为qsort的比较转换结构的指针

Question

So I'm using qsort() in c and my compare function usually consist of creating a pointer of the correct type and assigning it the value from the arguments. is it possible to just cast the pointer from the arguments without creating a new one? if so, what am i doing wrong?

struct date_t{
    int time;
    int dob;
};

/* the old working function
int cmpfunc (const void * one, const void * two) {
    struct date_t *itemtwo=two;
    struct date_t *itemone=one;
return itemone->time-itemtwo->time;
}
*/

int cmpfunc (const void * one, const void * two) {
    return (struct date_t*)(one)->time - (struct date_t*)two->time;
}

i'm getting :

main.c:17:30: warning: dereferencing 'void *' pointer
  return (struct date_t*)(one)->time - (struct date_t*)two->time;
                          ^~
main.c:17:30: error: request for member 'time' in something not a structure or union

EDIT:

i got it to compile with

int cmpfunc (struct date_t *one, struct date_t *two) {
    return one->time - two->time;
}

but still, how would i do it with casts?

Answer 1

The type cast operator () has lower precedence than the pointer-to-member operator -> . So this:

(struct date_t*)(one)->time

Is the same as this:

(struct date_t*)((one)->time)

You need to parenthesize what is casted, then you can dereference the pointer.

int cmpfunc (const void * one, const void * two) {
    return ((const struct date_t*)(one))->time - ((const struct date_t*)two)->time;
}

Also note that the casted pointers are const to be consistent with the original pointers.

Answer 2

According to the handy precedence table , the cast operation has a lower precedence than the structure member access operator -> . So when doing (struct date_t*)(one)->time , first the member time is accessed (and failed, as one is a void* and has no such a member). And only then the cast is performed on the result. Instead you should force the precedence by using parentheses in appropriate places like:

... ((struct date_t*)one)->time ...