Passing pointer to a function

[jewel87]

Hello,
I have a question about pointers.
Could anyone please explain to me the following:
I have a dynamically allocated array of structures like:

struct Employee{......};
 
Employee **pEnteredEmployee;
 
int main()
{
....
pEnteredEmployee = new Employee *[EntryNumber];
    for (int i=0; i < EntryNumber; i++)
        pEnteredEmployee[i] = new Employee;
....
 
 

when I call a function say

 
   InputDataFromFile(EntryNumber, pEnteredEmployee);
 
 

if I understand it correctly, the pEnteredEmploye e is passing to function the value it holds i.e. the address of the first pointer in the array,
but if the function is for example:

  void InputDataFromFile(int EntryNumber, Employee *EnteredEmployee[])
{
      for (int i=0; i < EntryNumber; i++)
      {    InputFile >> EnteredEmployee[i]->Name;
           InputFile >> EnteredEmployee[i]->Surname;
           InputFile >> EnteredEmployee[i]->Grad_Year;
           InputFile >> EnteredEmployee[i]->Average_Mark;
           InputFile >> EnteredEmployee[i]->Salary;
           InputFile >> EnteredEmployee[i]->University;
       }
}

through Employee *EnteredEmploye e[] does it receive the address of the pointer, its value or it converts it to the value this pointer points to due to the dereferencing symbol * i.e. the address of the first structure?
Why I cannot write in the function body like this:

       InputFile >> *EnteredEmployee[i].Name;

I am just trying to understand how exactly it is working.

Thanks a lot!

[Ganon11]

Why I cannot write in the function body like this:

       InputFile >> *EnteredEmployee[i].Name;

I am just trying to understand how exactly it is working.

Thanks a lot!

The reason you cannot do this is because the '.' operator takes precedence over the '*' operator. It is trying to find the Name attribute of a pointer in the above example. You can rewrite it as (*EnteredEmploy ee[i]).Name, and it will work correctly - however, note that this is the reason the '->' operator was introduced. (*Name).attribu te and Name->attribute mean the exact same thing - use whichever one you prefer.

[jewel87]

Thank you very much, now I got it!

And what about the first part of the question? Could someone explain that to me or maybe give a good link with theory on that?
Thanks.

[jewel87]

And one more thing,
I've tried to do the same program using only one pointer to an array of structures, and it works correctly.
I changed the previous code to:

Employee *pEnteredEmployee;
 
void InputDataFromFile(int EntryNumber, Employee EnteredEmployee[])
{
    for (int i=0; i < EntryNumber; i++)
    {           InputFile >> EnteredEmployee[i].Name;
    InputFile >> EnteredEmployee[i].Surname;
    InputFile >> EnteredEmployee[i].Grad_Year;
    InputFile >> EnteredEmployee[i].Average_Mark;
    InputFile >> EnteredEmployee[i].Salary;
    InputFile >> EnteredEmployee[i].University;
   }
};
 
int main()
{
....
pEnteredEmployee = new Employee [EntryNumber];    
 
.....
}

What is the point then of the previous one, if we can use one pointer instead of their array and still allocate memory dinamically?
Or I don't understand anything?
The previous method was suggested in the task for the program.

[Ganon11]

if I understand it correctly, the pEnteredEmploye e is passing to function the value it holds i.e. the address of the first pointer in the array,
but if the function is for example:

  void InputDataFromFile(int EntryNumber, Employee *EnteredEmployee[])
{
      for (int i=0; i < EntryNumber; i++)
      {    InputFile >> EnteredEmployee[i]->Name;
           InputFile >> EnteredEmployee[i]->Surname;
           InputFile >> EnteredEmployee[i]->Grad_Year;
           InputFile >> EnteredEmployee[i]->Average_Mark;
           InputFile >> EnteredEmployee[i]->Salary;
           InputFile >> EnteredEmployee[i]->University;
       }
}

through Employee *EnteredEmploye e[] does it receive the address of the pointer, its value or it converts it to the value this pointer points to due to the dereferencing symbol * i.e. the address of the first structure?

The function is accepting an array of pointers to Employees - that is, it receives a pointer with the memory address of a pointer with the memory address of some Employee object. Thus, you essentially have a pointer to a pointer.

And one more thing,
I've tried to do the same program using only one pointer to an array of structures, and it works correctly.
I changed the previous code to:

Employee *pEnteredEmployee;
 
void InputDataFromFile(int EntryNumber, Employee EnteredEmployee[])
{
    for (int i=0; i < EntryNumber; i++)
    {           InputFile >> EnteredEmployee[i].Name;
    InputFile >> EnteredEmployee[i].Surname;
    InputFile >> EnteredEmployee[i].Grad_Year;
    InputFile >> EnteredEmployee[i].Average_Mark;
    InputFile >> EnteredEmployee[i].Salary;
    InputFile >> EnteredEmployee[i].University;
   }
};
 
int main()
{
....
pEnteredEmployee = new Employee [EntryNumber];    
 
.....
}

What is the point then of the previous one, if we can use one pointer instead of their array and still allocate memory dinamically?
Or I don't understand anything?
The previous method was suggested in the task for the program.

Using double pointers (Employee** name) is usually used for dynamically allocated 2D arrays. This is because you can allocate any number of Employee* pointers in the original array, and each of these pointers can point to any number of Employee objects. I've never seen double pointers used for anything else, though.

[jewel87]

So, I understand if my original task is just to allocate a dynamic array of structures Employee depending on number of entries in the file, the method with only one pointer to the array of structures will do...

Thank you very much for the reply!