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January 01, 2021 Associative Array Question | ||||
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I'm missing something about the way associative arrays are allocated. In the example below, case 1 and case 2 seem to be the same, a type indexed by a long. Case 2 fails with a Range violation. Can someone explain the difference? I found a work around (or the correct way) in case 3. struct Project { string date; string name; } // case 1 string[long] a1; a1[10] = "testing a1"; foreach (f; a1) writeln ("name ", f); // case 2 Project[long] a2; a2[10].name = "testing a2"; foreach (f; a2) writeln ("name ", f.name); // case 3 Project*[long] a3; a3[10] = new Project; a3[10].name = "testing a3"; foreach (f; a3) writeln ("name ", f.name); |
January 01, 2021 Re: Associative Array Question | ||||
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Posted in reply to Chris Bare | On Friday, 1 January 2021 at 01:43:50 UTC, Chris Bare wrote: > a1[10] = "testing a1"; this actually constructs a new thing since it is a straight x = y assignment. > a2[10].name = "testing a2"; But this looks up something first. It doesn't construct a2[10], it looks it up first to fetch the name member... hence the range violation since it isn't constructed yet. > a3[10] = new Project; and this again will construct since it is a straight x = y again. So you could also do a2[10] = Person() then do the lookup of name |
January 01, 2021 Re: Associative Array Question | ||||
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Posted in reply to Adam D. Ruppe | On 12/31/20 8:54 PM, Adam D. Ruppe wrote:
> On Friday, 1 January 2021 at 01:43:50 UTC, Chris Bare wrote:
>> a1[10] = "testing a1";
>
> this actually constructs a new thing since it is a straight x = y assignment.
>
>> a2[10].name = "testing a2";
>
> But this looks up something first. It doesn't construct a2[10], it looks it up first to fetch the name member... hence the range violation since it isn't constructed yet.
>
>
>> a3[10] = new Project;
>
> and this again will construct since it is a straight x = y again.
>
>
>
> So you could also do
>
> a2[10] = Person()
>
> then do the lookup of name
or a2[10] = Person(null, "testing a2")
-Steve
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