In C++/CLI, how do I declare and call a function with an 'out' parameter?

Simon_15371 0
Viewed 47506

I have a function which parses one string into two strings. In C# I would declare it like this:

void ParseQuery(string toParse, out string search, out string sort)
{
    ...
}

and I'd call it like this:

string searchOutput, sortOutput;
ParseQuery(userInput, out searchOutput, out sortOutput);

The current project has to be done in C++/CLI. I've tried

using System::Runtime::InteropServices;

...

void ParseQuery(String ^ toParse, [Out] String^ search, [Out] String^ sort)
{
    ...
}

but if I call it like this:

String ^ searchOutput, ^ sortOutput;
ParseQuery(userInput, [Out] searchOutput, [Out] sortOutput);

I get a compiler error, and if I call it like this:

String ^ searchOutput, ^ sortOutput;
ParseQuery(userInput, searchOutput, sortOutput);

then I get an error at runtime. How should I declare and call my function?

pass-by-reference c++-cli .net
3 Answers

C++/CLI itself doesn't support a real 'out' argument, but you can mark a reference as an out argument to make other languages see it as a real out argument.

You can do this for reference types as:

void ReturnString([Out] String^% value)
{
   value = "Returned via out parameter";
}

// Called as
String^ result;
ReturnString(result);

And for value types as:

void ReturnInt([Out] int% value)
{
   value = 32;
}

// Called as
int result;
ReturnInt(result);

The % makes it a 'ref' parameter and the OutAttribute marks that it is only used for output values.

Bert_Huijben_20940

It's not supported. The closest you can get is ref

Granted you can fake it, but you lose a compile time check.

user_12005580
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