Solve Exercise 2-04: Squeeze v2

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diff --git a/ch2/2-04_squeeze-v2.c b/ch2/2-04_squeeze-v2.c
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+#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;
+}