c# 8 switch expression multiple cases with same result

问题

How does a switch-expression needs to be written to support multiple cases returning the same result?

With c# prior to version 8 a switch may be written like so:

var switchValue = 3;
var resultText = string.Empty;
switch (switchValue)
{
    case 1:
    case 2:
    case 3:
        resultText = "one to three";
        break;
    case 4:
        resultText = "four";
        break;
    case 5:
        resultText = "five";
        break;
    default:
        resultText = "unkown";
        break;
}

When I am using the c# version 8 expression syntax its like so:

var switchValue = 3;
var resultText = switchValue switch
{
    1 => "one to three",
    2 => "one to three",
    3 => "one to three",
    4 => "four",
    5 => "five",
    _ => "unknown",
};

So my question is: How to make the cases 1,2 and 3 to just one switch-case-arm so the value doesn`t need to be repeated?

Update per suggestion from "Rufus L":

For my given example this works.

var switchValue = 3;
var resultText = switchValue switch
{
    var x when (x >= 1 && x <= 3) => "one to three",
    4 => "four",
    5 => "five",
    _ => "unknown",
};

But its not exactly what I want to accomplish. This is still only one case (with a filter condition), not multiple cases yielding to the same right-hand result.

回答1:

I got around to installing it, but I have not found a way to specify multiple, separate case labels for a single switch section with the new syntax.

However, you can create a new variable that captures the value and then use a condition to represent the cases that should have the same result:

var resultText = switchValue switch
{
    var x when
        x == 1 ||
        x == 2 ||
        x == 3 => "one to three",
    4 => "four",
    5 => "five",
    _ => "unknown",
};

This is actually more concise if you have many cases to test, because you can test a range of values in one line:

var resultText = switchValue switch
{
    var x when x > 0 && x < 4 => "one to three",
    4 => "four",
    5 => "five",
    _ => "unknown",
};

回答2:

Sadly, this appears to be a shortcoming in the switch-expression syntax, relative to the switch-statement syntax. As other posters have suggested, the rather clumsy var syntax is your only real option.

So you might have been hoping you could write:

switchValue switch {
    Type1 t1:
    Type2 t2:
    Type3 t3 => ResultA,
    Type4 t4 => ResultB,
    Type5 t5 => ResultC
};

Instead you will need to write the rather awkward code below, with typename sprayed about:

switchValue switch {
    var x when x is Type1 || x is Type2 || x is Type 3 => ResultA,
    Type4 t4 => ResultB,
    Type5 t5 => ResultC
};

In such a simple example, you can probably live with this awkwardness. But more complicated example are much less liveable with. In fact my examples are actually a simplification of an example drawn from our own code base, where I was hoping to convert a switch-statement, with roughly six outcomes but over a dozen type-cases, into a switch-expression. And the result was clearly less readable than the switch-statement.

My view is that if the switch-expression needs shared outcomes and is more than a few lines long, then you are better off sticking to a switch-statement. Boo! It's more verbose but probably a kindness to your teammates.

ResultType tmp;
switch (switchValue) {
    case Type1 t1:
    case Type2 t2:
    case Type3 t3:
        tmp = ResultA;
        break;
    case Type4 t4:
        tmp = ResultB;
        break;
    case Type5 t5:
        tmp = ResultC;
        break;
};
return tmp;

来源：https://stackoverflow.com/questions/56676260/c-sharp-8-switch-expression-multiple-cases-with-same-result

标签

c#-8.0

switch-expression