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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;
}
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