C# Dictionary, what is the source of TryGetValue error in this simple code

Question

 Dictionary<string, int> test = new Dictionary<string, int>(); test.Add("dave", 12); test.Add("john", 14); int v; test.TryGetValue("dave", out int v) { Console.WriteLine(v); }

This simple C# code gives "Best Overload Method Match has some Invalid Arguments" error. Can you please tell me the source of the error? Thanks.

What version of the compiler are you using? VS2017 likes that code just fine, once I fix the syntax error. Please post complete, valid example code that reliably reproduces the issue in some specific, named version of the compiler. — 15ee8f99-57ff-4f92-890c-b56153
– 15ee8f99-57ff-4f92-890c-b56153, Commented Oct 12, 2017 at 20:21
Are you missing an if statement around test.TryGetValue? — Mr.Nimelo
– Mr.Nimelo, Commented Oct 12, 2017 at 20:22
@haim770 That error will be "invalid expression term int", not anything about the overload. — 15ee8f99-57ff-4f92-890c-b56153
– 15ee8f99-57ff-4f92-890c-b56153, Commented Oct 12, 2017 at 20:23
I am using Visual Studio 2012. I still receive the error despite the edits mentioned below. Thank you. — SoylentGreen
– SoylentGreen, Commented Oct 12, 2017 at 20:27

15ee8f99-57ff-4f92-890c-b56153 · Accepted Answer · 2017-10-12 20:33:48Z

OP is in VS2012, not using C#7.

First, get rid of int in the parameter list. It can't be there in your version of C#.

Second, put a semicolon after the TryGetValue() call...

int v; test.TryGetValue("dave", out v); Console.WriteLine(v);

Or put it in an if:

int v; if (test.TryGetValue("dave", out v)) { Console.WriteLine(v); }

Sean Sherman · Accepted Answer · 2017-10-12 20:23:06Z

Is "value" a variable you already declared or did you leave the example from the intellisense for TryGetValue? Pretty sure it's the latter case. edit: OR it's a new version of C# feature... This writes out 12 for v:

 Dictionary<string, int> test = new Dictionary<string, int>(); test.Add("dave", 12); test.Add("john", 14); int v; test.TryGetValue("dave", out v); { Console.WriteLine(v); }

As others pointed out, your version uses this: blogs.msdn.microsoft.com/dotnet/2016/08/24/…
Here's a fiddle demonstrating it with the Roslyn compiler: dotnetfiddle.net/eblkGk

Akaino · Accepted Answer · 2017-10-12 20:25:01Z

You got either a typo or a misunderstanding of

TryGetValue()

There is no need for the code block where your writeline sits in. Just end your line of code and do your writeLine.

test.TryGetValue("dave", out int value); // <---- notice the ; Console.WriteLine(value);

EDIT: Or, as Mr.Nimelo suggests, there might be an if statement missing like so:

if test.TryGetValue("dave", out int value) { Console.WriteLine(value); }

Youna · Accepted Answer · 2017-10-12 20:26:33Z

Don't you miss an if in your snippet, nope?

 Dictionary<string, int> test = new Dictionary<string, int>(); test.Add("dave", 12); test.Add("john", 14); // missing if there? test.TryGetValue("dave", out int value) { Console.WriteLine(value); }

My cheap 2 cents with a bit of refactoring...:

 var test = new Dictionary<string, int> {{"dave", 12}, {"john", 14}}; if (test.TryGetValue("dave", out var value)) { Console.WriteLine(value); } Console.ReadKey();

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