A trap of switch pattern

[LeaFrock]

Without debugging, tell me the result of x.

int c = 3;

int x = c % 3 switch
{
    0 => c / 3,
    1 => (c - 4) / 3 + 2,
    _ => (c - 2) / 3 + 1,
};

[KennethHoff]

I'd imagine it'd be 1, but I know that the % operator is kind of weird.

[HaloFour]

The result is 0.

Seems to be an issue of precedence, switch is evaluated before %.

[HaloFour]

Here's the documentation on operator precedence:

https://learn.microsoft.com/en-us/dotnet/csharp/language-reference/operators/#operator-precedence

[KyouyamaKazusa0805]

You should treat switch as a unary expression so this expression will be treated as follows:

int c = 3;
int x = c % (3 switch
{
    0 => c / 3,
    1 => (c - 4) / 3 + 2,
    _ => (c - 2) / 3 + 1,
});

switch expression is an "entire" expression, meaning you cannot split a switch expression into two individual parts (3 and { ... }).

Therefore the value is 0 (3 switch { _ => (c - 2) / 3 + 1 } -> 1 / 3 + 1 -> 1, 3 % 1 -> 0).

---

There's a potential rule for operators: The precedence of binary expressions is lower than unary expressions, which means you should firstly calculate for the result of an unary, and then binary if they are mixed without braces.

From this rule, the expression won't be considered as (c % 3) switch, because c % 3 is a binary expression, which is lower than 3 switch {...}.

Here is just one exception: range operator x..y is before unary expressions switch and with. This is because the type of two operands must be an int value or an expression that can be implicitly converted into an int value. This rule limits the usage on operands.
In addition, operands in range operators can be omitted. This is the main reason why it is before all binary operators and we should add an extra brace if we use +, - and other binary operators.

[LeaFrock]

A great 'Show and tell' 👍

[RasheedHendawi]

Switch will work first *

because the precedence of unary expressions is higher than binary.