A way to get the value that is used in a ternary operator without creating a new variable

[luizfernandonb]

Usually when we use a ternary operator, we want to return the value that was evaluated, and we usually create a variable that already stores this value, and using a variable to store a temporary value is unnecessary (at least in my view), so mine proposal is to create a keyword that takes the value that was analyzed in the ternary operator

Generally, we would do it like this:

var random = new Random();

int nextInt = random.Next();

Console.WriteLine(nextInt % 0 is 0 ? $"The number {nextInt} is an even number" : $"The number {nextInt} is an odd number");

What I want to propose is a keyword that prevents us from creating a new variable to store a value that we will compare and probably display later, as was done in the example above, thus saving time

According to what I thought, it would be something +- like:

var random = new Random();

Console.WriteLine(random.Next() % 0 is 0 ? $"The number {ternaryvalue} is an even number" : $"The number {ternaryvalue} is an odd number");

Don't mind the name ternaryvalue, it's just a symbol for what I mean, and also because I couldn't think of a better name, but anyway, it's pretty simple, if random.Next() generates any value, like 4318, instead of storing this value in another variable, the keyword ternaryvalue takes the value 4318 and displays it on the screen normally

I believe I have not been clear enough, any questions, just ask

[333fred]

What you're looking for isn't just the value of the ternary, you're asking for the left-hand side of an expression inside the ternary, which is much more complex and not something we can really generally purpose in C#. You could do something like this, but it's not elegant:

var random = new Random();
Console.WriteLine(random.Next is var nextInt && (nextInt % 0 is 0) ? $"The number {nextInt} is an even number" : $"The number {nextInt} is an odd number");

I honestly wouldn't recommend this though: your first example is much clearer.

[luizfernandonb]

The value that is being compared in the example you gave is not random.Next(), it's random.Next() % 0. You can tell because % is not a comparison operator.

You're right, I think we've reached the point @333fred said, that it would be complicated to implement this in C#

[luizfernandonb]

What I ultimately want is just to get random.Next(), but apparently it's pretty impossible

[jnm2]

No possibility stands out to me that would not also be a super subtle set of rules, making it hard to understand when reading the code. That's why explicit variable creation feels better even though you have to type more and invent a name to describe the value you're capturing.

[CyrusNajmabadi]

random.Next() is var ternvalue && ...

[theunrepentantgeek]

@luizfernandonb wrote

What I ultimately want is just to get random.Next(), but apparently it's pretty impossible

It's not impossible - and you don't need to wait years for a new, complicated, hard-to-predict, impossible-to-discover, and difficult-to-document feature to be debated, relitigated, championed, designed, implemented, tested, and released.

int nextInt = random.Next();

Using a temporary variable is the right solution.

The LDM have been quite clear in the past that making C# "more concise" is an explicit non-goal.
Saving keystrokes at the cost of impairing readability is a step backwards, not forwards.

[michaelcheers]

var random = new Random();
Console.WriteLine((defer random.Next()) % 2 == 0 ? $"The number {it} is an even number" : $"The number {it} is an odd number");

[CyrusNajmabadi]

Defer makes me think: happens at the end.

[jnm2]

capture instead of defer?

If it was capture...@, wanting (capture random.Next()) % 2 == 0 ? $"The number {@} is an even number" : ... reminds me vaguely of wanting people.Where(@Age >= 65).Select(@Name).

[michaelcheers]

How about

with(random.Next())(@ % 2 == 0 ? $"The number {@} is an even number" : $"The number {@} is an odd number")

Or similarly:

(v => v % 2 == 0 ? $"The number {v} is an even number" : $"The number {v} is an odd number")(random.Next())

which is also not currently valid but is in javascript.

[huoyaoyuan]

Or similarly:

(v => v % 2 == 0 ? $"The number {v} is an even number" : $"The number {v} is an odd number")(random.Next())

which is also not currently valid but is in javascript.

This should be valid soon with lambda improvement feature.

[jnm2]

I thought you would have to do ((int v) => ...)(random.Next()).

[huoyaoyuan]

using a variable to store a temporary value is unnecessary

I would say that using a variable is necessary. If you are defining much more variables, you should consider creating a helper method.

And for this specific case, random.Next() has side effect and I will avoid to call it in nested expression.