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authorzlg <zlg@zlg.space>2013-03-22 11:41:52 -0500
committerzlg <zlg@zlg.space>2013-03-22 11:41:52 -0500
commit999a2d4dcf3d918e37f76aae249b7b23f14bdd7d (patch)
treea0f4b452c5e5126d6971c986fb2f49ebd0dc10f8
parentSolve Exercise 2-3: Hex to integer converter (diff)
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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;
+}