Solve Exercise 2-07: invert()

1 files changed, 38 insertions, 0 deletions

diff --git a/ch2/2-07_invert.c b/ch2/2-07_invert.c
new file mode 100644
index 0000000..244f854
--- /dev/null
+++ b/ch2/2-07_invert.c
@@ -0,0 +1,38 @@
+#include <stdio.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 2-07: Write a function invert(x,p,n) that returns x with the n bits
+ * that begin at position p inverted (i.e. 1 changed into 0 and vice-versa),
+ * leaving the others unchanged.
+ *
+ * Answer: The trick is to start with ~0 to get all 1s, then left-shift n
+ * times. After that, flip again to get n number of 1s. Next, left shift the
+ * appropriate number of times needed to line up the mask of 1s to the proper
+ * location. Apply that mask to x with XOR and bam, job done!
+ *
+ * Here's a step by step illustration of invert(205, 5, 4), assuming 16-bit
+ * unsigned integers:
+ *
+ * 1111111111111111     ~0
+ * 1111111111110000     << 4
+ * 0000000000001111     ~
+ * 0000000000011110     << (5 - 4 == 1)
+ *
+ * Mask created. Now let's take our original number...
+ * 0000000011001101     205
+ *            ####      ^ ...and apply the mask!
+ * 0000000011010011     End Result (211)
+ */
+
+unsigned invert(unsigned x, unsigned p, unsigned n) {
+	return x ^ ((~(~0 << n)) << (p - n));
+}
+
+int main() {
+	printf("invert(205, 5, 4) returns %d\n", invert(205, 5, 4));
+	printf("11001101 flips 110[0110]1 to produce...\n11010011\n");
+	printf("invert(1876, 7, 3) returns %d\n", invert(1876, 7, 3));
+	printf("11101010100 flips 1110[101]0100 to produce...\n11100100100\n");
+	return 0;
+}