c# generic with slight difference for types?

Question

Notice the two extensions, one for float, one for Vector3.

Notice there\'s only a slight difference in the var( call.

In c# could these be writt

Answer 1

You can do it if you make an extra Func<T,T> that performs transformation before calling the var action (which you should rename, because var is a C# keyword).

Here is one approach that you could take:

public static IEnumerator Tweeng<T>(
    this float duration
,   System.Action<T> varAction
,   T aa
,   T zz
) {
    Func<T,T,float,T> transform = MakeTransform<T>();
    float sT = Time.time;
    float eT = sT + duration;
    while (Time.time < eT) {   
        float t = (Time.time-sT)/duration;
        varAction(transform(aa, zz, t));
        yield return null;
    }
    varAction(zz);
}

private static Func<T,T,float,T> MakeTransform<T>() {
    if (typeof(T) == typeof(float)) {
        Func<float, float, float, float> f = Mathf.SmoothStep;
        return (Func<T,T,float,T>)(Delegate)f;
    }
    if (typeof(T) == typeof(Vector3)) {
        Func<Vector3, Vector3, float, Vector3> f = Vector3.Lerp;
        return (Func<T,T,float,T>)(Delegate)f;
    }
    throw new ArgumentException("Unexpected type "+typeof(T));
}

It can even be done inline:

public static IEnumerator DasTweeng<T>( this float duration, System.Action<T> vary, T aa, T zz )
    {
    float sT = Time.time;
    float eT = sT + duration;

    Func<T,T,float,T> step;

    if (typeof(T) == typeof(float))
        step = (Func<T,T,float,T>)(Delegate)(Func<float, float, float, float>)Mathf.SmoothStep;
    else if (typeof(T) == typeof(Vector3))
        step = (Func<T,T,float,T>)(Delegate)(Func<Vector3, Vector3, float, Vector3>)Vector3.Lerp;
    else
        throw new ArgumentException("Unexpected type "+typeof(T));

    while (Time.time < eT)
        {
        float t = (Time.time-sT)/duration;
        vary( step(aa,zz, t) );
        yield return null;
        }
    vary(zz);
    }

Perhaps a more natural idiom is

    Delegate d;

    if (typeof(T) == typeof(float))
        d = (Func<float, float, float, float>)Mathf.SmoothStep;
    else if (typeof(T) == typeof(Vector3))
        d = (Func<Vector3, Vector3, float, Vector3>)Vector3.Lerp;
    else
        throw new ArgumentException("Unexpected type "+typeof(T));

    Func<T,T,float,T> step = (Func<T,T,float,T>)d;

Answer 2

You can define your method as follows:

public static IEnumerator Tweeng<T>(this float duration,
         System.Action<T> var, T aa, T zz, Func<T,T,float,T> thing)
{
    float sT = Time.time;
    float eT = sT + duration;

    while (Time.time < eT)
    {
        float t = (Time.time - sT) / duration;
        var(thing(aa, zz, t));
        yield return null;
    }

    var(zz);
}

And then using it:

float a = 5;
float b = 0;
float c = 0;
a.Tweeng(q => {}, b, c, Mathf.SmoothStep);

Or:

float a = 0;
Vector3 b = null;
Vector3 c = null;
a.Tweeng(q => {}, b, c, Vector3.Lerp);

Alternatively, if you want to get rid of the method passing, you can have simple overloads to handle it:

public static IEnumerator Tweeng(this float duration, System.Action<float> var, float aa, float zz)
{
    return Tweeng(duration, var, aa, zz, Mathf.SmoothStep);
}
public static IEnumerator Tweeng(this float duration, System.Action<Vector3> var, Vector3 aa, Vector3 zz)
{
    return Tweeng(duration, var, aa, zz, Vector3.Lerp);
}

private static IEnumerator Tweeng<T>(this float duration,
         System.Action<T> var, T aa, T zz, Func<T,T,float,T> thing)
{
    float sT = Time.time;
    float eT = sT + duration;

    while (Time.time < eT)
    {
        float t = (Time.time - sT) / duration;
        var(thing(aa, zz, t));
        yield return null;
    }

    var(zz);
}

And then using it:

float a = 5;
float b = 0;
float c = 0;
a.Tweeng(q => {}, b, c);

Or:

float a = 0;
Vector3 b = null;
Vector3 c = null;
a.Tweeng(q => {}, b, c);

---

Stub methods to compile in LINQPad/without unity:

public class Mathf { public static float SmoothStep(float aa, float zz, float t) => 0; }
public class Time { public static float time => DateTime.Now.Ticks; }
public class Vector3 { public static Vector3 Lerp(Vector3 aa, Vector3 zz, float t) => null; }

Answer 3

I liked the Tweeng thing, but why extend float if the Coroutine can only be used on MonoBehaviours? You should do the extensions for MonoBehaviour so, for example I did a extension to do Interpolation:

public static void _Interpolate(this MonoBehaviour monoBehaviour, float duration,
    Action<float, bool> callback, float from, float to, Interpolator interpolator)
{
    monoBehaviour.StartCoroutine(ExecuteInterpolation(interpolator, duration, callback, from, to));
}

So I just Started a Coroutine inside the extension:

private static IEnumerator ExecuteInterpolation(Interpolator interpolator, float duration,
    Action<float, bool> callback, float from, float to)
{
    float sT = Time.time;
    float eT = sT + duration;
    bool hasFinished = false;
    while (Time.time < eT)
    {
        float t = (Time.time - sT) / duration;
// ----> my logic here with callback(to, false)
        yield return null;
    }

    hasFinished = true;
    callback(to, hasFinished);
}

Note that I have a boolean to say that the interpolation finished, this happens because is not best practice to rely on float comparison to check end of a flow, if it rounds for the max result the result before the last we are going to have the call back for the last interaction called twice.