• C

Using qsort()

Hi

I'm having trouble using qsort(). I've tried for a while now & just cant master it.

My code (I've left my debugs in)

int compare( const void *a, const void *b)
      {
      printf("OK C\n");
                return *(char*)a - *(char*)b;
      }

void sorting()
      {
        char *a="string";
        printf("OK A\n");
        qsort(a,strlen(a),1,compare);
        printf("OK B\n");
                 }

The compiler keeps spitting out  OKA OKC OKC then  a segmentation fault(core dumped).

Any ideas??
Benjamin_BarrettAsked:
Who is Participating?
 
PaulCaswellConnect With a Mentor Commented:
Hi Benjamin_Barrett,

I found it, and that's exactly what I would have suggested.

The problem is that:

char * s = "A String";

puts the string in protected memory and therefore the first attempt that qsort makes to swap two of the characters results in an access violation.

Your method makes the string into an array of bytes that is changeable.

Paul
0
 
Benjamin_BarrettAuthor Commented:
I forgot to add that the segmentation fault suggests to me that its a pointer issue,  somewhere in compare() .....
0
 
PaulCaswellCommented:
Hi Benjamin_Barrett,

Just in case, you may like to try:

                return *((char*)a) - *((char*)b);

I cant see anything else wrong. I'll try it.

Paul
0
Improve Your Query Performance Tuning

In this FREE six-day email course, you'll learn from Janis Griffin, Database Performance Evangelist. She'll teach 12 steps that you can use to optimize your queries as much as possible and see measurable results in your work. Get started today!

 
Benjamin_BarrettAuthor Commented:
Thanks for your quick response Paul

I tried your suggestion but I still get the same output.

I'll keep trying

Thanks

Benjamin
0
 
Benjamin_BarrettAuthor Commented:
Well I tried changing the declaration & intialization of a(the string to sort) to....

char a[10] = "string";

instead of

char *a="string";


And it works!

Clearly this is a pointer issue...Im a Java programmer trying to learn better  C,  so  pointers are a bit heavy going for me.

Any advice on why this worked & how to get it working with a pointer as an argument?
0
 
PaulCaswellCommented:
Hi Benjamin_Barrett,

FYI: To turn a char [] into a char*, just use &array[0].

Paul
0
 
Benjamin_BarrettAuthor Commented:
Excellent, thats why "OK C" was outputted twice before the fault

Ok, can you help me abit more....

If I have a pointer as an arg, how do I convert it to a non pointer?

I've tried doing declaring a string then making it equal the pointer...

char *a ="test";

char b[5] = a; || char b[5] = &a;

but no dice.

Any suggestions?

Just to let you know Im disappearing for lunch now, so no stress.

Benjamin
0
 
Benjamin_BarrettAuthor Commented:
Brilliant Paul!

You only need to access the address of the first element of the array to get the whole thing....

Thank you kindly for your help.

Benjamin

PS We posted @ the same time before, so I just got your last reply
0
Question has a verified solution.

Are you are experiencing a similar issue? Get a personalized answer when you ask a related question.

Have a better answer? Share it in a comment.

All Courses

From novice to tech pro — start learning today.