虚空型方法表达式主体成员允许非空类型的表达式*如果使用其他方法*

Question

I'm not sure what to make of this, I was playing with expression-bodied members in C#, and I noticed something odd.

Essentially, if you create a method expression-bodied member which has a void return type, you can still use a non-void return type method as the expression.

Example:

public void MethodA() { }
public void MethodB() => MethodA();

Works as expected.

public string MethodA() { return "MethodA"; }
public void MethodB() => MethodA();

Compiles just fine (I haven't run it to test, but the fact that it compiles confuses me). This confuses me, as in my head I see:

public void MethodB() { return MethodA(); }

Which is obviously invalid, as the return statement actually returns a value, which isn't allowed in a void method.

However, what's odd is that the following does not compile:

public void MethodC() => "MethodC";

And gives the "Only assignment..." error.

Why is this happening? Is this intended, or is it some sort of bug? Is there something I'm missing? Are expression bodied members allowed to just throw-away the result of the expression if it's another method?

Answer 1

This confuses me, as in my head I see:

 public void MethodB() { return MethodA(); } 

That is not actually correct; The actual syntax generated is this:

public void MethodB() { MethodA(); }

And that is perfectly fine. The return value is just ignored.

The "MethodC" code fragment isn't a method which result can be voided. It is an actual value and that needs to be assigned according to C# syntax. The same is true for regular statements: putting "MethodC"; on an empty line fails.