Directly yield other IEnumerable

[lucasteles]

Today if we have a yieldable IEnumerable function that needs to consume and return values from another IEnumerable we need to explicitly enumerate it and return:

using System;
using System.Linq;
using System.Collections.Generic;

IEnumerable<int> RandomInt(int qtd)
{
   for(var i = 0; i<=qtd; i++)
       yield return Random.Shared.Next(qtd);
}

IEnumerable<int> Values(int qtd)
{
    foreach(var n in Enumerable.Range(0, qtd))
        yield return n;
    
    foreach(var n in RandomInt(qtd))
        yield return n;
}

Console.WriteLine(string.Join(",",Values(5)));

My proposal idea is a syntax to make this simpler, inspired by how F# deals with the same problem:

open System.Linq
open System

let randomInt max = seq {
     for i = 0 to max do
       yield System.Random.Shared.Next max
}

// explicitly enumerate
let valuesExplit max = seq {
   for i in Enumerable.Range(0, max) do
     yield i
     
   for i in randomInt(max) do
     yield i
}

// auto unwrap 
let values max = seq {
     // note the "!"
     yield! Enumerable.Range(0, max)
     yield! randomInt(max)
}

Console.WriteLine(String.Join(",", valuesExplit(5)));
Console.WriteLine(String.Join(",", values(5)));

The "!" is what makes the yield automatically enumerate and return from other enumerable.

This syntax is not csharp-ish, so I thought of something like:

IEnumerable<int> RandomInt(int qtd)
{
   for(var i = 0; i<=qtd; i++)
       yield return Random.Shared.Next(qtd);
}

IEnumerable<int> Values(int qtd)
{
    yield Enumerable.Range(0, qtd);
    yield RandomInt(qtd);
}

This could make yieldable and LINQ to clue seamless and is a nice syntax sugar

[HaloFour]

See also:

#378
#5303

[TonyValenti]

This has been begged for so many times... and denied. :-(

(I'd really like this too!)