import std.stdio;

void main()
{
    ubyte[] a1 = new ubyte[65];
    ubyte[65] a2;

    writeln("a1.sizeof = ", a1.sizeof); // prints 16
    writeln("a2.sizeof = ", a2.sizeof); // prints 65
}

Why a1.sizeof is 16?

On Thursday, 11 June 2015 at 20:09:38 UTC, Adel Mamin wrote:
> import std.stdio;
>
> void main()
> {
>     ubyte[] a1 = new ubyte[65];
>     ubyte[65] a2;
>
>     writeln("a1.sizeof = ", a1.sizeof); // prints 16
>     writeln("a2.sizeof = ", a2.sizeof); // prints 65
> }
>
> Why a1.sizeof is 16?

ubyte[] is a slice, which is actually a struct. It's more or less the same as:

struct Slice(T)
{
    T* ptr;
    size_t length;
}

So sizeof returns the size of the struct, not the size of the data that its ptr member points to.

ubyte[65] is a static array, which is just a big block of data on the stack. That's why it returns the expected value for sizeof. To create a slice of a static array, use the slice operator:

writeln(sizeof(a2[])); //Prints 16

On Thursday, 11 June 2015 at 20:09:38 UTC, Adel Mamin wrote:
> Why a1.sizeof is 16?

sizeof is tied to *type*, not a variable. (I kinda wish a1.sizeof was prohibited, forcing you to say typeof(a1).sizeof so it is clear but whatever).

A dynamic array's size is the length variable plus the pointer variable.

A static array's size is the content itself.


If you want the size of the content in bytes, best way is to do (cast(ubyte[]) a1[]).length or somethign like that - use the length property instead of sizeof.