user3514538 - 1 year ago 67

C++ Question

I can write operator new for one-dimensional array as follows:

`int n{3};`

new int[n];

It allocates at least

`sizeof(int) * n`

`int n{3};`

new int[n][3]; //ok

new int[n][n]; //error;

Why such restrictions are take place? Are there any difficulties to determine, that it is at least

`sizeof(int) * n * n`

Recommended for you: Get network issues from **WhatsUp Gold**. **Not end users.**

Answer Source

The C++ type system does not include arrays with runtime bound. This is a very complicated thing to do, considering that it will have implications for templates and overload resolution. There have been proposals but none has progressed to being accepted for standardization.

So `T[n]`

is not a valid type. However it can be used in a `new`

-expression because there is a special case for it. The `new`

-expression can be either:

`new X`

, where`X`

is a type`new T[n]`

, where`T`

is a type and`n`

is not a constant expression.

Note that both cases are needed because `T[n]`

is not a type but we want to allow that in a `new`

-expression.

The second point needs a little bit more explanation. It actually uses the C++ infix notation, so if `T`

is an array or function type, the `[n]`

will be in a different place. For example `new int[n][3]`

is OK , which is the same as `typedef int T[3]; new T[n]`

. But `new int[3][n]`

is not.

If we did allow `new int[3][n]`

, what would the return type be? `int (*)[n]`

is not part of the C++ type system as mentioned earlier.

Recommended from our users: **Dynamic Network Monitoring from WhatsUp Gold from IPSwitch**. ** Free Download**