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by Jiří {x2} Činčura

What’s the maximum number of arguments for method in C# and in .NET?

7 Mar 2022 2 mins .NET, C#

Silly question, right? Whatever the limit is, it is surely beyond what one should practically ever write. Right? But as with mine other explorations of limits, I’ll try it anyway.

Let’s start with C# or IL respectively. For C# compiler to call method and pass an argument it needs to generate ldarg (or ldarg.0, ldarg.1, etc.). The documentation states, that the argument is unsigned int16. Thus, it should be impossible to go over this limit. Let’s try it.

const int Max = 65537;

class Test
	public static void Huge(<#= string.Join(", ", Enumerable.Range(0, Max).Select(x => $"byte arg{x}")) #>)
	{ }

	public static void CallHuge()
		Huge(<#= string.Join(", ", Enumerable.Range(0, Max).Select(_ => "0")) #>);
		Console.WriteLine("It works!");

This simple T4 template generates method with 65537 arguments. The compilation fails with System.ArgumentOutOfRangeException: Specified argument was out of the range of valid values. (Parameter 'sequenceNumber'), as expected. Using 65536 works.

OK, all the pieces fit together. But can I actually call such method? That’s where the CallHuge from the template comes into picture. Trying to run such program results in System.InvalidProgramException: Common Language Runtime detected an invalid program.. Fair enough. But where’s the limit then? With a bit of bisection I ended up on 8192. That might look like a weird number, but it’s actually 2^13 or also 65536/8.

And there you have it. Roslyn compiler will compile method with 65536 arguments. Runtime version 6.0.2 executes only with 8192.

Profile Picture Jiří Činčura is .NET, C# and Firebird expert. He focuses on data and business layers, language constructs, parallelism, databases and performance. For almost two decades he contributes to open-source, i.e. FirebirdClient. He works as a senior software engineer for Microsoft. Frequent speaker and blogger at