CODEWITHSUNDEEP
c++pointers
user5162957
user5162957

Reputation:

What does expression *(int*)a mean?

int compare (const void * a, const void * b)
{
  return ( *(int*)a - *(int*)b );
}

I don't understand expression *(int*)a?

Upvotes: 0

Views: 308

Answers (3)

Mukit09
Mukit09

Reputation: 3399

(int*)a converts void pointer to int type pointer. *(int*)a takes the value that the pointer points to.

Upvotes: 0

Some programmer dude
Some programmer dude

Reputation: 409176

A void * pointer is a generic pointer, it can point to anything, but as the compiler doesn't know exactly what it's pointing to you have to tell the compiler what exactly it's pointing to. This "telling" is what the cast does: (int *) a tells the compiler that a is actually pointing to an int.

Then the code is simply using the dereference operator (unary *) to get the value of where a is pointing and using that in a normal subtraction expression.

Upvotes: 8

cadaniluk
cadaniluk

Reputation: 15229

(int*)a

casts a to int*. Simply dereferencing a doesn't work because *a would have type void, which is not allowed.

Furthermore, this is somewhat unsafe here as a is passed as const and the caller expects *a not to be changed. OK, a isn't changed indeed but such an explicit conversion should be a rarity.

The additional dereferencing "gets" the value a points to. In the end, compare returns 0 if both data are equal.

Upvotes: 4

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