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authorzlg <zlg@zlg.space>2013-07-15 06:38:40 -0500
committerzlg <zlg@zlg.space>2013-07-15 06:38:40 -0500
commit990ad145ea2f1013dbe6aefa0533b9700c331b60 (patch)
treec068fd19788ba148f5b6a34030fe43e7f928dc6a /ch4
parentSolve Exercise 4-11: getop() without ungetch() (diff)
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Solve Exercise 4-12: Recursive itoa()
Diffstat (limited to 'ch4')
-rw-r--r--ch4/4-12_recursive-itoa.c72
1 files changed, 72 insertions, 0 deletions
diff --git a/ch4/4-12_recursive-itoa.c b/ch4/4-12_recursive-itoa.c
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+++ b/ch4/4-12_recursive-itoa.c
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+#include <stdio.h>
+#include <string.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 4-12: Adapt the ideas of printd() to write a recursive version of
+ * itoa; that is, convert an integer into a string by calling a recursive
+ * routine.
+ *
+ * Answer: Making itoa() recursive isn't too much work; it helps knowing how to
+ * use static variables; they make it possible to increment the position in the
+ * string without using an external variable. Additionally, you'll need to
+ * understand flow control to prevent the sub-routines from executing further
+ * than needed. As a result, itoa() gets called multiple times, but if it
+ * calls itself, it immediately returns so it won't venture further. This is
+ * to make sure i is reset to zero only after the number is finished, so the
+ * next call starts at the beginning of the string.
+ *
+ * Normally, recursive functions shouldn't worry about state, but it's necessary
+ * in this version of itoa.
+ */
+
+void itoa(int, char[]);
+void reverse(char[]);
+
+int main(void) {
+ char foo[40] = "";
+
+ itoa(829048, foo);
+ printf("%s\n", foo);
+ itoa(-4021, foo);
+ printf("%s\n", foo);
+
+ return 0;
+}
+
+void itoa(int num, char target[]) {
+ static int i = 0;
+ static int neg = 0;
+
+ if (num < 0) {
+ neg = 1;
+ num = -num;
+ }
+
+ if (num /= 10 > 0) {
+ target[i++] = (num % 10) + '0';
+ num /= 10;
+ itoa(num, target);
+ return;
+ } else {
+ if (neg == 1) {
+ neg = 0;
+ target[i++] = '-';
+ target[i] = '\0';
+ } else {
+ target[i] = '\0';
+ }
+ reverse(target);
+ }
+ i = 0;
+}
+
+void reverse(char s[]) {
+ int c, i, j;
+
+ for (i = 0, j = strlen(s)-1; i < j; i++, j--) {
+ c = s[i];
+ s[i] = s[j];
+ s[j] = c;
+ }
+}