From 999a2d4dcf3d918e37f76aae249b7b23f14bdd7d Mon Sep 17 00:00:00 2001
From: zlg <zlg@zlg.space>
Date: Fri, 22 Mar 2013 11:41:52 -0500
Subject: Solve Exercise 2-04: Squeeze v2

---
 ch2/2-04_squeeze-v2.c | 51 +++++++++++++++++++++++++++++++++++++++++++++++++++
 1 file changed, 51 insertions(+)
 create mode 100644 ch2/2-04_squeeze-v2.c

(limited to 'ch2')

diff --git a/ch2/2-04_squeeze-v2.c b/ch2/2-04_squeeze-v2.c
new file mode 100644
index 0000000..2c3b56a
--- /dev/null
+++ b/ch2/2-04_squeeze-v2.c
@@ -0,0 +1,51 @@
+#include <stdio.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 2-4: Write an alternate version of squeeze(s1, s2) that deletes
+ * each character in s1 that matches any character in the string s2.
+ *
+ * Answer: This one is fairly easy, considering the fun trick that's covered
+ * in the passage before this exercise with the unary operators ++ and --. C's
+ * interesting behavior with these operators allows the programmer to write
+ * shorter, faster logic.
+ *
+ * That said, I couldn't find a way to iterate through s2[] and s1[] in a
+ * single loop, and I needed to use a flag. There may be a more clever way to
+ * solve this.
+ *
+ */
+
+// It'd be better to make this return a pointer (or string), but the book
+// hasn't covered it yet!
+void squeeze(char s1[], char s2[]) {
+	int i, j, k, match;
+
+	for (i = j = 0; s1[i] != '\0'; i++) {
+		// I don't see a way to do this without a flag
+		match = 0;
+		for (k = 0; s2[k] != '\0'; k++) {
+			if (s1[i] == s2[k]) {
+				match = 1;
+				break;
+			}
+		}
+		// check our flag. if there wasn't a match, j needs to match i's value
+		if (!match) {
+			s1[j++] = s1[i];
+		}
+	}
+
+	s1[j] = '\0';
+}
+
+int main() {
+	char foo[16] = "foobarbaz";
+	char bar[16] = "boz";
+
+	squeeze(foo, bar);
+
+	printf("%s\n", foo); // Should read "fara"
+
+	return 0;
+}
-- 
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