1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
|
#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;
}
|
