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authorzlg <zlg@zlg.space>2013-03-27 05:00:43 -0500
committerzlg <zlg@zlg.space>2013-03-27 05:00:43 -0500
commit239a885d441cdb3585c42b594ed9e87b37b33c62 (patch)
treec37a632687906d0227eb775ee5fc3ba8ab26b606 /ch2
parentSolve Exercise 2-04: Squeeze v2 (diff)
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Solve Exercise 2-5: The any() function
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-rw-r--r--ch2/2-05_any-func.c44
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diff --git a/ch2/2-05_any-func.c b/ch2/2-05_any-func.c
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+#include <stdio.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 2-5: Write the function any(s1,s2), which returns the first
+ * location of the string s1 where any character from the string s2 occurs, or
+ * -1 if s1 contains no characters from s2. (The standard library function
+ * strpbrk does the same job but returns a pointer to the location.)
+ *
+ * Answer: Similarly to squeeze(), any() uses two loops to do its job, but
+ * instead of changing the string, we're just returning the index of the first
+ * character found. If it's not found, it'll return a negative number. This
+ * is a common practice in functions that return integers since it gives
+ * programmers an easy case to check for: if any() returns anything less than
+ * zero, the characters in s2 weren't found.
+ *
+ */
+
+int any(char s1[], char s2[]) {
+ int i, j;
+ for (i = j = 0; s1[i] != '\0'; ++i) {
+ while (s2[j++] != '\0') {
+ if (s1[i] == s2[j]) {
+ return i;
+ }
+ }
+ j = 0;
+ }
+ return -1;
+}
+
+int main() {
+ char foo[80] = "argle bargle";
+ char bar[80] = "toobz";
+ int result;
+
+ if ((result = any(foo, bar)) != -1) {
+ printf("Found a match at position %d!\n", result);
+ } else {
+ printf("No match found.\n");
+ }
+ // Should return a match at index 6.
+ return 0;
+}