Deferred execution in C#

Question

How could I implement my own deferred execution mechanism in C#?

So for instance I have:

string x = DoFoo();

Is it possible to perform some magic so that DoFoo does not execute until I "use" x?

Answer 1

You can use lambdas/delegates:

Func<string> doit = () => DoFoo();
//  - or -
Func<string> doit = DoFoo;

Later you can invoke doit just like a method:

string x = doit();

---

I think the closest you can get is something like this:

Lazy<string> x = DoFoo;

string y = x; // "use" x

With a definition of Lazy<T> similar to this (untested):

public class Lazy<T>
{
    private readonly Func<T> func;
    private bool hasValue;
    private T value;

    public Lazy(Func<T> func)
    {
        this.func = func;
        this.hasValue = false;
    }

    public static implicit operator Lazy<T>(Func<T> func)
    {
        return new Lazy<T>(func);
    }

    public static implicit operator T(Lazy<T> lazy)
    {
        if (!lazy.hasValue)
        {
            lazy.value = lazy.func();
            lazy.hasValue = true;
        }
        return lazy.value;
    }
}

Unfortunately, it seems that the compiler's type inferencing algorithms can't auto-infer the type of the Func<T> and so can't match it to the implicit conversion operator. We need to explicitly declare the delegate's type, which makes the assignment statements more verbose:

// none of these will compile...
Lazy<string> x = DoFoo;
Lazy<string> y = () => DoFoo();
Lazy<string> z = delegate() { return DoFoo(); };

// these all work...
Lazy<string> a = (Func<string>)DoFoo;
Lazy<string> b = (Func<string>)(() => DoFoo());
Lazy<string> c = new Func<string>(DoFoo);
Lazy<string> d = new Func<string>(() => DoFoo());
Lazy<string> e = new Lazy<string>(DoFoo);
Lazy<string> f = new Lazy<string>(() => DoFoo);