How to pass a variable into a function and return it modified?

cry_wifi 0
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I've been learning c# for about three weeks now so please bear with me. I got through some online courses with no problem but I've come upon what I think is a very basic, but very real barrier to my understanding of how to use the code to actually write programs.

int inputOutput = 1;
Console.WriteLine(inputOutput); //Prints 1
static void IncrementTest(int inputInput, out int intputOutput)
{
    intputOutput = inputInput++;
}
IncrementTest(inputOutput, out inputOutput);
Console.WriteLine(inputOutput); //Also prints 1?

If using out parameters is not the way forward, what is?

out methods c#
2 Answers

The out keyword is for parameters that are used for output only. If you want to use a parameter for input and output then you use ref. When a parameter is unadorned, it is considered input-only, e.g.

public static void Main()
{
    var number = 1;
    Console.WriteLine("Outside Method before: " + number);
    Method(number);
    Console.WriteLine("Outside Method after: " + number);
}

public static void Method(int inputOnly)
{
    Console.WriteLine("Inside Method before: " + inputOnly);
    inputOnly++;
    Console.WriteLine("Inside Method after: " + inputOnly);
}

Output:

Outside Method before: 1
Inside Method before: 1
Inside Method after: 2
Outside Method after: 1

When a parameter is declared out, it is considered output-only. This code:

public static void Main()
{
    var number = 1;
    Console.WriteLine("Outside Method before: " + number);
    Method(out number);
    Console.WriteLine("Outside Method after: " + number);
}

public static void Method(out int outputOnly)
{
    Console.WriteLine("Inside Method before: " + outputOnly);
    outputOnly++;
    Console.WriteLine("Inside Method after: " + outputOnly);
}

will not compile because any existing value for an out parameter is ignored and it is considered uninitialised at the beginning of the method. You MUST assign a value to an out parameter before using it and, if you don't use it, before the end of the method.

public static void Main()
{
    var number = 1;
    Console.WriteLine("Outside Method before: " + number);
    Method(out number);
    Console.WriteLine("Outside Method after: " + number);
}

public static void Method(out int outputOnly)
{
    outputOnly = 10;
    Console.WriteLine("Inside Method before: " + outputOnly);
    outputOnly++;
    Console.WriteLine("Inside Method after: " + outputOnly);
}

Output:

Outside Method before: 1
Inside Method before: 10
Inside Method after: 11
Outside Method after: 11

When a parameter is declared ref, it is considered input-output, e.g.

public static void Main()
{
    var number = 1;
    Console.WriteLine("Outside Method before: " + number);
    Method(ref number);
    Console.WriteLine("Outside Method after: " + number);
}

public static void Method(ref int inputOutput)
{
    Console.WriteLine("Inside Method before: " + inputOutput);
    inputOutput++;
    Console.WriteLine("Inside Method after: " + inputOutput);
}

Output:

Outside Method before: 1
Inside Method before: 1
Inside Method after: 2
Outside Method after: 2
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