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authorzlg <zlg@zlg.space>2013-04-23 02:09:09 -0500
committerzlg <zlg@zlg.space>2013-04-23 02:09:09 -0500
commite6b7015c5dd0431aaaf40256d1d2db3cda6769bb (patch)
treed1eae9a443ffc2b3975a1b102bb77700e1a984dd /ch3
parentSolve Exercise 3-5: itob (diff)
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Solve Exercise 3-6: itoa (3 arg version)
Diffstat (limited to '')
-rw-r--r--ch3/3-06_itoa3.c67
1 files changed, 67 insertions, 0 deletions
diff --git a/ch3/3-06_itoa3.c b/ch3/3-06_itoa3.c
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+++ b/ch3/3-06_itoa3.c
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+#include <stdio.h>
+#include <string.h>
+#include <limits.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 3-6: Write a version of itoa that accepts three arguments instead
+ * of two. The third argument is a minimum field width; the converted number
+ * must be padded with blanks on the left if necessary to make it wide enough.
+ *
+ * Answer: This is trivial. Process the number, then check the value of i. If
+ * it's less than the field width, add a blank until i is equal to the field
+ * width and terminate your string.
+ */
+
+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;
+ }
+}
+
+void itoa(int n, char s[], unsigned fw) {
+ int i, sign, min; // Add 'min' for later use
+
+ if ((sign = n) < 0) {
+ /* Detect this and add one so it can properly be made positive */
+ if (n == INT_MIN) {
+ n += 1;
+ min = 1;
+ }
+ n = -n;
+ }
+ i = 0;
+ do {
+ /* If it's the first iteration of the loop and we've established
+ * that n == INT_MIN, we need to add one to the resulting string for it
+ * to be displayed properly. */
+ if (i == 0 && min == 1) {
+ s[i++] = n % 10 + '1';
+ } else {
+ s[i++] = n % 10 + '0';
+ }
+ } while ((n /= 10) > 0);
+ if (sign < 0) {
+ s[i++] = '-';
+ }
+ while (i < fw) {
+ s[i++] = ' ';
+ }
+ s[i] = '\0';
+ reverse(s);
+}
+
+int main() {
+ int i;
+ char foo[16] = "";
+ int tests[5] = {-25, 409689, 8, -1000, 135};
+ for (i = 0; i < 5; i++) {
+ itoa(tests[i], foo, 5);
+ printf("Pad to 5 spaces! '%s'\n", foo);
+ }
+ return 0;
+}