CODEWITHSUNDEEP
cfunction-pointersqsort
user2925557
user2925557

Reputation: 85

Function Pointer Variable

I'm trying to create a function variable that point to a function that can be passed to the qsort() function, and I received an "assignment type mismatch" error when I tried to run the below codes.

int compareFunc (const void * a, const void * b)
{
   //codes
}

int main(void) {
  int *ptr;     //create a function pointer variable that point to compareFunc
  ptr = &compareFunc   //Initialize function pointer
  //codes
  return 0;
}

Could someone please tell me what I did wrong and explain what exactly is const void *?

Upvotes: 0

Views: 418

Answers (5)

alk
alk

Reputation: 70931

Do a man qsort and you'll get (besides more):

void qsort(void *base, size_t nmemb, size_t size,
int(*compar)(const void *, const void *));

The emphasised part obviously has to be what you are looking for.

Upvotes: 0

jester
jester

Reputation: 3489

Your declaration of function pointer is incorrect. You are declaring a regular pointer. The syntax of a function pointer declaration is :

<return_type> (* func_ptr_Variable_name) ( [params...] )

The function pointer should be :

int main(void){
   int (*ptr)(const void *,const void *);
   ptr = compareFunc;   //'&' is not needed here, function name is automatically converted to its address 
   ...
   return 0;
}

const void * simply means that the data pointed to by the void * pointer cannot be modified in the function.

Upvotes: 0

hawk
hawk

Reputation: 1847

Your declaration of pointer isn't a function pointer. It's just a regular pointer. It should be changed to declare a function pointer as follows.

int (*fptr) (const void *a, const void *b);
fptr = compareFunc;

For function pointers you never have to take the address explicitly using the & operator.

Also const void * means that the object/data pointed to by the pointer will not be modified in that function. So compiler will raise an error whenever you perform a write operation on the value the pointer points to.

Upvotes: 2

Sinn
Sinn

Reputation: 274

a function pointer is declared like this:

return_type (* variable_name)(params)

So in your case:

int (*cfptr)(const void * a, const void * b);
cfptr = compareFunc;

Upvotes: 1

Yu Hao
Yu Hao

Reputation: 122383

Because int * is a type of object pointer, not a function pointer.

You can declare it as int (*ptr)(const void *,const void *);, or use typedef to simplify declaring function pointers:

typedef int (*comptr)(const void *a, const void *b);
int compareFunc (const void * a, const void * b)
{
   //codes
}

int main(void) {
  comptr ptr;     
  ptr = compareFunc;  
  //codes
  return 0;
}

Also note that instead of ptr = &compareFunc;, you can omit the & because the function name is converted to a function pointer automatically.

Upvotes: 0

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