Execute LambdaExpression and get returned value as object

Question

Is there a clean way to do this?

Expression<Func<int, string>> exTyped = i => "My int = " + i;
LambdaExpression lambda = exTyped;

//later on:

object input = 4;
object result = ExecuteLambdaSomeHow(lambda, input);
//result should be "My int = 4"

This should work for different types.

Why not just `var func = new Func(x => { return string.Format("My int = {0}", x); });`? — Arian Motamedi, Jul 31 '13 at 22:02
@Kevin Makes sense, otherwise a simple `string.Format()` would do it :) — Arian Motamedi, Jul 31 '13 at 22:04
Why do you want to use late binding (`DynamicInvoke`). If it is possible, I would try to avoid this (because you have no compile time check; thus it is very much possible you get runtime errors). — Styxxy, Jul 31 '13 at 22:05
@Styxxy excellent point! I've updated my answer accordingly. — Kevin, Jul 31 '13 at 22:27

Kevin · Accepted Answer · 2013-07-31T22:24:17.390

44

Sure... you just need to compile your lambda and then invoke it...

object input = 4;
var compiledLambda = lambda.Compile();
var result = compiledLambda.DynamicInvoke(input);

Styxxy brings up an excellent point... You would be better served by letting the compiler help you out. Note with a compiled expression as in the code below input and result are both strongly typed.

var input = 4;
var compiledExpression = exTyped.Compile();
var result = compiledExpression(input);

edited Jul 31 '13 at 22:24

answered Jul 31 '13 at 22:01

Kevin

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`compiledLambda.Invoke(input);` may be a better choice here if the exact type is known as @Styxxy pointed out. `Invoke` is faster than `DynamicInvoke` due to less reflection going on, see http://stackoverflow.com/questions/12858340/difference-between-invoke-and-dynamicinvoke – May 05 '16 at 21:28

Execute LambdaExpression and get returned value as object

1 Answers1