C＃out结构参数

Question

I have this code in C# but I have a problem whit this code:

struct myStruct
{
  public string sOne;
  public string sTwo;
}

public static int ChangeStruct(out myStruct[] arrmyStruct)
{
  arrmyStruct= new myStruct[256];
  arrSNChildrenStruct[0].sOne= "";
  arrSNChildrenStruct[0].sTwo= "";

  return 0;
}

But when I build, I have this error: Inconsistent accessibility: parameter type 'out ........ is less accessible than method .....

What's wrong? Thanks

Answer 1

Make a public struct myStruct instead of internal struct. Or make ChangeStruct() private if you only use it locally.

Answer 2

This has nothing to do with it being an out parameter, or an array. You'd get the same error with:

public static void ChangeStruct(myStruct foo)

Your method is public, but your struct is internal (the default accessibility for any top-level type) or private if it's a nested type. That means that any caller external to your assembly should have access to the method... but can't possibly understand the method signature. C# doesn't allow you to declare methods which refer to types which can't be seen by all possible callers.

Options:

• Make the method internal or private
• Make the struct public

Other notes:

• Your naming is very unconventional. Name your types according to their meaning, and make it PascalCased. Drop the "arr" prefix from your variable names.
• Public fields are usually a bad idea, as are mutable structs.