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#include <stdio.h>
#include <string.h>
#include <limits.h>
/* The C Programming Language: 2nd Edition
*
* Exercise 3-4: In a two's complement number representation, our version of
* itoa does not handle the largest negative number, that is, the value of n
* equal to -(2^wordsize-1). Explain why not. Modify it to print that value
* correctly, regardless of the machine on which it runs.
*
* Answer: The reason it doesn't work is because zero (0) takes up one of the
* possible numbers. When the number n is INT_MIN, the operation `n = -n` to
* make it positive doesn't work, which makes the math later on moot. I fixed
* this by detecting that it's the lowest possible number, then adding one.
* This makes the resulting positive number the largest possible positive
* number. In order to make the string appear right, I spend the first
* iteration of the do/while loop detecting INT_MIN and adding '1' to the
* character that results from the math. The string then outputs properly.
*/
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[]) {
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++] = '-';
}
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++) {
itoa(tests[i], st);
printf("%12d in string form is %12s\n", tests[i], st);
}
return 0;
}
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