How can I do that? I though something like this:

auto arr = [1, 2, 3, 4];
arr = arr[0 .. 0];
arr ~= 6; // does this cause reallocation?

assumeSafeAppend() wouldn't work in this case because I don't know the number of items that is going to be added to the array. I thought into setting the length property to 0. Does this free the memory block of the array? if it did reuse the existing memory block in further appends, would do what I want

On Sunday, 7 February 2021 at 21:31:11 UTC, Jack wrote:
> assumeSafeAppend() wouldn't work in this case because I don't know the number of items that is going to be added to the array.

I don't think that matters. assumeSafeAppend seems appropriate for your need.

On Sunday, 7 February 2021 at 21:34:22 UTC, Adam D. Ruppe wrote:
> On Sunday, 7 February 2021 at 21:31:11 UTC, Jack wrote:
>> assumeSafeAppend() wouldn't work in this case because I don't know the number of items that is going to be added to the array.
>
> I don't think that matters. assumeSafeAppend seems appropriate for your need.

I think it would be fine except it assumes the number of items of the array to doesn't grow, it rather overwritten new elements

from docs:

"Use this only when it is certain there are no elements in use beyond the array in the memory block. If there are, those elements will be overwritten by appending to this array."

On Sunday, 7 February 2021 at 21:40:12 UTC, Jack wrote:
> I think it would be fine except it assumes the number of items of the array to doesn't grow, it rather overwritten new elements
>
> from docs:
>
> "Use this only when it is certain there are no elements in use beyond the array in the memory block. If there are, those elements will be overwritten by appending to this array."

That's referring to a case like this:

int[] a = [1, 2, 3, 4, 5];
int[] b = a[0 .. 3];

Normally, if you were to do

b ~= 6;

it would allocate a new array for b, leaving a alone.

a == [1, 2, 3, 4, 5];
b == [1, 2, 3, 6];

a.ptr != b.ptr because b got reallocated.



If you assumeSafeAppended there though, the b ~= 6 would reuse the remainder of the block.

b.assumeSafeAppend();
b ~= 6;

a == [1, 2, 3, 6, 5];
b == [1, 2, 3, 6];

a.ptr == b.ptr; // assumeSafeAppend meant no reallocation




So the "elements in use beyond the array in the memory block" are referring to the variable a in the example still having [4, 5] at the end. Since that 4 gets overwritten by the 6, if you weren't prepared for this, it can be a surprising bug.

The docs also say this is "undefined behavior" simply because the 4 won't *always* get overwritten by the 6. well, in this case the 4 is always overwritten by the 6, but say I wanted to append [6, 7, 8], then it might reallocate anyway because the memory block is only big enough for two new elements, not three. If that happened, that 4 would stay put.

So the array is allowed to grow as much as it wants, just in some cases it will overwrite existing data and in other cases it will realloc a new slice to make room.

On Sunday, 7 February 2021 at 21:55:34 UTC, Adam D. Ruppe wrote:
> On Sunday, 7 February 2021 at 21:40:12 UTC, Jack wrote:
>> I think it would be fine except it assumes the number of items of the array to doesn't grow, it rather overwritten new elements
>>
>> from docs:
>>
>> "Use this only when it is certain there are no elements in use beyond the array in the memory block. If there are, those elements will be overwritten by appending to this array."
>
> That's referring to a case like this:
>
> int[] a = [1, 2, 3, 4, 5];
> int[] b = a[0 .. 3];
>
> Normally, if you were to do
>
> b ~= 6;
>
> it would allocate a new array for b, leaving a alone.
>
> a == [1, 2, 3, 4, 5];
> b == [1, 2, 3, 6];
>
> a.ptr != b.ptr because b got reallocated.
>
>
>
> If you assumeSafeAppended there though, the b ~= 6 would reuse the remainder of the block.
>
> b.assumeSafeAppend();
> b ~= 6;
>
> a == [1, 2, 3, 6, 5];
> b == [1, 2, 3, 6];
>
> a.ptr == b.ptr; // assumeSafeAppend meant no reallocation
>
>
>
>
> So the "elements in use beyond the array in the memory block" are referring to the variable a in the example still having [4, 5] at the end. Since that 4 gets overwritten by the 6, if you weren't prepared for this, it can be a surprising bug.
>
> The docs also say this is "undefined behavior" simply because the 4 won't *always* get overwritten by the 6. well, in this case the 4 is always overwritten by the 6, but say I wanted to append [6, 7, 8], then it might reallocate anyway because the memory block is only big enough for two new elements, not three. If that happened, that 4 would stay put.
>
> So the array is allowed to grow as much as it wants, just in some cases it will overwrite existing data and in other cases it will realloc a new slice to make room.

got it, thank you very much!