Why do we not use a (*) when accessing the value of a dynamic array? [duplicate]

Question

Our professor taught us to use an indirection operator (*) when accessing a dynamic variable. Hence, I fail to understand why the code is written like this:

cout << arr[i];

And not like this:

cout << *arr[i];

Here is the entire code:

#include <iostream>

using namespace std;

int main()
{
    int size;
    int *arr;

    cout << "Please enter size: ";
    cin >> size;

    arr = new int[size];

    cout << "Enter " << size << " items" << endl;
    
    for (int i = 0; i < 5; i++)
    {
        cin >> arr[i];
    }

    cout << "You entered: ";
    for (int i = 0; i < 5; i++)
    {
        cout << arr[i] << " ";
    }
}

Just as a note, you are missing the delete[] at the end. If you dont want to write it, you can use std::unique_ptr<T[]> — Pablochaches
– Pablochaches, Commented Jan 9, 2021 at 1:12

Chris Uzdavinis · Accepted Answer · 2021-01-09 00:32:44Z

1

For arrays like you have:

arr = new int[size];

The syntax:

arr[5];

Is equivalent to this:

*(arr+5)

It's just nicer looking.

answered Jan 9, 2021 at 0:32

Chris Uzdavinis

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2 Comments

SoronelHaetir Over a year ago

And the professor I had for CS101 (introduction to C) in the mid-90s would knock off points if you ever used the [] or -> operators. And he never gave what I consider a convincing argument as to why, just "because I can".

user4581301 Over a year ago

World's full of strange people, @SoronelHaetir . Warning to those that follow. Because of the rules outlined in the post above, you can take a pointer to a single object and use it as an array to very unfortunate effects. Say you have int x; and then you int * xp = &x; to get a pointer to x. Now you can xp[1] = 42; and write over memory that is one integer past x. Not a good idea. Syntactically correct, so it compiles, but logically incorrect. This is a nice example: ideone.com/n4ZcaP. IDEOne tells you you screwed up. Don't count on this level of niceness. C++ is cruel.

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