asked on

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_Barrett

ASKER

I forgot to add that the segmentation fault suggests to me that its a pointer issue, somewhere in compare() .....

PaulCaswell

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

Benjamin_Barrett

ASKER

Thanks for your quick response Paul

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

I'll keep trying

Thanks

Benjamin

Benjamin_Barrett

ASKER

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?

ASKER CERTIFIED SOLUTION

PaulCaswell

membership

This solution is only available to members.

To access this solution, you must be a member of Experts Exchange.

Start Free Trial

PaulCaswell

Hi Benjamin_Barrett,

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

Paul

Benjamin_Barrett

ASKER

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

Benjamin_Barrett

ASKER

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