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Declaration of ref/out parameters in lambdas without typename #303
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In this particular case you don't need to write any of the parameter stuff, it's just: TryParse<int> parse1 = Int32.TryParse; |
This would be a worthwhile improvement to type inference, in my view. As |
I'm not sure I understand what you are proposing. Would you care to elaborate a bit more? |
@paulomorgado Of course! Let us see a code below: delegate T Parse<T>(string text);
delegate bool TryParse<T>(string text, out T result);
static void Main() {
// We can create an instance of Parse<int> like this:
Parse<int> parseHex1 = (string text) => Int32.Parse(text, NumberStyles.HexNumber);
// or like this (and, I think, this way is more simple and more readable):
Parse<int> parseHex2 = text => Int32.Parse(text, NumberStyles.HexNumber);
// To create an instance of TryParse<int> delegate
// we must explicitly specify a types of arguments in a lambda expression:
TryParse<int> tryParseHex1 = (string text, out int result) => Int32.TryParse(text, NumberStyles.HexNumber, null, out result);
// And we can not now use a sintax like this:
TryParse<int> tryParseHex2 = (text, out result) => Int32.TryParse(text, NumberStyles.HexNumber, null, out result);
} Why this is meaningful:
|
👍 |
I ink you are missing the fact that, although you can't declare by reference type parameters in C#, when a parameter is expressed as of being of type ref T (or out T, which is the same for the CLR - just extra validation from the compiler), the type is, actually, &T, which is not the same as T. |
@paulomorgado Excuse me, can you explain what do you mean? Why in you point of view TryParse<int> tryParseHex1 = (string text, out int result) => Int32.TryParse(text, NumberStyles.HexNumber, null, out result); is correct (it is valid C# code) and TryParse<int> tryParseHex2 = (text, out result) => Int32.TryParse(text, NumberStyles.HexNumber, null, out result); is not? In a both examples, types of arguments exactly the same. But, in the second line, it calculated by the compiler, not specified by user explicitly. |
I was trying to understand your issue, and I think I got it: the compiler should be able to infer the types from usage when there are out parameters. Is that it? |
@paulomorgado Exactly! I'm sorry for my bad and poor English. |
👍 |
+1 thanks |
👍 |
Issue moved to dotnet/csharplang #338 via ZenHub |
Hello!
I have a little suggestion for C# language.
Let us have delegate like this:
and we want to create an instance:
All is OK. What about ref/out parameters in a delegate?
when we want to create an instance…
…we shoud to specify types on parameters.
Why is this required? What about a syntax below:
?
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