• Adel Mamin (9/9) Jun 11 2015 import std.stdio;
• Meta (14/23) Jun 11 2015 ubyte[] is a slice, which is actually a struct. It's more or less
• Adam D. Ruppe (10/11) Jun 11 2015 sizeof is tied to *type*, not a variable. (I kinda wish a1.sizeof

"Adel Mamin" <adel mm.st> writes:

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?

"Meta" <jared771 gmail.com> writes:

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

"Adam D. Ruppe" <destructionator gmail.com> writes:

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.