Creating Struct for an output of a program.

[Mengu]

On Tuesday, 9 January 2018 at 07:57:19 UTC, Vino wrote:
> Hi All,
>
>  Request your help on how to create a struct with the output of 
> the below program.
>
> Program:
> import std.algorithm: all, map, filter;
> import std.stdio: File, writeln;
> import std.typecons: Tuple, tuple;
> import std.container.array;
> import std.string: split, strip;
> import std.uni: isWhite, toLower;
> import std.range: chunks;
>
> void main () {
> Array!string TableData, StructureData;
> auto Meta = 
> File("C:\\Users\\bheev1\\Desktop\\Current\\Script\\Others\\meta\\meta.txt", "r");
> auto MetaData = Array!(Tuple!(string, string))(Meta.byLineCopy()
> .filter!(line => !line.all!isWhite)
> .map!(a => a.split(":"))
> .map!(a => tuple(a[0].toLower.strip, a[1].toLower.strip)));
> foreach (line; MetaData[]) { TableData.insertBack(line[0]); 
> StructureData.insertBack(line[1]); }
> for(int i = 0; i < TableData[].length; i++ ) {
> auto S1 = StructureData[i].split(",").chunks(3);
> auto S2 = S1.map!(a => tuple(a[0],a[1],a[2]));
> for(int z =0; z < S2.length; z++)
> { writefln("%-8s %;s", S2[z][1] , S2[z][0]); }
> }
> }
>
> Output:
> string name;
> string country;
> int age;
>
>
> Need to create as struct using the output
>
> struct Layout {
> {
> string name;
> string country;
> int age;
>
> }
>
> From,
> Vino.B

if S2 consists of data for Layout struct, then you can simply do:

     auto S2 = S1.map!(a => Layout(a[0], a[1], a[2]));

which will give you a range of Layout.