问题
C# 8 introduced nullable reference types which is a very cool feature. Now if you expect to get nullable value you have to write so-called guards:
object? value = null;
if (value is null)
{
throw new ArgumentNullException();
}
...
These can be a bit repetetive. What I am wondering is it possible to avoid writing this type of code for every variable but have a guard-type static void function that throws exception if value is null or just returns if value is not null. Or is this too hard for compiler to infer? Especially if it's external library/package?
回答1:
There are a few things you can do.
You can use [DoesNotReturnIf(...)]
in your guard method, to indicate that it throws if a particular condition is true or false, for example:
public static class Ensure
{
public static void True([DoesNotReturnIf(false)] bool condition)
{
if (!condition)
{
throw new Exception("!!!");
}
}
}
Then:
public void TestMethod(object? o)
{
Ensure.True(o != null);
Console.WriteLine(o.ToString()); // No warning
}
This works because:
[DoesNotReturnIf(bool)]
: Placed on a bool parameter. Code after the call is unreachable if the parameter has the specified bool value
Alternatively, you can declare a guard method like this:
public static class Ensure
{
public static void NotNull([NotNull] object? o)
{
if (o is null)
{
throw new Exception("!!!");
}
}
}
And use it like this:
public void TestMethod(object? o)
{
Ensure.NotNull(o);
Console.WriteLine(o.ToString()); // No warning
}
This works because:
[NotNull]
: For outputs (ref/out parameters, return values), the output will not be null, even if the type allows it. For inputs (by-value/in parameters) the value passed is known not to be null when we return.
SharpLab with examples
Of course, the real question is why you want to do this. If you don't expect value
to be null
, then declare it as object?
, rather than object
-- that's the point of having NRTs.
来源:https://stackoverflow.com/questions/58411013/possibility-of-external-functions-as-nullable-guards