countUntil to print all the index of a given string.

[Vino]

On Tuesday, 20 February 2018 at 08:50:27 UTC, Jonathan M Davis 
wrote:
> On Tuesday, February 20, 2018 08:44:37 aberba via 
> Digitalmars-d-learn wrote:
>> On Sunday, 18 February 2018 at 15:23:14 UTC, Cym13 wrote:
>> > On Sunday, 18 February 2018 at 14:48:59 UTC, Cym13 wrote:
>> >> [...]
>> >
>> > Just thought of a much better/simpler solution for that last 
>> > case that also doesn't force you to read all data (which 
>> > might
>> >
>> > be impossible when dealing with infinite ranges):
>> >     import std.range;
>> >     import std.algorithm;
>> >
>> >     a[]
>> >
>> >      .enumerate                       // get tuples (index,
>> >
>> > value)
>> >
>> >      .filter!(t => t[1] == "Test2")   // keep only if value 
>> > ==
>> >
>> > "Test2"
>> >
>> >      .map!(t => t[0])                 // keep only the index
>> >
>> > part
>> >
>> >      .writeln;
>> >
>> > Completely lazy.
>>
>> How does one detect an operation as lazy or not?  Is the some 
>> compile-time or runtime check for that?
>>
>> My guess is by referring to the docs function signature.
>
> You have to read the docs or read the source code, though in 
> general, functions that return a range type that wraps the 
> original range tend to be lazy, whereas if a function returns 
> the original range type or an array, then it's clearly not lazy.
>
> - Jonathan M Davis

Hi All,

  Thank you very much, the provide solution work's for the given 
example, so how can we achieve the same for the below code


import std.stdio;
import std.container;
import std.algorithm;

void main () {
auto a = Array!string("Test1", "Test2", "Test3", "Test1", 
"Test2");
auto b = Array!string("Test1", "Test2", "Test3");

foreach(i; b[]) {
writeln(SList!int(a[].countUntil(i))[]); }
}

Output
[0]
[1]
[2]

Required
[0,3]
[1,4]
[2]

From,
Vino.B