incompatible pointer type in C

Question

so I\'m trying to pass a type double * to a function that accepts void ** as one of the parameters. This is the warning that I am getting.

Answer 1

int main( void )
{
    //  Local Declaration
    double *target;

   //   Statement
   success = dequeue(queueIn, (void**)&target);
}

Use it like this.

Answer 2

In your prototype declaration , you said second argument as void** ,so you have to type cast double** to void**. Instead of this line success = dequeue(queueIn, &target);.

Call like this success = dequeue(queueIn,(void**) &target);

Answer 3

Even though all other pointer types can be converted to and from void * without loss of information, the same is not true of void ** and other pointer-to-pointer types; if you dereference a void ** pointer, it needs to be pointing at a genuine void * object¹.

In this case, presuming that dequeue() is returning a single pointer value by storing it through the provided pointer, to be formally correct you would need to do:

int main( void )
{
    void *p;
    double *target;

    success = dequeue(queueIn, &p);
    target = p;

When you write it like this, the conversion from void * to double * is explicit, which allows the compiler to do any magic that's necessary (even though in the overwhelmingly common case, there's no magic at all).

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^{1. ...or a char *, unsigned char * or signed char * object, because there's a special rule for those.}