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authorzlg <zlg@zlg.space>2013-04-20 07:33:45 -0500
committerzlg <zlg@zlg.space>2013-04-20 07:33:45 -0500
commitd8c9ddf97c92d9dd3146a478c04b96b996d6d5c4 (patch)
treec4ad5a36e4803a0237c6ff0e875b02034ef4098e /ch3
parentSolve Exercise 3-4: itoa improved (diff)
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Solve Exercise 3-5: itob
Diffstat (limited to 'ch3')
-rw-r--r--ch3/3-05_itob.c85
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diff --git a/ch3/3-05_itob.c b/ch3/3-05_itob.c
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+++ b/ch3/3-05_itob.c
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+#include <stdio.h>
+#include <string.h>
+#include <limits.h>
+
+/* The C Programming Language: 2nd Edition
+ *
+ * Exercise 3-5: Write the function itob(n,s,b) that converts the integer n
+ * into a base b character representation in the string s. In particular,
+ * itob(n,s,16) formats n as a hexadecimal integer in s.
+ *
+ * Answer: Retool the itoa() function written for 3-4 to take an additional
+ * argument, then add a few more checks and account for ASCII's 7-character
+ * distance between '9' and 'A', which makes bases higher than 16 function
+ * without a large switch statement.
+ */
+
+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 itob(int n, char s[], unsigned b) {
+ int i, sign, min, rem; // 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;
+ if (b > 1) {
+ 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. */
+ rem = n % b;
+ if (rem > 9) {
+ rem += 7;
+ }
+ if (i == 0 && min == 1) {
+ s[i++] = rem + '1';
+ } else {
+ s[i++] = rem + '0';
+ }
+ } while ((n /= b) > 0);
+ /* This is annoying, but to make hex numbers right, you have to account
+ * for the +1 added to the final number causing bit flipping */
+ if (min == 1) {
+ for (i = 0; s[i] != '\0'; i++) {
+ if (s[i] == '-') {
+ continue;
+ }
+ if (i != 0 && s[i - 1] == '0' + (b + 7)) {
+ s[i] += 1;
+ s[i - 1] = '0';
+ }
+ }
+ }
+ }
+ if (sign < 0) {
+ s[i++] = '-';
+ }
+ s[i] = '\0';
+ reverse(s);
+}
+
+int main() {
+ int tests[5] = {INT_MIN, INT_MAX, -300, 172, 38478235};
+ char st[101] = "";
+ int i;
+
+ for (i = 0; i < 5; i++) {
+ itob(tests[i], st, 16);
+ printf("%12d in string form is %12s\n", tests[i], st);
+ }
+ return 0;
+}