"in" operator gives a pointer result from a test against an Associative Array?

May 10

Andy Valencia

May 10

Meta

May 10

May 10

May 10

May 10

May 10

tst7.d(6): Error: cannot implicitly convert expression `e in this.members` of type `bool*` to `bool` tst7.d(15): Error: template instance `tst7.Foo!uint` error instantiating I'm getting this for this bit of source (trimmed from the bigger code). I switched to this.members.get(e, false) and that works fine, but I'm still curious: struct Foo(T) { bool[T] members; bool has(T e) { return (e in this.members); } } void main() { import std.stdio : writeln; auto t = Foo!uint(); writeln(t.has(123)); }

On Friday, 10 May 2024 at 00:18:16 UTC, Andy Valencia wrote:

tst7.d(6): Error: cannot implicitly convert expression e in this.members of type bool* to bool
tst7.d(15): Error: template instance tst7.Foo!uint error instantiating

I'm getting this for this bit of source (trimmed from the bigger code). I switched to this.members.get(e, false) and that works fine, but I'm still curious:

struct Foo(T) {
bool[T] members;

bool
has(T e) {
    return (e in this.members);
}

}

void
main()
{
import std.stdio : writeln;

auto t = Foo!uint();
writeln(t.has(123));

}

Yes. The reason for this is that it avoids having to essentially do the same check twice. If in returned a bool instead of a pointer, after checking for whether the element exists (which requires searching for the element in the associative array), you'd then have to actually get it from the array, which would require searching again. Returning a pointer to the element if it exists (or null if it doesn't) cuts this down to 1 operation.

May 10

Re: "in" operator gives a pointer result from a test against an Associative Array?

Posted by Steven Schveighoffer
in reply to Andy Valencia

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Steven Schveighoffer

Posted in reply to Andy Valencia

Permalink

On Friday, 10 May 2024 at 01:00:09 UTC, Andy Valencia wrote:

On Friday, 10 May 2024 at 00:40:01 UTC, Meta wrote:

Looking at Programming in D section 28.5, I'm guessing that pointer versus null is treated as the appropriate boolean value when consumed by an "if" test. So that example is getting a pointer to a string, or null, but the example looks exactly as the same as if it had directly gotten a bool.

Yes, we say that a type has "truthiness" if it can be used in a condition (while, if, assert, etc).

For a pointer, null is considered "false", whereas any other value is considered "true". So you can use statements like if(key in aa) to test for membership. A very nice idiom is to check if a key is in an associative array, and if so, use the value that it maps to:

if(auto v = key in aa) {
   // use *v as the value here
}

You can change your code to return (e in this.members) !is null;

-Steve

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