On Sunday, 5 November 2023 at 18:36:40 UTC, Ctn-Dev wrote:
I wrote this earlier:
[...]
`if` runs when both "One" and "Two" are in the given array as
intended, but its conditional statement looks verbose. Is there
a more concise way of getting the same result?
If sorting the arrays is an option, you can use `setIntersection`
from `std.algorithm.setops`, which returns the overlap between
two sorted ranges:
```d
import std.stdio;
import std.algorithm.searching: count;
import std.algorithm.setops: setIntersection;
import std.algorithm.sorting: sort;
string[] numeric_traits_1 = [];
string[] numeric_traits_2 = ["Two"];
string[] numeric_traits_3 = ["One"];
string[] numeric_traits_4 = ["One", "Two"];
void main() {
void numeric_traits_contains(string[] numeric) {
sort(numeric); // required by setIntersection
size_t howMany = setIntersection(numeric, ["One",
"Two"]).count;
if (howMany == 2) {
writeln("Array contains 'One' and 'Two'.");
} else if(howMany == 1) {
writeln("Array contains 'One' or 'Two'.");
} else {
writeln("Array contains neither 'One' nor 'Two'");
}
}
numeric_traits_contains(numeric_traits_1);
numeric_traits_contains(numeric_traits_2);
numeric_traits_contains(numeric_traits_3);
numeric_traits_contains(numeric_traits_4);
}
```
