Pointer version of itoa

Question

Rewrite appropiate programs from earlier chapters and exercises with pointers instead of array indexing. Good posibilities include getline(Chapter 1 and 4), atoi, itoa, and their variants(Chapters 2, 3, and 4), reverse(Chapter 3), and strindex and getop(Chapter 4)

Here is my solution:

static char *hidden(int n, char *s) {
    char *p;

    if(n / 10) {
        p = hidden(n / 10, s);
    }
    else {
        p = s;

    }
    *p++ = n % 10 + '0';
    *p = '\0';

    return p;
}

char *itoa(int n, char *s) {
    if(n < 0) {
        *s = '-';
        hidden(-n, s + 1);
    }
    else {
        hidden(n, s);
    }

    return s; /* pointer to first char of s*/
}

itoa is a wrapper for the function hidden. The function hidden returns a pointer to the next free slot in the array s. At the deepest level of recursion it returns a pointer to the first element (the else branch) of the array.

I think that this is the most relevant function that needs pr, the other ones are quite simple. If you want to see them: drive.google.com/… — cristid9
– cristid9, Commented Feb 5, 2014 at 14:48
Please give a better name (and some explanatory comments) to your hidden function. It's hard to tell what the function does. — Brennan Vincent
– Brennan Vincent, Commented Feb 5, 2014 at 15:25
There is absolutely no reason to use recursion for this. It is hard to read, inefficient and dangerous. I think the classic K&R implementation of this algorithm is hard to beat in terms of efficiency. (But of course, K&R code is always an unreadable mess, filled with dangerous programming practice, so that snippet would need a code review of its own.) — Lundin
– Lundin, Commented Feb 6, 2014 at 14:04

ChrisW · Accepted Answer · 2014-02-05 16:17:19Z

2

The standard version of itoa takes a parameter named base (which you have assumed is 10).

You probably don't need a second hidden function; this would do instead:

if (n < 0)
{
    *s++ = '-';
    n = -n;
}
// ... itoa implementation continues here ...

Recursing is clever; you could also probably do it with two loops instead (untested code ahead):

i = 1;
while ((i * 10) <= n)
    i *= 10;
for (; i != 0; i /= 10)
{
    int x = n / i;
    *p++ = (char)(x % 10 + '0');
    n -= (x * i);
}
assert(n == 0);

answered Feb 5, 2014 at 16:17

ChrisW

13.1k1 gold badge35 silver badges76 bronze badges

\$\begingroup\$ There is no standard version of itoa. As in: how it should behave isn't specified anywhere. \$\endgroup\$

Lundin
– Lundin

2014-02-06 13:34:35 +00:00
Commented Feb 6, 2014 at 13:34

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