Questions about a college project - c

Question

I have
the struct:
typedef struct Rental {
int nDays;
float kmsDriven;
char carLicensePlate[LICENSE_PLATE_LENGTH+1];
char *clientName;
char chargingCategory;
} Rental;
Different -Rental type- structs are stored and accessed via a dynamically allocated array of pointers (here is a part of the project):
int main (){
Rental *rentals;
int max_num;
printf("Give a number of rentals you would like to store and manage: ");
scanf("%d", &max_num);
rentals=(Rentals *)malloc(max_num * (sizeof(Rental)))
This is what I have thought of so far but I can't understand it completely...so:
I'm having trouble understanding how *rentals can be an array. I mean shouldn't I declare it at least this way: Rental *rentals[];? I know that if I compile the above code I will see an error...but why?
I've read numerous posts here in Stack Overflow about doing this with double pointers (Rental **rentals;) but the code other people have posted is often very hard for me to read (I don't know all the functions etc. etc.)
Let's say I have the object rentals[0] which will be a pointer towards rentals. If I wanted to pass the struct to a function, should I write:
variable=function(*arguments*... , Rental *rentals[0]);?

Answer 1

rentals is a pointer, not an array, but it is a pointer to the first (zeroth) element of a block of max_num structures, so it can be treated as an array in that you can use rentals[n] to refer to the nth element of the array.
This is not a question and hence it is unanswerable.
Let's say I have the object rentals[0] which will be a pointer towards rentals. If I wanted to pass the struct to a function, should I write: variable=function(*arguments*... , Rental *rentals[0]);?
rentals[0] is not a pointer; it is a struct Rental or Rental.
If you want to pass the structure to the function, you write:
variable = function(…args…, rentals[0]);
If you want to pass a pointer to the structure to the function, you write:
variable = function(…args…, &rentals[0]);
or:
variable = function(…args…, rentals);
These pass the same address to the function.
You should be error checking the call to scanf() to make sure you got a number, and you should error check the number you got (it should be strictly positive, not zero or negative), and you should error check the value returned by malloc().

Answer 2

When you declare an array (for example char buffer[10]; the variable is actually pointing to that array. Pointers and arrays are very close together. In fact when you have a pointer where you store an array of data (just like your case with malloc) you can do something like pointer[0] and pointer[1] to get the correct element.
With a pointer in order to access an element you'd normally use *(pointer +1) to get the element on position 1, this is exactly the same as pointer[1].
When you want to pass a struct in an array, you can either give it by value like this:
void function(struct mystruct var)
{
//...
}
int main()
{
struct mystruct var;
function(var);
}
Or by reference (passing the address instead of the data - this is ideal if your structs are big in size) :
void function(struct mystruct *var)
{
//...
}
int main()
{
struct mystruct var;
function(&var);
}
By using an array, you can do it like this (still by reference):
void function(struct mystruct *var)
{
//...
}
int main()
{
struct mystruct var[10];
function(&var[0]);
}
And using a pointer (to an array) :
void function(struct mystruct *var)
{
//...
}
int main()
{
struct mystruct *var;
var = malloc( sizeof(struct mystruct) *10 );
//This will pass the address of the whole array (from position 0)
function(&var);
//This will pass the address of the selected element
function(&var[0]);
}
As you can see, declaring an array or a pointer is almost the same, expect that you have to initialize the pointer-array yourself (with malloc) and as with anything created with malloc you have to free it yourself too.