I have been looking at this first draft by Rikki for a few days now. We have discussed it many times before:

• First Draft: Member of Operator
• Type Inference for Struct/Enum Literals
• The enum type inference problem is easily solved with identifier types

Finally we have this. Because it is the best way to solve the example below!

alias DaysoftheWeekinEnglish D;

void main()
{
  // with (Days)
  auto days = [
      "Pazar"     : D.Sunday,
      "Pazartesi" : D.Monday,
      "Salı"      : D.Tuesday,
      "Çarşamba"  : D.Wednesday,
      "Perşembe"  : D.Thursday,
      "Cuma"      : D.Friday,
      "Cumartesi" : D.Saturday
  ];
  assert(days["Cuma"] == DaysoftheWeekinEnglish.Friday);
  assert(days["Pazar"] == D.Sunday);

In this example, if you open the with() scope, everything works inside the scope and the object is created correctly. However, outside the scope, the object does not actually exist!

However, Rikki wrote that the proposed draft for this example would not be a solution. However, with() looks nice but should be used with caution.

SDB@79

On Friday, 13 September 2024 at 18:49:06 UTC, Salih Dincer wrote:

alias DaysoftheWeekinEnglish D;

void main()
{
  // with (Days)
  auto days = [
      "Pazar"     : D.Sunday,
      "Pazartesi" : D.Monday,
      "Salı"      : D.Tuesday,
      "Çarşamba"  : D.Wednesday,
      "Perşembe"  : D.Thursday,
      "Cuma"      : D.Friday,
      "Cumartesi" : D.Saturday
  ];
  assert(days["Cuma"] == DaysoftheWeekinEnglish.Friday);
  assert(days["Pazar"] == D.Sunday);

This could also be solved by allowing with() to be used as an expression, not just a statement:

void main()
{
    auto days = with (Days) [
        "Pazar" : Sunday,
        // ...
    ];
}

As for scope extension, C# has extended scope statements:
This is normal scoped statement:

with(Days){ ... }

This is extended scoped statement:

{
with Days;
...
}

Lol, thank you, markdown.

with(Days){ ... }

{
	with Days;
	...
}

On Monday, 16 September 2024 at 08:47:51 UTC, Kagamin wrote:

with(Days){ ... }

{
	with Days;
	...
}

Moving on to C++, there is no direct operator to shorten enum member names. However, there are several methods to make enum members more readable. You can shorten long enum names using typedef or using:

#include <iostream>
using namespace std;

//* <--For other solution remove the first character
enum Days
{
  Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
};

typedef Days Day;

/*/
//  Namespace Usage:
//  You can place enum members inside
//  a namespace to use shorter names:


namespace Days
{
  enum Day
  {
    Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
  };
}

using namespace Days;//*/

int main()
{
    Day today = Sunday;

    cout << "Hello " << today;

    return 0;
}

From these examples, we see that C++ is proactive in solving the problem.

SDB@79

On Saturday, 14 September 2024 at 01:01:06 UTC, Paul Backus wrote:

In fact, if the D parser is not sufficient, you can make your own parser using a simple regular expression! Dconf'24 brought me such a perspective:

import std.regex, std.stdio;

enum Days
{
  Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
};

void main()
{
  auto jsonStr = `{
      "Pazar": Sunday,
      "Pazartesi": Monday,
      "Salı": Tuesday,
      "Çarşamba": Wednesday,
      "Perşembe": Thursday,
      "Cuma": Friday,
      "Cumartesi": Saturday,
  }`;

  Days[string] Item;

  enum regex = `"(\w+)": (\w+),`;
  foreach (matches; jsonStr.matchAll(regex))
  {
    string key = matches.captures[1];
    string value = matches.captures[2];
    writefln("\"%s\": %s", key, value);
    Item[key] = value.indexOf!Days;
  }
  Item.writeln;
}

auto indexOf(E)(string value)
{
  import std.conv : to;

  E count;
  while (count < count.max)
  {
    if (count.to!string == value) break;
    ++count;
  }
  return count;
}

Prints:

/*
"Pazar": Sunday
"Pazartesi": Monday
"Salı": Tuesday
"Çarşamba": Wednesday
"Perşembe": Thursday
"Cuma": Friday
"Cumartesi": Saturday
["Cumartesi":Saturday, "Pazartesi":Monday, "Pazar":Sunday, "Salı":Tuesday, "Çarşamba":Wednesday, "Perşembe":Thursday, "Cuma":Friday]

Process finished.
*/

SDB@79

On Monday, 16 September 2024 at 08:49:39 UTC, Kagamin wrote:

with(Days){ ... }

{
	with Days;
	...
}

By far the most elegant solution. Especially as in most cases, the extra scope is either already there or wouldn’t be needed. In a switch statement, it can even go before the first case:

switch (day)
{
    with Days;
case Monday:
    …;
    break;
case Tuesday:
    …;
    break;
}

Still, I think for switch in particular, if a label is a qualified identifier QI and doesn’t resolve, it should try typeof(SwitchedExpression).QI. Solves most cases and is flexible.

If you have an extra scope to apply, it already works:

with(Days)
switch(day)
{
case Monday:
     …;
     break;
case Tuesday:
     …;
     break;
}

On Wednesday, 18 September 2024 at 16:30:41 UTC, Kagamin wrote:

with(Days)
switch(day)
{
case Monday:
     …;
     break;
case Tuesday:
     …;
     break;
}

Is there a with() in C#? It seems like there isn't one; record has a with!

SDB@79

On Saturday, 14 September 2024 at 01:01:06 UTC, Paul Backus wrote:

void main()
{
    auto days = with (Days) [
        "Pazar" : Sunday,
        // ...
    ];
}

https://github.com/dlang/dmd/pull/13776