Struct pointer cast doubt

Question

I can’t understand the meaning of this cast: the function will execute and returns a type type_t, which is a typedef for void*.

Then is made a cast for header_t*, which is a struct, but I can’t understand how this occurs cast. When I make a cast like this, the function’s return value becomes a type header_t*.

How does the cast values for abstract data type?

header_t *orig_hole_header = (header_t *)lookup_ordered_array(iterator, heap->index);

u32int orig_hole_pos = (u32int)orig_hole_header;

Answer 1

First of all

header_t* is a pointer to the type header_t.

Answer

The pointers for the void type void* are pointers for any type. They are used when the programmer does not know what kind of information he will receive. That is, if it is a int, char...

As such these can be converted to any type of pointer.

int i = 5;
void *p = &i;
*p = 4;//Valor de i agora é 4

Why the cast (header_t *) ??

For no reason. At least in C it is possible to cast void* for any type of pointer, not adding any necessary information to the programmer (until it makes the readability of the code difficult), but in C++ it is mandatory to cast.

Example of void type return*

malloc your return is void* because the programmer can set whatever size he wants, so it didn’t make sense for him to have a type with defined size.

Function cast problem returning void*

If you forget to include the header file stdlib.h and call the function malloc you will have problems because in C the standard type is int so would be considered int malloc() and the cast would only hide Warning depending on the compiler.