如何编写强类型的lambda表达式？

Question

I want to write a lambda expression within an inline if statement. But inline if statement must have strong type results.

MyType obj = someObj.IsOk ? null : () => {
   MyType o = new MyType(intVal);
   o.PropertyName = false;
   return o;
};

Of course this doesn't work, because lambda expression isn't strongly typed. I thought of using Func<intVal, MyType> delegate, to make it strong type.

But how do I use this Func<> inside inline if? Is it at all possible of would I have to define my own function outside and use it in inline if statement?

Answer 1

It has nothing to do with the lambda's typing here. You are trying to return either null or (a function taking no arguments and returning a MyType) but you are telling the compiler that the result of that statement is not a function, but just a MyType. I think what you want to do is

MyType obj = someObj.IsOk ? null : new MyType(intVal);

Answer 2

Even with the more complicated code, you can use an object initializer expression:

MyType obj = someObj.IsOk ? null : new MyType(intVal) { ProName = false };

If you really want to use a lambda though, you could write:

MyType obj = someObj.IsOk ? null : ((Func<MyType>) (() => {
   MyType o = new MyType(intVal);
   o.ProName = false;
   return o;
}))();

However, this is frankly a nightmare of brackets and casts. You can make it simpler with a helper method:

public T Execute(Func<T> func)
{
    return func();
}

...

MyType obj = someObj.IsOk ? null : Execute(() => {
   MyType o = new MyType(intVal);
   o.ProName = false;
   return o;
});

Answer 3

If you have something like this ...

  var obj = (someObj.IsOK) ? null : () => {
             return new MyType(intVal) { PropName =false }; }

You will get the error ...

"There is no explicit conversion between null and lambda expression."

The reason for that is discussed in this SO thread .

Mark is correct on what you're trying to do with the code sample, except you can set the property in like as well like this ...

var obj = someObj.IsOk ? null : new MyType(intVal) { PropName = false };