problem with incompatible pointer type

When compiling one of my function, it gives me a warning message
"passing arg 4 of 'qsort' from incompatible pointer type"
what does this mean?

here's some of the codes that's related

Code:

---

int compare(int *key1, int *key2)
{
 return (*key1 - *key2);
}

void mysort(int *keyArray, int nel, int width, int (*compare)(int *, int *))
{
 qsort(keyArray, nel, width, compare);
}

---

and I have initialised
int *keyArray;
in my main function

did I pass in some wrong arguments into the library function qsort?


Code tags added by kermi3

Code:

---

#include <stdlib.h>
void qsort( void *base,
            size_t num,
            size_t width,
            int (*compar) ( const void *,
                            const void *) );

---

Note that the function ptr in the 4th parm takes void pointers, you are using int pointers.

From the manual, here's an example of a char* compare function. I'll leave it to you to change it to int*:

Code:

---

int compare( const void *op1, const void *op2 )
  {
    const char **p1 = (const char **) op1;
    const char **p2 = (const char **) op2;
    return( strcmp( *p1, *p2 ) );
  }

---

Code:

---

int compare( const void *op1, const void *op2 )
  {
    const char **p1 = (const char **) op1;
    const char **p2 = (const char **) op2;
    return( strcmp( *p1, *p2 ) );
  }

---

When I tried that, I get warnings for
const char **p1 = (const char **)op1; and op2

"declaration of 'op1' shadows a parameter"


Code tags added by kermi3

An int version (untested)

Code:

---

int compare(void *p1, void *p2)
{
  int *key1 = (int *) p1;
  int *key2 = (int *) p2;
  return (*key1 - *key2);
}

---

Go from there...

(my prev post was only a sample for you to review, not to actually use directly in your code).

thanks
now when I want to use the function in my main function
I get parse error

I tried to do mysort(keyArray, 50, 30, compare(const void *,const void *);

and mysort is

Code:

---

void mysort(int *keys, int nel, int width, int (*compare)(const void *, const void *) )
{
qsort(keys, nel, width, compare);
return;
}

---

Code tags added by kermi3

Quote:

---

Originally posted by SIKCAR
thanks
now when I want to use the function in my main function
I get parse error

I tried to do mysort(keyArray, 50, 30, compare(const void *,const void *);

and mysort is
void mysort(int *keys, int nel, int width, int (*compare)(const void *, const void *) )
{
qsort(keys, nel, width, compare);
return;
}

---

count your parentheses

sorry, I missed a ) when I typed it here, but that wasn't the problem

basically I get
parse error before 'const'

in an example I've got, they were comparing char
but they could do

Code:

---

mysort(charArray, 50, sizeof(char), compare);

---

I've been trying to do that, but get the error I've got in the first place

OK, here's some complete code. There's a different in the compare function to the one I posted earlier, both of it's parameters are declared as const void* now.

Code:

---

#include <stdio.h>
#include <stdlib.h>

void printarray(int a[], int size)
{
    int i;
    for (i = 0; i < size; i++)
        printf ("%d\n", a[i]);
}

int compare(const void *p1, const void *p2)
{
  int *key1 = (int *) p1;
  int *key2 = (int *) p2;
  return (*key1 - *key2);
}
   
int main(void)
{
    int myarray[] = {5, 4, 1, 9, 5, 10};
    int num_elements = sizeof(myarray)/sizeof(myarray[0]);

    puts("Before");
    printarray(myarray, num_elements);
    qsort (myarray, num_elements, sizeof(int), compare);
    puts("After");
    printarray(myarray, num_elements);
   
    return 0;
}

/* Program output
Before
5
4
1
9
5
10
After
1
4
5
5
9
10
*/

---