Hi

The pointer (.ptr) of an array is the address of the first array element. But what exactly is the address of the array? And how is it used?

  auto myArray = [5, 10, 15, 20, 25, 30, 35];
  writeln("myArray.ptr: ", myArray.ptr); // myArray.ptr: 7FFA95F29000
  writeln("&myArray[0]: ", &myArray[0]); // &myArray[0]: 7FFA95F29000
  writeln("&myArray: ", &myArray); // &myArray: 7FFE4B576B10
  writeln("*&myArray: ", *&myArray); // *&myArray: [5, 10, 15, 20, 25, 30, 35]

On Thursday, 27 October 2016 at 16:13:34 UTC, David wrote:
> Hi
>
> The pointer (.ptr) of an array is the address of the first array element. But what exactly is the address of the array? And how is it used?
>
>   auto myArray = [5, 10, 15, 20, 25, 30, 35];
>   writeln("myArray.ptr: ", myArray.ptr); // myArray.ptr: 7FFA95F29000
>   writeln("&myArray[0]: ", &myArray[0]); // &myArray[0]: 7FFA95F29000
>   writeln("&myArray: ", &myArray); // &myArray: 7FFE4B576B10
>   writeln("*&myArray: ", *&myArray); // *&myArray: [5, 10, 15, 20, 25, 30, 35]

Sorry that went to quickly ;-)

I observe for example that the even if the pointer is moved to another address the address of the (dynamic) array stays constant:

  auto shrink = myArray[0 .. $-1];
  writeln("shrink.ptr: ", shrink.ptr);
  writeln("&shrink: ", &shrink);

Note: myArray and shrink have the same ptr but a different address

  shrink ~= 100;
  writeln("shrink.ptr: ", shrink.ptr);
  writeln("&shrink: ", &shrink);

Note: After appending, shrink changes its ptr but its address stays the same.

Thanks for your help in advance.

On Thursday, October 27, 2016 16:13:34 David via Digitalmars-d-learn wrote:
> Hi
>
> The pointer (.ptr) of an array is the address of the first array element. But what exactly is the address of the array? And how is it used?
>
>    auto myArray = [5, 10, 15, 20, 25, 30, 35];
>    writeln("myArray.ptr: ", myArray.ptr); // myArray.ptr:
> 7FFA95F29000
>    writeln("&myArray[0]: ", &myArray[0]); // &myArray[0]:
> 7FFA95F29000
>    writeln("&myArray: ", &myArray); // &myArray: 7FFE4B576B10
>    writeln("*&myArray: ", *&myArray); // *&myArray: [5, 10, 15,
> 20, 25, 30, 35]

A dynamic array looks something kind of like this:

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

So, if you take the address of a dynamic array, you're basically taking the address of a struct on the stack, whereas the address in ptr is the address in memory where the data is (be it GC-allocated memory, malloc-ed memory, or a static array on the stack somewhere).

Similarly, if you have a class reference, and you take its address, you're taking the address of the reference, not the class object that it points to. e.g.

class MyClass
{
    string foo;
}

MyClass mc;
auto addr = &mc;

addr is the address of mc on the stack, whereas mc itself is null.

- Jonathan M Davis

> A dynamic array looks something kind of like this:
>
> struct DynamicArray(T)
> {
>     size_t length;
>     T* ptr;
> }
>
> So, if you take the address of a dynamic array, you're basically taking the address of a struct on the stack, whereas the address in ptr is the address in memory where the data is (be it GC-allocated memory, malloc-ed memory, or a static array on the stack somewhere).
>
> Similarly, if you have a class reference, and you take its address, you're taking the address of the reference, not the class object that it points to. e.g.
>
> class MyClass
> {
>     string foo;
> }
>
> MyClass mc;
> auto addr = &mc;
>
> addr is the address of mc on the stack, whereas mc itself is null.
>
> - Jonathan M Davis

Many thanks for your speedy and clear answer Jonathan! That makes even sense to me :-)