Built-in array with variable size?

Question

Chapter 2.3.2 of The C++ Programming Language lists this constructor:

class Vector {
public:
    Vector(int s) :elem{new double[s]}, sz{s} { }
private:
    double* elem;
    int sz;
};

As far as I know, the array size must be a constant expression, and s isn't. Is this legal? If so, why?

"the array size must be a constant expression" Since when? — Nicol Bolas
– Nicol Bolas, Commented Feb 20, 2022 at 19:33
You can dynamically allocate an array of a size which is not a constant expression. This is the reason why new[] was invented in the first place. A *declared * array size must be a constant, but new is not a declaration. — n. m. could be an AI
– n. m. could be an AI, Commented Feb 20, 2022 at 19:35
The size specified in a new expression is not required to be a constant expression. Also, in your code, elem is a pointer, not an array. — Peter
– Peter, Commented Feb 20, 2022 at 22:09

Ted Klein Bergman · Accepted Answer · 2022-02-20 19:42:36Z

2

Yes, it's legal because the array is allocated at runtime with the 'new' operator.

If you want to allocate an array at compile-time, you must provide a const int, or constant expression.

int count = 0;
cin >> count; 
int* a = new int[count];  // This is dynamic allocation happen at runtime.
int b[6];                 // This is static allocation and it happen at compile-time.

edited Feb 20, 2022 at 19:42

Ted Klein Bergman

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answered Feb 20, 2022 at 19:39

Huzifa Terkawi

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1 Comment

user17732522 Over a year ago

At block scope int b[6]; is not static allocation. It would be automatic storage duration, "allocated" when the scope is entered, not at compile-time.

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