• Hendrik Renken (18/18) Oct 09 2008 Hi,
• torhu (18/42) Oct 09 2008 "data" is a dynamic array reference, so STRUCT is really just this:
• Hendrik Renken (5/26) Oct 10 2008 return s;
• Jarrett Billingsley (7/19) Oct 10 2008 Yes, the contents of *s* will be copied (that is, the array
• Steven Schveighoffer (18/46) Oct 10 2008 the pointer will be copied. If you want a copy of the data, try this:

Hendrik Renken <funsheep gmx.net> writes:

Hi,

i have a question regarding array-initialization.


consider:

struct STRUCT
{
     byte[] data;
}


STRUCT foo(byte[] myData)
{
     STRUCT* s = new STRUCT(myData);
}


do i now create a struct with an array on the heap and then drop the 
array and assign to the array-pointer the array "myData"? If yes, how 
can i avoid the creation and deletion of the array "data" in the first 
place? Am i right, that the variable "data" only holds a pointer to the 
real array?

regards,
hendrik

torhu <no spam.invalid> writes:

Hendrik Renken wrote:
 Hi,
 
 i have a question regarding array-initialization.
 
 
 consider:
 
 struct STRUCT
 {
      byte[] data;
 }
 
 
 STRUCT foo(byte[] myData)
 {
      STRUCT* s = new STRUCT(myData);
 }
 
 
 do i now create a struct with an array on the heap and then drop the 
 array and assign to the array-pointer the array "myData"? If yes, how 
 can i avoid the creation and deletion of the array "data" in the first 
 place? Am i right, that the variable "data" only holds a pointer to the 
 real array?

"data" is a dynamic array reference, so STRUCT is really just this:

struct STRUCT
{
   size_t length;
   byte* ptr;
}


So "data" is just a number of elements and a pointer to the first one. 
When you do "new STRUCT(myData)", you're only allocating room for that. 
  There's no room for array contents  allocated, just the reference.

If you do writeln(STRUCT.sizeof); it will always say 8 on a 32-bit 
system, same goes for data.sizeof.


Your example should probably look like this, as there's no point in heap 
allocating small structs:


STRUCT foo(byte[] myData)
{
      STRUCT s = STRUCT(myData);
}

Hendrik Renken <funsheep gmx.net> writes:

 "data" is a dynamic array reference, so STRUCT is really just this:
 
 struct STRUCT
 {
   size_t length;
   byte* ptr;
 }


This explains everything. Thanks.


  So "data" is just a number of elements and a pointer to the first one.
 When you do "new STRUCT(myData)", you're only allocating room for that. 
  There's no room for array contents  allocated, just the reference.
 
 If you do writeln(STRUCT.sizeof); it will always say 8 on a 32-bit 
 system, same goes for data.sizeof.
 
 


let me modify your example:

 STRUCT foo(byte[] myData)
 {
      STRUCT s = STRUCT(myData);

        return s;
 }


will the content of s now be copied when it is returned?


 Your example should probably look like this, as there's no point in heap
 allocating small structs:

what means small? below how many bytes?

"Jarrett Billingsley" <jarrett.billingsley gmail.com> writes:

On Fri, Oct 10, 2008 at 4:42 AM, Hendrik Renken <funsheep gmx.net> wrote:

 let me modify your example:

 STRUCT foo(byte[] myData)
 {
     STRUCT s = STRUCT(myData);

       return s;
 }


 will the content of s now be copied when it is returned?

Yes, the contents of *s* will be copied (that is, the array
reference), but *not* the array data itself.

 Your example should probably look like this, as there's no point in heap
 allocating small structs:

 what means small? below how many bytes?

It's subjective, but a struct that's only 4-8 bytes can be easily
passed around in registers and there is therefore no need to
heap-allocate it.  Of course on a 64-bit platform even a 16-byte
struct is small.

"Steven Schveighoffer" <schveiguy yahoo.com> writes:

"Hendrik Renken" wrote
 "data" is a dynamic array reference, so STRUCT is really just this:

 struct STRUCT
 {
   size_t length;
   byte* ptr;
 }


 This explains everything. Thanks.


  So "data" is just a number of elements and a pointer to the first one.
 When you do "new STRUCT(myData)", you're only allocating room for that. 
 There's no room for array contents  allocated, just the reference.

 If you do writeln(STRUCT.sizeof); it will always say 8 on a 32-bit 
 system, same goes for data.sizeof.


 let me modify your example:

 STRUCT foo(byte[] myData)
 {
      STRUCT s = STRUCT(myData);

        return s;
 }


 will the content of s now be copied when it is returned?

the pointer will be copied.  If you want a copy of the data, try this:

STRUCT s = STRUCT(myData.dup);

This duplicates the data before setting the pointer.  array.dup is a shallow 
copy, so if you are duping an array of pointers, only the pointers get 
copied, not the data they point to.  If you want a 'deep' copy, you have to 
loop and do it yourself at this point.

 Your example should probably look like this, as there's no point in heap
 allocating small structs:

 what means small? below how many bytes?

All heap allocations are a minimum of 16 bytes.  So if you are allocating a 
struct that just contains a pointer and a length (i.e. an array), that is an 
8-byte struct on a 32 bit system.  That means 100% overhead...  Plus heap 
allocations require grabbing a global thread mutex, finding a suitable 
allocation block, possibly running a garbage collect cycle.  They are slow, 
much slower than using a stack allocated struct.

Of course if you are copying the data, a dup is going to do a heap 
allocation, so not much is gained by putting your struct on the stack. 
However, you probably don't *need* to use the heap for the struct data, I'd 
recommend against it.

-Steve