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up vote 2 down vote favorite

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.

share|improve this question
    
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/… –  Ionut Grt. Feb 5 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 Feb 5 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 Feb 6 at 14:04
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1 Answer

up vote 2 down vote accepted

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);
share|improve this answer
    
There is no standard version of itoa. As in: how it should behave isn't specified anywhere. –  Lundin Feb 6 at 13:34
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