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

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?

CodePudding user response：

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

CodePudding user response：

int is a Value types

You can use the ref keywords

But I recommend that you deepen the difference between Value types to Reference types.