pointer array to structure in C++

Question

int noOfEmployee = 0;
cout << "Enter no. of Employee" << endl;
cin >> noOfEmployee;

Ereg = new EMPLOYEE[noOfEmployee];

string defaultName = "Emloyee";

for(int i = 0; i < noOfEmployee; i++) {
    Ereg->regno = i + 1;
    Ereg->name = defaultName;
}

for(int i = 0; i < noOfEmployee; i++) {
    cout << Ereg->regno << "\t"
         << Ereg->name << endl;
}

delete [] Ereg; //segmentation Fault if [] missed

output is:

Enter no. of employee
5
5    Employee
5    Employee
5    Employee
5    Employee

How could access array elements in this or do something like this

Ereg[i]->regno = i;
Ereg[i]->name = defaultName;

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

Your output repeats itself because you never use Ereg as an array, only as a pointer to EMPLOYEE.

How could access array elememts in this or do something like this
Ereg[i]->regno = i;
Ereg[i]->name = defaultName;

Use . instead of ->, because Ereg[i] is a struct, not a pointer to struct.

//segmentation Fault if [] missed

This is expected. You need [] because you allocated an array. Hence, delete[] must be used to avoid undefined behavior.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered May 27, 2016 at 2:45

Sergey Kalinichenko

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Comments

songyuanyao · Accepted Answer · 2016-05-27 02:51:53Z

1

Ereg points to the 1st element of array, so Ereg->regno or Ereg->name will always access the 1st element.

How could access array elements

You should

for(int i = 0; i < noOfEmployee; i++) {
    Ereg[i].regno = i + 1;
    Ereg[i].name = defaultName;
}

for(int i = 0; i < noOfEmployee; i++) {
    cout << Ereg[i].regno << "\t"
         << Ereg[i].name << endl;
}

See subscript operator.

BTW

//segmentation Fault if [] missed

You should use delete[] for array new[]d (and use delete for pointer newd).

edited May 27, 2016 at 2:51

answered May 27, 2016 at 2:44

songyuanyao

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Comments

Shiv Pratap Singh · Accepted Answer · 2016-05-27 03:44:45Z

0

Thanks for help,

There is one way of doing this i got just now.

As incrementing a structure pointer will move it towards next structure location so I used basic method.

(Ereg + i)->regno = i;
(Ereg + i)->name = defaultname;

And it worked just fine.

answered May 27, 2016 at 3:44

Shiv Pratap Singh

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Comments

Shiv Pratap Singh · Accepted Answer · 2016-05-27 04:11:38Z

0

This is some kind of fun....

Also I can do this to use as a array.

As we know

*(Ereg + i) = Ereg[i]
// this is how array works, so I can write this

(Ereg + i) as (&Erg[i])

And this also worked fine.

answered May 27, 2016 at 4:11

Shiv Pratap Singh

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