Play with elements of Array of Structures in C++

Question

I have just started learning c++ and I am stuck over here. I made a structure

...
struct student{
string name;
int roll;
};
...

And then, I called it globally. student s1[20]; Inside my main function, I did something like this

...
int count=0;
student *p;
p=s1;
cout<<"Enter the Number of records, you wish to consider.";
cin>>arraySize;
p = new student[arraySize];
while (count!=arraySize) {
    cout<<"Enter name.";
    cin>>*(p+count)->name; //Error comes over here
}
...

The error message, I get is Indirection requires pointer operand. Can some one Please help me out with this ??

cin>>*(p+count)->name; change to this cin >> (p + count)->name; *(p + count)->name; resultin in dereference a var that not an address and cause error, you just want to fill the var, so just do it directly. — AchmadJP, Feb 12 '16 at 13:09

Cory Kramer · Accepted Answer · 2016-02-12T13:11:02.880

5

You could either use pointer arithmetic as you are doing

(p+count)->name

or more canonically use [] to index into the array

p[count].name

The reason you are getting an error is because you are combining * which dereferences your pointer with -> which also dereferences your pointer. You could just do one or the other

(p+count)->name
(*(p+count)).name

Though as I said for readability I would prefer using index notation []

edited Feb 12 '16 at 13:11

answered Feb 12 '16 at 12:19

Cory Kramer

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`*(p+count).name` is incorrect. `.` has higher precedence than `*`. – anukul Feb 12 '16 at 13:05
`(*(p+count).)name` is also incorrect. The closing bracket should be before the `.`, not after it. – anukul Feb 12 '16 at 13:09
1

@AnukulSangwan Still only half-way through my coffee this morning :) should be squared away now – Cory Kramer Feb 12 '16 at 13:11
If I ever saw `(*(foo)).bar`, I wouldn't be alright. – erip Feb 12 '16 at 16:59

score 2 · Answer 2 · edited May 23 '17 at 10:28

2

Using *(p+count)->name makes no sense.

Picking up from here, foo->bar calls method bar on the object pointed by pointer foo.

So, you should either do (p+count)->name or (*(p+count)).name.

Note the brackets, they're necessary as . has greater precedence than *.

edited May 23 '17 at 10:28

Community

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answered Feb 12 '16 at 12:52

anukul

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Play with elements of Array of Structures in C++

2 Answers2