Solve Exercise 2-5: The any() function

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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;
+}