Associativity of '*' operator in a 'new' statement in C++

Question

I have a user-defined class Person and I want to create a variable-sized array (size taken at runtime) of Person pointers. The following code works as expected.

Person **arr = new Person* [size];

However, I didn't understand exactly whom the * operator is associated with. I tried putting paranthesis like:

new ((Person*)[size])
new (Person (*)[size])

Only the second option works.

This seems counterintuitive as in a regular static declaration, Person* arr[10] would mean an array of Person* type (Array of Person pointers) (associativity with Person).
While Person (*arr)[10] would mean a pointer to an array of Person type (associativity with arr).

Edit: As pointed out by @Jan Schultke I realised that new Person* [size] and new (Person (*)[size]) don't have the same effect. One returns Person** and the other returns Person (**)[size]. Also, the second does not allow size to be a variable.

Again, I know I can use vector for dynamic arrays. I am learning OOP methodology lately, focussing on the concepts of programming language C++, and my aim is to get on grips with what goes on under the hood, rather than simply using a library.

and I want to create a variable-sized array -- In C++, this is spelled std::vector. — PaulMcKenzie
– PaulMcKenzie, Commented May 1 at 6:01
In current C++, just assume that if you have to type new (outside of datastructure internals) you're not using C++ as it is intended to be used. Even though many "teachers" still teach you to use "new". — Pepijn Kramer
– Pepijn Kramer, Commented May 1 at 6:27

Jan Schultke · Accepted Answer · 2025-05-01 05:58:09Z

Grammatically, what follows the new keyword is a new-type-id, which is a limited form of abstract-declarator. When you write new Person*, the principles are the same as if you wrote:

using n = Person*;
// or
void f(Person*);
// or
auto f() -> Person*;

To read abstract-declarators, imagine that there is a variable name in the declaration, like:

Person *p;

Therefore, the * makes the type a pointer, and new Person* is allocating a new pointer.

new Person*[size] allocates a new array size 10 of pointers (not a pointer to an array) because [size] has greater precedence in declarators than *.

Other cases

new (Person (*)[size]) is allocating a new pointer to a Person[size] array. What you're doing here is using the alternative new-expression syntax where you provide a type-id in parentheses, and that works like:

using T = Person(*)[size];
new T;

new ((Person*)[size]) is simply gibberish. It's as if you wrote a C-style cast in a type, like:

using T = (Person*)[size];

While Person (*arr)[10] would mean a pointer to an array of Person type (associativity with arr).

I'm not sure why you're surprised by the behavior. Person(*arr)[10] declares a pointer to an array, and new (Person(*)[10]) allocates a new pointer to an array. The effect of parentheses is identical in both cases.

wohlstad · Accepted Answer · 2025-05-01 06:53:28Z

4

The other answer already answers the immediate question, involving raw C arrays.

However - in C++ it is recommended to use std::vector for dynamic sized arrays.

std::vector offers many advantages over raw C arrays, including automatic memory management.
It also solved the syntax awkwardness that probably confused you:

An "array" of pointers to Person (initialized with size elements) would be:
```
std::vector<Person*> arr(size);
```
A pointer to an "array" of Persons would be:
```
std::vector<Person> * pArr; 
```

A side note:
In C++ it is recommended to avoid using raw new/delete in general.
Most situations where you would need it could be solved by using standard containers (like std::vector) or smart pointers.

edited May 1 at 6:53

answered May 1 at 6:19

wohlstad

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