It seems a bit silly that I have to write

int[] a = [1, 2, 300, -29];
int[] b;
b.length = 4;
b[] = a[] * 2;
writeln(b);

to do what I would expect

int[] a = [1, 2, 300, -29];
writeln(a[] * 2);

to do. What am I not understanding?

On Saturday, 6 January 2018 at 23:08:14 UTC, Lily wrote:
> What am I not understanding?

Where do you expect it to put the result?

On 01/06/2018 03:08 PM, Lily wrote:
> It seems a bit silly that I have to write
> 
> int[] a = [1, 2, 300, -29];
> int[] b;
> b.length = 4;
> b[] = a[] * 2;
> writeln(b);
> 
> to do what I would expect
> 
> int[] a = [1, 2, 300, -29];
> writeln(a[] * 2);
> 
> to do. What am I not understanding?

So, apparently a[] * 2 is not an expression in D. It looks like such array-wise operations require a left-hand side with memory for the results.

The reason must be for performance. If a[]*2 were an expression, the runtime would have to allocate memory and put the results there. Assigning that memory then to b[] would require an additional copy. Current rule puts the burden on the programmer to find the memory. No need to allocate if there's memory already.

However, idiomatic D delays arrays as much as possible. For example, there is no need for destination memory for this example:

import std.stdio;
import std.algorithm;

void main() {
    int[] a = [1, 2, 300, -29];
    writeln(a.map!(x => 2 * x));
}

Ali

On Saturday, 6 January 2018 at 23:17:53 UTC, Ali Çehreli wrote:
> So, apparently a[] * 2 is not an expression in D.

> The reason must be for performance. If a[]*2 were an expression, the runtime would have to allocate memory and put the results there. Assigning that memory then to b[] would require an additional copy. Current rule puts the burden on the programmer to find the memory. No need to allocate if there's memory already.

That makes sense, thanks!