setting an array of pointers to pointers, point to a pointer in another array

Question

The program I am working on uses qsort, and my own compare function, to sort a group of words read in through standard input. I am putting each character into an array of pointers to characters as below.

The array now has a bunch of characters, for example - "\0The\n\0brown\n\0fox\n\0is\n\0lazy\n"

I am trying to create another array of pointers to pointers where each element of this new array points to the first letter (this case the null character) of each word. So element 0 points to the first \0 and element 1 points to the next \0. I'm not sure if there is a small syntax error I have or if I have the wrong idea, but something keeps going wrong because the output is never in the correct order. Code below:

int buffersize = 2048;
int count = 0;
char* p = (char*) malloc(sizeof(char) * buffersize);
int c;
do{
    c = getchar();
    p[count++] = (char)c;
    if (count == buffersize)
    {
        p = (char*) realloc(p, buffersize * 2);
        buffersize *= 2;
    }
}while (c != EOF);
p[count-1] = '\n';
int i = 0;
int a = 1;
char ** pp = (char**) malloc(sizeof(char*) * count);
pp[0] = &p[0];
for (i; i < count; i++)
{
    if (p[i] == '\n')
    {
        while (p[i+1] == '\n')
        {i++;}
        if ( i != (count-1) )
        {
            pp[a++] = &p[i+1];
        }
    }
}
qsort (pp, (a-1), sizeof(char*), compare);

My compare function

int rot13cmp (const void* c, const void* d)
{
    const char* a = (const char*)c;
    const char* b = (const char*)d;
    if (a[0] == '\0' && b[0] == '\t')
    {
        return -1;
    }
    else if (a[0] == '\t' && b[0] == '\0')
    {
        return 1;
    }
    int k = 0;
    for (;;k++)
    {
        if (a[k] == '\n' && b[k] != '\n')
            return -1;
        if (a[k] != '\n' && b[k] == '\n')
            return 1;
        if (a[k] == '\n' && b[k] == '\n')
            return 0;
        int one = (int)a[k];
        int two = (int)b[k];
        int difference = a[k] - b[k];
        if (difference != 0)
            return difference;
    }
}

Jonathan Leffler · Accepted Answer · 2014-02-09 02:57:15Z

1

Your compare function is incorrect. If the sequence being sorted is a sequence of pointers, then the addresses passed to your compare are addresses of the pointers; not addresses in the pointers.

replace this:

const char* a = (const char*)c;
const char* b = (const char*)d;

with this:

const char * const* lhs = c;
const char * const* rhs = d;
const char* a = *lhs;
const char* b = *rhs;

or simplify as desired. The rest of your function should work (at least as well as you wrote it, I never checked it for accuracy except to say that both one and two are not used and should be removed, and your function should have a outer-most return 0; to avoid an undefined result if the strings are identical).

edited Feb 9, 2014 at 2:57

Jonathan Leffler

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answered Feb 9, 2014 at 1:46

WhozCraig

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4 Comments

David Scheibe Over a year ago

Thank you. That fixed most of my problems! I'll come back if anything else occurs.

WhozCraig Over a year ago

Np You say thanks on this site by up-voting answers or marking them as "the" solution. See the arrows and green-check next to the answer above.

David Scheibe Over a year ago

Haha I don't have enough reputation to upvote, however I did do the green-check!

WhozCraig Over a year ago

@DavidScheibe No worries. For questions unrelated to this specific problem (which is hopefully resolved) just post a new question. Make sure to include what the problem is, steps to reproduce, code that reproduces, and what you're expecting vs. what you're seeing. It is the best shot at getting solid eyes and good answers. The more effort you put in to solving it yourself first, the more likely someone will step up to assist. trust me on that.

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