaces all lower-order bits to 1. So for example:
  *
  * 110100 (52) minus 1 is
  * 110011 (51)
@@ -22,6 +22,18 @@
  * Why? Because the 2nd operand (51) is only masking off two fields, the 3rd
  * and 4th (which are zeros). The rightmost 1-bit in the 1st operand (52) is
  * in that mask, so it goes poof.
+ *
+ * Oh, and here's the in-book version of bitcount():
+ *
+ * int bitcount(unsigned x) {
+ *     int b;
+ *     for (b = 0; x != 0; x >>= 1) {
+ *         if (x & 01) {
+ *             b++;
+ *         }
+ *     }
+ *     return b;
+ * }
  */
 
 unsigned bitcount(unsigned x) {
diff --git a/ch2/2-10_lower.c b/ch2/2-10_lower.c
new file mode 100644
index 0000000..fe0288b
--- /dev/null
+++ b/ch2/2-10_lower.c
@@ -0,0 +1,20 @@
+#include <stdio.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 2-10: Rewrite the function 'lower', which converts upper case
+ * letters to lower case, with a conditional expression instead of if-else.
+ *
+ * Answer: The tertiary ?: operators also _evaluate_, so they can be used in
+ * a lot of different places.
+ */
+
+int lower(int c) {
+	return (c >= 'A' && c <= 'Z') ? c + 'a' - 'A' : c;
+}
+
+int main() {
+	char foo = 'F';
+	printf("The follow letter should be lowercase: %c\n", lower(foo));
+	return 0;
+}